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Cayley–Hamilton theorem
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Arthur Cayley, F.R.S. (1821–1895) is widely regarded as Britain's leading pure mathematician of the 19th century. Cayley in 1848 went to Dublin to attend lectures on quaternions by Hamilton, their discoverer. Later Cayley impressed him by being the second to publish work on them.[1] Cayley proved the theorem for matrices of dimension 3 and less, publishing proof for the two-dimensional case.[2][3] As for n × n matrices, Cayley stated “..., I have not thought it necessary to undertake the labor of a formal proof of the theorem in the general case of a matrix of any degree”.
William Rowan Hamilton (1805–1865), Irish physicist, astronomer, and mathematician, first foreign member of the American National Academy of Sciences. While maintaining opposing position about how geometry should be studied, Hamilton always remained on the best terms with Cayley.[1]

Hamilton proved that for a linear function of quaternions there exists a certain equation, depending on the linear function, that is satisfied by the linear function itself.[4][5][6]
In linear algebra, the Cayley–Hamilton theorem (named after the mathematicians Arthur Cayley and William Rowan Hamilton) states that every square matrix over a commutative ring (such as the real or complex field) satisfies its own characteristic equation.
If A is a given n×n matrix and In is the n×n identity matrix, then the characteristic polynomial of A is defined as[7]
p ( λ ) = det ( λ I n − A ) , {\displaystyle p(\lambda )=\det(\lambda I_{n}-A)~,}
where det is the determinant operation and λ is a scalar element of the base ring. Since the entries of the matrix are (linear or constant) polynomials in λ, the determinant is also an n-th order monic polynomial in λ. The Cayley–Hamilton theorem states that substituting the matrix A for λ in this polynomial results in the zero matrix,
p ( A ) = O . {\displaystyle p(A)=O.}
The powers of A, obtained by substitution from powers of λ, are defined by repeated matrix multiplication; the constant term of p(λ) gives a multiple of the power A0, which is defined as the identity matrix. The theorem allows An to be expressed as a linear combination of the lower matrix powers of A. When the ring is a field, the Cayley–Hamilton theorem is equivalent to the statement that the minimal polynomial of a square matrix divides its characteristic polynomial.
The theorem was first proved in 1853[8] in terms of inverses of linear functions of quaternions, a non-commutative ring, by Hamilton.[4][5][6] This corresponds to the special case of certain 4 × 4 real or 2 × 2 complex matrices. The theorem holds for general quaternionic matrices.[9][nb 1] Cayley in 1858 stated it for 3 × 3 and smaller matrices, but only published a proof for the 2 × 2 case.[2] The general case was first proved by Frobenius in 1878.[10]
Contents
1 Examples
1.1 1×1 matrices
1.2 2×2 matrices
2 Applications
2.1 Determinant and inverse matrix
2.2 n-th Power of matrix
2.3 Matrix functions
2.4 Algebraic Number Theory
3 Proving the theorem in general
3.1 Preliminaries
3.1.1 Adjugate matrices
3.2 A direct algebraic proof
3.3 A proof using polynomials with matrix coefficients
3.4 A synthesis of the first two proofs
3.5 A proof using matrices of endomorphisms
3.6 A bogus "proof": p(A) = det(AIn − A) = det(A − A) = 0
4 Abstraction and generalizations
5 See also
6 Remarks
7 Notes
8 References
9 External links
Examples[edit]
1×1 matrices[edit]
For a 1×1 matrix A = (a1,1), the characteristic polynomial is given by p(λ) = λ − a, and so p(A) = (a) − a1,1 = 0 is obvious.
2×2 matrices[edit]
As a concrete example, let
A = ( 1 2 3 4 ) . {\displaystyle A={\begin{pmatrix}1&2\\3&4\end{pmatrix}}.}
Its characteristic polynomial is given by
p ( λ ) = det ( λ I 2 − A ) = det ( λ − 1 − 2 − 3 λ − 4 ) = ( λ − 1 ) ( λ − 4 ) − ( − 2 ) ( − 3 ) = λ 2 − 5 λ − 2. {\displaystyle p(\lambda )=\det(\lambda I_{2}-A)=\det {\begin{pmatrix}\lambda -1&-2\\-3&\lambda -4\end{pmatrix}}=(\lambda -1)(\lambda -4)-(-2)(-3)=\lambda ^{2}-5\lambda -2.}
The Cayley–Hamilton theorem claims that, if we define
p ( X ) = X 2 − 5 X − 2 I 2 , {\displaystyle p(X)=X^{2}-5X-2I_{2},}
then
p ( A ) = A 2 − 5 A − 2 I 2 = ( 0 0 0 0 ) . {\displaystyle p(A)=A^{2}-5A-2I_{2}={\begin{pmatrix}0&0\\0&0\\\end{pmatrix}}.}
We can verify by computation that indeed,
A 2 − 5 A − 2 I 2 = ( 7 10 15 22 ) − ( 5 10 15 20 ) − ( 2 0 0 2 ) = ( 0 0 0 0 ) . {\displaystyle A^{2}-5A-2I_{2}={\begin{pmatrix}7&10\\15&22\\\end{pmatrix}}-{\begin{pmatrix}5&10\\15&20\\\end{pmatrix}}-{\begin{pmatrix}2&0\\0&2\\\end{pmatrix}}={\begin{pmatrix}0&0\\0&0\\\end{pmatrix}}.}
For a generic 2×2 matrix,
A = ( a b c d ) , {\displaystyle A={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}},}
the characteristic polynomial is given by p(λ) = λ2 − (a + d)λ + (ad − bc), so the Cayley–Hamilton theorem states that
p ( A ) = A 2 − ( a + d ) A + ( a d − b c ) I 2 = ( 0 0 0 0 ) ; {\displaystyle p(A)=A^{2}-(a+d)A+(ad-bc)I_{2}={\begin{pmatrix}0&0\\0&0\\\end{pmatrix}};}
which is indeed always the case, evident by working out the entries of A2.
Applications[edit]
Determinant and inverse matrix[edit]
See also: Determinant § Relation to eigenvalues and trace, and Characteristic polynomial § Properties
For a general n×n invertible matrix A, i.e., one with nonzero determinant, A−1 can thus be written as an (n − 1)-th order polynomial expression in A: As indicated, the Cayley–Hamilton theorem amounts to the identity
p ( A ) = A n + c n − 1 A n − 1 + ⋯ + c 1 A + ( − 1 ) n det ( A ) I n = O . {\displaystyle p(A)=A^{n}+c_{n-1}A^{n-1}+\cdots +c_{1}A+(-1)^{n}\det(A)I_{n}=O.}
The coefficients ci are given by the elementary symmetric polynomials of the eigenvalues of A. Using Newton identities, the elementary symmetric polynomials can in turn be expressed in terms of power sum symmetric polynomials of the eigenvalues:
s k = ∑ i = 1 n λ i k = tr  ( A k ) , {\displaystyle s_{k}=\sum _{i=1}^{n}\lambda _{i}^{k}=\operatorname {tr} (A^{k}),}
where tr (Ak) is the trace of the matrix Ak. Thus, we can express ci in terms of the trace of powers of A.
In general, the formula for the coefficients ci is given in terms of complete exponential Bell polynomials as [nb 2]
c n − k = ( − 1 ) k k ! B k ( s 1 , − 1 ! s 2 , 2 ! s 3 , … , ( − 1 ) k − 1 ( k − 1 ) ! s k ) . {\displaystyle c_{n-k}={\frac {(-1)^{k}}{k!}}B_{k}(s_{1},-1!s_{2},2!s_{3},\ldots ,(-1)^{k-1}(k-1)!s_{k}).}
In particular, the determinant of A corresponds to c0. Thus, the determinant can be written as a trace identity
det ( A ) = 1 n ! B n ( s 1 , − 1 ! s 2 , 2 ! s 3 , … , ( − 1 ) n − 1 ( n − 1 ) ! s n ) . {\displaystyle \det(A)={\frac {1}{n!}}B_{n}(s_{1},-1!s_{2},2!s_{3},\ldots ,(-1)^{n-1}(n-1)!s_{n}).}
Likewise, the characteristic polynomial can be written as
− ( − 1 ) n det ( A ) I n = A ( A n − 1 + c n − 1 A n − 2 + ⋯ + c 1 I n ) , {\displaystyle -(-1)^{n}\det(A)I_{n}=A(A^{n-1}+c_{n-1}A^{n-2}+\cdots +c_{1}I_{n}),}
and, by multiplying both sides by A−1 (note −(−1)n = (−1)n−1), one is led to an expression for the inverse of A as a trace identity,
A − 1 = ( − 1 ) n − 1 det A ( A n − 1 + c n − 1 A n − 2 + ⋯ + c 1 I n ) , = 1 det A ∑ k = 0 n − 1 ( − 1 ) n + k − 1 A n − k − 1 k ! B k ( s 1 , − 1 ! s 2 , 2 ! s 3 , … , ( − 1 ) k − 1 ( k − 1 ) ! s k ) . {\displaystyle {\begin{aligned}A^{-1}&={\frac {(-1)^{n-1}}{\det A}}(A^{n-1}+c_{n-1}A^{n-2}+\cdots +c_{1}I_{n}),\\&={\frac {1}{\det A}}\sum _{k=0}^{n-1}(-1)^{n+k-1}{\frac {A^{n-k-1}}{k!}}B_{k}(s_{1},-1!s_{2},2!s_{3},\ldots ,(-1)^{k-1}(k-1)!s_{k}).\end{aligned}}}
For instance, the first few Bell polynomials are B0 = 1, B1(x1) = x1, B2(x1, x2) = x2
1 + x2, and B3(x1, x2, x3) = x3
1 + 3 x1x2 + x3.
Using these to specify the coefficients ci of the characteristic polynomial of a 2×2 matrix yields
c 2 = B 0 = 1 , {\displaystyle c_{2}=B_{0}=1,}
c 1 = − 1 1 ! B 1 ( s 1 ) = − s 1 = − tr  ( A ) , {\displaystyle c_{1}={\frac {-1}{1!}}B_{1}(s_{1})=-s_{1}=-\operatorname {tr} (A),}
c 0 = 1 2 ! B 2 ( s 1 , − 1 ! s 2 ) = 1 2 ( s 1 2 − s 2 ) = 1 2 ( ( tr  ( A ) ) 2 − tr  ( A 2 ) ) . {\displaystyle c_{0}={\frac {1}{2!}}B_{2}(s_{1},-1!s_{2})={\frac {1}{2}}(s_{1}^{2}-s_{2})={\frac {1}{2}}((\operatorname {tr} (A))^{2}-\operatorname {tr} (A^{2})).}
The coefficient c0 gives the determinant of the 2×2 matrix, c1 minus its trace, while its inverse is given by
A − 1 = − 1 det A ( A + c 1 I 2 ) = − 2 ( A − tr  ( A ) I 2 ) ( tr  ( A ) ) 2 − tr  ( A 2 ) . {\displaystyle A^{-1}={\frac {-1}{\det A}}(A+c_{1}I_{2})={\frac {-2(A-\operatorname {tr} (A)I_{2})}{(\operatorname {tr} (A))^{2}-\operatorname {tr} (A^{2})}}.}
It is apparent from the general formula for cn-k, expressed in terms of Bell polynomials, that this expression, ½((trA)2 − tr(A2)), always gives the coefficient cn−2 of λn−2 in the characteristic polynomial of any n×n matrix; so, for a 3×3 matrix A, the statement of the Cayley–Hamilton theorem can also be written as
A 3 − ( tr  A ) A 2 + 1 2 ( ( tr  A ) 2 − tr  ( A 2 ) ) A − det ( A ) I 3 = O , {\displaystyle A^{3}-(\operatorname {tr} A)A^{2}+{\frac {1}{2}}\left((\operatorname {tr} A)^{2}-\operatorname {tr} (A^{2})\right)A-\det(A)I_{3}=O,}
where the right-hand side designates a 3×3 matrix with all entries reduced to zero. Likewise, this determinant in the n = 3 case, is now
det ( A ) = 1 3 ! B 3 ( s 1 , − 1 ! s 2 , 2 ! s 3 ) = 1 6 ( s 1 3 + 3 s 1 ( − s 2 ) + 2 s 3 ) = 1 6 ( ( tr  A ) 3 − 3 tr  ( A 2 ) ( tr  A ) + 2 tr  ( A 3 ) ) . {\displaystyle {\begin{aligned}\det(A)&={\frac {1}{3!}}B_{3}(s_{1},-1!s_{2},2!s_{3})={\frac {1}{6}}(s_{1}^{3}+3s_{1}(-s_{2})+2s_{3})\\&={\tfrac {1}{6}}\left((\operatorname {tr} A)^{3}-3\operatorname {tr} (A^{2})(\operatorname {tr} A)+2\operatorname {tr} (A^{3})\right).\end{aligned}}}
This expression gives the negative of coefficient cn−3 of λn−3 in the general case, as seen below.
Similarly, one can write for a 4×4 matrix A,
A 4 − ( tr  A ) A 3 + 1 2 ( ( tr  A ) 2 − tr  ( A 2 ) ) A 2 − 1 6 ( ( tr  A ) 3 − 3 tr  ( A 2 ) ( tr  A ) + 2 tr  ( A 3 ) ) A + det ( A ) I 4 = O , {\displaystyle A^{4}-(\operatorname {tr} A)A^{3}+{\tfrac {1}{2}}{\bigl (}(\operatorname {tr} A)^{2}-\operatorname {tr} (A^{2}){\bigr )}A^{2}-{\tfrac {1}{6}}{\bigl (}(\operatorname {tr} A)^{3}-3\operatorname {tr} (A^{2})(\operatorname {tr} A)+2\operatorname {tr} (A^{3}){\bigr )}A+\det(A)I_{4}=O,}
where, now, the determinant is cn−4,
1 24 ( ( tr  A ) 4 − 6 tr  ( A 2 ) ( tr  A ) 2 + 3 ( tr  ( A 2 ) ) 2 + 8 tr  ( A 3 ) tr  ( A ) − 6 tr  ( A 4 ) ) , {\displaystyle {\tfrac {1}{24}}\left((\operatorname {tr} A)^{4}-6\operatorname {tr} (A^{2})(\operatorname {tr} A)^{2}+3(\operatorname {tr} (A^{2}))^{2}+8\operatorname {tr} (A^{3})\operatorname {tr} (A)-6\operatorname {tr} (A^{4})\right),}
and so on for larger matrices. The increasingly complex expressions for the coefficients ck is deducible from Newton's identities or the Faddeev–LeVerrier algorithm.
Another method for obtaining these coefficients ck for a general n×n matrix, provided no root be zero, relies on the following alternative expression for the determinant,
p ( λ ) = det ( λ I n − A ) = λ n exp  ( tr  ( log  ( I n − A / λ ) ) ) . {\displaystyle p(\lambda )=\det(\lambda I_{n}-A)=\lambda ^{n}\exp(\operatorname {tr} (\log(I_{n}-A/\lambda ))).}
Hence, by virtue of the Mercator series,
p ( λ ) = λ n exp  ( − tr  ∑ m = 1 ∞ ( A λ ) m m ) , {\displaystyle p(\lambda )=\lambda ^{n}\exp \left(-\operatorname {tr} \sum _{m=1}^{\infty }{({A \over \lambda })^{m} \over m}\right),}
where the exponential only needs be expanded to order λ−n, since p(λ) is of order n, the net negative powers of λ automatically vanishing by the C–H theorem. (Again, this requires a ring containing the rational numbers.) The coefficients of λ can be directly written in terms of complete Bell polynomials by comparing this expression with the generating function of the Bell polynomial.
Differentiation of this expression with respect to λ allows determination of the generic coefficients of the characteristic polynomial for general n, as determinants of m×m matrices,[nb 3]
c n − m = ( − 1 ) m m ! | tr  A m − 1 0 ⋯ tr  A 2 tr  A m − 2 ⋯ ⋮ ⋮ ⋮ tr  A m − 1 tr  A m − 2 ⋯ ⋯ 1 tr  A m tr  A m − 1 ⋯ ⋯ tr  A | . {\displaystyle c_{n-m}={\frac {(-1)^{m}}{m!}}{\begin{vmatrix}\operatorname {tr} A&m-1&0&\cdots \\\operatorname {tr} A^{2}&\operatorname {tr} A&m-2&\cdots \\\vdots &\vdots &&&\vdots \\\operatorname {tr} A^{m-1}&\operatorname {tr} A^{m-2}&\cdots &\cdots &1\\\operatorname {tr} A^{m}&\operatorname {tr} A^{m-1}&\cdots &\cdots &\operatorname {tr} A\end{vmatrix}}~.}
n-th Power of matrix[edit]
The Cayley–Hamilton theorem always provides a relationship between the powers of A (though not always the simplest one), which allows one to simplify expressions involving such powers, and evaluate them without having to compute the power An or any higher powers of A.
As an example, for A = ( 1 2 3 4 ) {\displaystyle A={\begin{pmatrix}1&2\\3&4\end{pmatrix}}} the theorem gives
A 2 = 5 A + 2 I 2 . {\displaystyle A^{2}=5A+2I_{2}\,.}
Then, to calculate A4, observe
A 3 = ( 5 A + 2 I 2 ) A = 5 A 2 + 2 A = 5 ( 5 A + 2 I 2 ) + 2 A = 27 A + 10 I 2 {\displaystyle A^{3}=(5A+2I_{2})A=5A^{2}+2A=5(5A+2I_{2})+2A=27A+10I_{2}}
A 4 = A 3 A = ( 27 A + 10 I 2 ) A = 27 A 2 + 10 A = 27 ( 5 A + 2 I 2 ) + 10 A = 145 A + 54 I 2 . {\displaystyle A^{4}=A^{3}A=(27A+10I_{2})A=27A^{2}+10A=27(5A+2I_{2})+10A=145A+54I_{2}\,.}
Likewise,
A − 1 = A − 5 I 2 2 . {\displaystyle A^{-1}={\frac {A-5I_{2}}{2}}~.}
Notice that we have been able to write the matrix power as the sum of two terms. In fact, matrix power of any order k can be written as a matrix polynomial of degree at most n - 1, where n is the size of a square matrix. This is an instance where Cayley–Hamilton theorem can be used to express a matrix function, which we will discuss below systematically.
Matrix functions[edit]
Given an analytic function
f ( x ) = ∑ k = 0 ∞ a k x k {\displaystyle f(x)=\sum _{k=0}^{\infty }a_{k}x^{k}}
and the characteristic polynomial p(x) of degree n of an n × n matrix A, the function can be expressed using long division as
f ( x ) = q ( x ) p ( x ) + r ( x ) , {\displaystyle f(x)=q(x)p(x)+r(x),}
where q(x) is some quotient polynomial and r(x) is a remainder polynomial such that 0 ≤ deg r(x) < n. By the Cayley–Hamilton theorem, replacing x by the matrix A gives p(A) = 0, so one has
f ( A ) = r ( A ) . {\displaystyle f(A)=r(A).}
Thus, the analytic function of matrix A can be expressed as a matrix polynomial of degree less than n.
Let the remainder polynomial be
r ( x ) = c 0 + c 1 x + ⋯ + c n − 1 x n − 1 . {\displaystyle r(x)=c_{0}+c_{1}x+\cdots +c_{n-1}x^{n-1}.}
Since p(λ) = 0, evaluating the function f(x) at the n eigenvalues of A, yields
f ( λ i ) = r ( λ i ) = c 0 + c 1 λ i + ⋯ + c n − 1 λ i n − 1 , f o r i = 1 , 2 , . . . , n . {\displaystyle f(\lambda _{i})=r(\lambda _{i})=c_{0}+c_{1}\lambda _{i}+\cdots +c_{n-1}\lambda _{i}^{n-1},\qquad \mathrm {for} \qquad i=1,2,...,n.}
This amounts to a system of n linear equations, which can be solved to determine the coefficients ci. Thus, one has
f ( A ) = ∑ k = 0 n − 1 c k A k . {\displaystyle f(A)=\sum _{k=0}^{n-1}c_{k}A^{k}.}
When the eigenvalues are repeated, that is λi = λj for some i ≠ j, two or more equations are identical; and hence the linear equations cannot be solved uniquely. For such cases, for an eigenvalue λ with multiplicity m, the first m – 1 derivative of p(x) vanishes at the eigenvalues. Thus, there are the extra m – 1 linearly independent solutions
d k f ( x ) d x k | x = λ = d k r ( x ) d x k | x = λ f o r k = 1 , 2 , … , m − 1 , {\displaystyle {\frac {\mathrm {d} ^{k}f(x)}{\mathrm {d} x^{k}}}{\Big |}_{x=\lambda }={\frac {\mathrm {d} ^{k}r(x)}{\mathrm {d} x^{k}}}{\Big |}_{x=\lambda }\qquad \mathrm {for} \qquad k=1,2,\ldots ,m-1,}
which, when combined with others, yield the required n equations to solve for ci.
Finding a polynomial that passes through the points (λi, f (λi)) is essentially an interpolation problem, and can be solved using Lagrange or Newton interpolation techniques, leading to Sylvester's formula.
For example, suppose the task is to find the polynomial representation of
f ( A ) = e A t w h e r e A = ( 1 2 0 3 ) . {\displaystyle f(A)=e^{At}\qquad \mathrm {where} \qquad A={\begin{pmatrix}1&2\\0&3\end{pmatrix}}.}
The characteristic polynomial is p(x) = (x - 1)(x - 3) = x2 - 4 x + 3, and the eigenvalues are λ = 1, 3. Let r(x) = c0 + c1x. Evaluating f(λ) = r(λ) at the eigenvalues, one obtains two linear equations et = c0 + c1 and e3t = c0 + 3 c1. Solving the equations yields c0 = (3 et - e3t)/2 and c1 = (e3t - et)/2. Thus, it follows that
e A t = c 0 I 2 + c 1 A = ( c 0 + c 1 2 c 1 0 c 0 + 3 c 1 ) = ( e t e 3 t − e t 0 e 3 t ) . {\displaystyle e^{At}=c_{0}I_{2}+c_{1}A={\begin{pmatrix}c_{0}+c_{1}&2c_{1}\\0&c_{0}+3c_{1}\end{pmatrix}}={\begin{pmatrix}e^{t}&e^{3t}-e^{t}\\0&e^{3t}\end{pmatrix}}.}
If, instead, the function were f(A) = sin At, then the coefficients would have been c0 = (3 sin t - sin 3t)/2 and c1 = (sin 3t - sin t)/2; hence
sin  ( A t ) = c 0 I 2 + c 1 A = ( sin  t sin  3 t − sin  t 0 sin  3 t ) . {\displaystyle \sin(At)=c_{0}I_{2}+c_{1}A={\begin{pmatrix}\sin t&\sin 3t-\sin t\\0&\sin 3t\end{pmatrix}}.}
As a further example, when considering
f ( A ) = e A t w h e r e A = ( 0 1 − 1 0 ) , {\displaystyle f(A)=e^{At}\qquad \mathrm {where} \qquad A={\begin{pmatrix}0&1\\-1&0\end{pmatrix}},}
then the characteristic polynomial is p(x) = x2 + 1, and the eigenvalues are λ = i, -i. As before, evaluating the function at the eigenvalues gives us the linear equations eit = c0 + i c1 and e−it = c0 - i c1; the solution of which gives, c0 = (eit + e−it)/2 = cos t and c1 = (eit - e−it)/2i = sin t. Thus, for this case,
e A t = ( cos  t ) I 2 + ( sin  t ) A = ( cos  t sin  t − sin  t cos  t ) , {\displaystyle e^{At}=(\cos t)I_{2}+(\sin t)A={\begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix}},}
which is a rotation matrix.
Standard examples of such usage is the exponential map from the Lie algebra of a matrix Lie group into the group. It is given by a matrix exponential,
exp : g → G ; t X ↦ e t X = ∑ n = 0 ∞ t n X n n ! = I + t X + t 2 X 2 2 + ⋯ , t ∈ R , X ∈ g . {\displaystyle \exp :{\mathfrak {g}}\rightarrow G;\qquad tX\mapsto e^{tX}=\sum _{n=0}^{\infty }{\frac {t^{n}X^{n}}{n!}}=I+tX+{\frac {t^{2}X^{2}}{2}}+\cdots ,t\in \mathbb {R} ,X\in {\mathfrak {g}}.}
Such expressions have long been known for SU(2),
e i ( θ / 2 ) ( n ^ ⋅ σ ) = I 2 cos  θ / 2 + i ( n ^ ⋅ σ ) sin  θ / 2 , {\displaystyle e^{i(\theta /2)({\hat {n}}\cdot \sigma )}=I_{2}\cos \theta /2+i({\hat {n}}\cdot \sigma )\sin \theta /2,}
where the σ are the Pauli matrices and for SO(3),
e i θ ( n ^ ⋅ J ) = I 3 + i ( n ^ ⋅ J ) sin  θ + ( n ^ ⋅ J ) 2 ( cos  θ − 1 ) , {\displaystyle e^{i\theta ({\hat {n}}\cdot \mathbf {J} )}=I_{3}+i({\hat {n}}\cdot \mathbf {J} )\sin \theta +({\hat {n}}\cdot \mathbf {J} )^{2}(\cos \theta -1),}
which is Rodrigues' rotation formula. For the notation, see rotation group SO(3)#A note on Lie algebra.
More recently, expressions have appeared for other groups, like the Lorentz group SO(3, 1),[11] O(4, 2)[12] and SU(2, 2),[13] as well as GL(n, R).[14] The group O(4, 2) is the conformal group of spacetime, SU(2, 2) its simply connected cover (to be precise, the simply connected cover of the connected component SO+(4, 2) of O(4, 2)). The expressions obtained apply to the standard representation of these groups. They require knowledge of (some of) the eigenvalues of the matrix to exponentiate. For SU(2) (and hence for SO(3)), closed expressions have recently been obtained for all irreducible representations, i.e. of any spin.[15]
Ferdinand Georg Frobenius (1849–1917), German mathematician. His main interests were elliptic functions differential equations, and later group theory.
In 1878 he gave the first full proof of the Cayley–Hamilton theorem.[10]
Algebraic Number Theory[edit]
The Cayley–Hamilton theorem is an effective tool for computing the minimal polynomial of algebraic integers. For example, given a finite extension Q [ α 1 , … , α k ] {\displaystyle \mathbb {Q} [\alpha _{1},\ldots ,\alpha _{k}]} of Q {\displaystyle \mathbb {Q} } and an algebraic integer α ∈ Q [ α 1 , … , α k ] {\displaystyle \alpha \in \mathbb {Q} [\alpha _{1},\ldots ,\alpha _{k}]} which is a non-zero linear combination of the α 1 n 1 ⋯ α k n k {\displaystyle \alpha _{1}^{n_{1}}\cdots \alpha _{k}^{n_{k}}} we can compute the minimal polynomial of α {\displaystyle \alpha } by finding a matrix representing the Q {\displaystyle \mathbb {Q} } -linear transformation
⋅ α : Q [ α 1 , … , α k ] → Q [ α 1 , … , α k ] {\displaystyle \cdot \alpha :\mathbb {Q} [\alpha _{1},\ldots ,\alpha _{k}]\to \mathbb {Q} [\alpha _{1},\ldots ,\alpha _{k}]}
If we call this transformation matrix A {\displaystyle A} , then we can find the minimal polynomial by applying the Cayley–Hamilton theorem to A {\displaystyle A} .[16]
Proving the theorem in general[edit]
As the examples above show, obtaining the statement of the Cayley–Hamilton theorem for an n×n matrix
A = ( a i j ) i , j = 1 n {\displaystyle A=(a_{ij})_{i,j=1}^{n}}
requires two steps: first the coefficients ci of the characteristic polynomial are determined by development as a polynomial in t of the determinant
p ( t ) = det ( t I n − A ) = | t − a 1 , 1 − a 1 , 2 ⋯ − a 1 , n − a 2 , 1 t − a 2 , 2 ⋯ − a 2 , n ⋮ ⋮ ⋱ ⋮ − a n , 1 − a n , 2 ⋯ t − a n , n | = t n + c n − 1 t n − 1 + ⋯ + c 1 t + c 0 , {\displaystyle {\begin{aligned}p(t)&=\det(tI_{n}-A)={\begin{vmatrix}t-a_{1,1}&-a_{1,2}&\cdots &-a_{1,n}\\-a_{2,1}&t-a_{2,2}&\cdots &-a_{2,n}\\\vdots &\vdots &\ddots &\vdots \\-a_{n,1}&-a_{n,2}&\cdots &t-a_{n,n}\end{vmatrix}}\\&=t^{n}+c_{n-1}t^{n-1}+\cdots +c_{1}t+c_{0},\end{aligned}}}
and then these coefficients are used in a linear combination of powers of A that is equated to the n×n null matrix:
A n + c n − 1 A n − 1 + ⋯ + c 1 A + c 0 I n = ( 0 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 0 ) . {\displaystyle A^{n}+c_{n-1}A^{n-1}+\cdots +c_{1}A+c_{0}I_{n}={\begin{pmatrix}0&\cdots &0\\\vdots &\ddots &\vdots \\0&\cdots &0\end{pmatrix}}.}
The left hand side can be worked out to an n×n matrix whose entries are (enormous) polynomial expressions in the set of entries ai,j of A, so the Cayley–Hamilton theorem states that each of these n2 expressions are equal to 0. For any fixed value of n these identities can be obtained by tedious but completely straightforward algebraic manipulations. None of these computations can show however why the Cayley–Hamilton theorem should be valid for matrices of all possible sizes n, so a uniform proof for all n is needed.
Preliminaries[edit]
If a vector v of size n happens to be an eigenvector of A with eigenvalue λ, in other words if A⋅v = λv, then
p ( A ) ⋅ v = A n ⋅ v + c n − 1 A n − 1 ⋅ v + ⋯ + c 1 A ⋅ v + c 0 I n ⋅ v = λ n v + c n − 1 λ n − 1 v + ⋯ + c 1 λ v + c 0 v = p ( λ ) v , {\displaystyle {\begin{aligned}p(A)\cdot v&=A^{n}\cdot v+c_{n-1}A^{n-1}\cdot v+\cdots +c_{1}A\cdot v+c_{0}I_{n}\cdot v\\[6pt]&=\lambda ^{n}v+c_{n-1}\lambda ^{n-1}v+\cdots +c_{1}\lambda v+c_{0}v=p(\lambda )v,\end{aligned}}}
which is the null vector since p(λ) = 0 (the eigenvalues of A are precisely the roots of p(t)). This holds for all possible eigenvalues λ, so the two matrices equated by the theorem certainly give the same (null) result when applied to any eigenvector. Now if A admits a basis of eigenvectors, in other words if A is diagonalizable, then the Cayley–Hamilton theorem must hold for A, since two matrices that give the same values when applied to each element of a basis must be equal.
A = X D X − 1 , D = diag  ( λ i ) , i = 1 , 2 , . . . , n {\displaystyle A=XDX^{-1},\quad D=\operatorname {diag} (\lambda _{i}),\quad i=1,2,...,n}
p A ( λ ) = | λ I − A | = {\displaystyle p_{A}(\lambda )=|\lambda I-A|=} product of eigenvalues of λ I − A = ∏ i = 1 n ( λ − λ i ) ≡ ∑ k = 0 n c k λ k {\displaystyle \lambda I-A=\prod _{i=1}^{n}(\lambda -\lambda _{i})\equiv \sum _{k=0}^{n}c_{k}\lambda ^{k}}
p A ( A ) = ∑ c k A k = X p A ( D ) X − 1 = X C X − 1 {\displaystyle p_{A}(A)=\sum c_{k}A^{k}=Xp_{A}(D)X^{-1}=XCX^{-1}}
C i i = ∑ k = 0 n c k λ i k = ∏ j = 1 n ( λ i − λ j ) = 0 , C i , j ≠ i = 0 {\displaystyle C_{ii}=\sum _{k=0}^{n}c_{k}\lambda _{i}^{k}=\prod _{j=1}^{n}(\lambda _{i}-\lambda _{j})=0,\qquad C_{i,j\neq i}=0}
∴ p A ( A ) = X C X − 1 = O . {\displaystyle \therefore p_{A}(A)=XCX^{-1}=O.}
Not all matrices are diagonalizable, but for matrices with complex coefficients many of them are: the set of diagonalizable complex square matrices of a given size is dense in the set of all such square matrices[17] (for a matrix to be diagonalizable it suffices for instance that its characteristic polynomial not have any multiple roots). Now if any of the n2 expressions that the theorem equates to 0 would not reduce to a null expression, in other words if it would be a nonzero polynomial in the coefficients of the matrix, then the set of complex matrices for which this expression happens to give 0 would not be dense in the set of all matrices, which would contradict the fact that the theorem holds for all diagonalizable matrices. Thus one can see that the Cayley–Hamilton theorem must be true.
While this provides a valid proof (for matrices over the complex numbers), the argument is not very satisfactory, since the identities represented by the theorem do not in any way depend on the nature of the matrix (diagonalizable or not), nor on the kind of entries allowed (for matrices with real entries the diagonalizable ones do not form a dense set, and it seems strange one would have to consider complex matrices to see that the Cayley–Hamilton theorem holds for them). We shall therefore now consider only arguments that prove the theorem directly for any matrix using algebraic manipulations only; these also have the benefit of working for matrices with entries in any commutative ring.
There is a great variety of such proofs of the Cayley–Hamilton theorem, of which several will be given here. They vary in the amount of abstract algebraic notions required to understand the proof. The simplest proofs use just those notions needed to formulate the theorem (matrices, polynomials with numeric entries, determinants), but involve technical computations that render somewhat mysterious the fact that they lead precisely to the correct conclusion. It is possible to avoid such details, but at the price of involving more subtle algebraic notions: polynomials with coefficients in a non-commutative ring, or matrices with unusual kinds of entries.
Adjugate matrices[edit]
All proofs below use the notion of the adjugate matrix adj(M) of an n×n matrix M, the transpose of its cofactor matrix.
This is a matrix whose coefficients are given by polynomial expressions in the coefficients of M (in fact, by certain (n − 1)×(n − 1) determinants), in such a way that the following fundamental relations hold,
adj  ( M ) ⋅ M = det ( M ) I n = M ⋅ adj  ( M ) . {\displaystyle \operatorname {adj} (M)\cdot M=\det(M)I_{n}=M\cdot \operatorname {adj} (M)~.}
These relations are a direct consequence of the basic properties of determinants: evaluation of the (i,j) entry of the matrix product on the left gives the expansion by column j of the determinant of the matrix obtained from M by replacing column i by a copy of column j, which is det(M) if i = j and zero otherwise; the matrix product on the right is similar, but for expansions by rows.
Being a consequence of just algebraic expression manipulation, these relations are valid for matrices with entries in any commutative ring (commutativity must be assumed for determinants to be defined in the first place). This is important to note here, because these relations will be applied below for matrices with non-numeric entries such as polynomials.
A direct algebraic proof[edit]
This proof uses just the kind of objects needed to formulate the Cayley–Hamilton theorem: matrices with polynomials as entries. The matrix t In −A whose determinant is the characteristic polynomial of A is such a matrix, and since polynomials form a commutative ring, it has an adjugate
B = adj  ( t I n − A ) . {\displaystyle B=\operatorname {adj} (tI_{n}-A).}
Then, according to the right-hand fundamental relation of the adjugate, one has
( t I n − A ) B = det ( t I n − A ) I n = p ( t ) I n . {\displaystyle (tI_{n}-A)B=\det(tI_{n}-A)I_{n}=p(t)I_{n}~.}
Since B is also a matrix with polynomials in t as entries, one can, for each i , collect the coefficients of ti in each entry to form a matrix B i of numbers, such that one has
B = ∑ i = 0 n − 1 t i B i . {\displaystyle B=\sum _{i=0}^{n-1}t^{i}B_{i}~.}
(The way the entries of B are defined makes clear that no powers higher than tn−1 occur). While this looks like a polynomial with matrices as coefficients, we shall not consider such a notion; it is just a way to write a matrix with polynomial entries as a linear combination of n constant matrices, and the coefficient t i has been written to the left of the matrix to stress this point of view.
Now, one can expand the matrix product in our equation by bilinearity
p ( t ) I n = ( t I n − A ) B = ( t I n − A ) ∑ i = 0 n − 1 t i B i = ∑ i = 0 n − 1 t I n ⋅ t i B i − ∑ i = 0 n − 1 A ⋅ t i B i = ∑ i = 0 n − 1 t i + 1 B i − ∑ i = 0 n − 1 t i A B i = t n B n − 1 + ∑ i = 1 n − 1 t i ( B i − 1 − A B i ) − A B 0 . {\displaystyle {\begin{aligned}p(t)I_{n}&=(tI_{n}-A)B\\&=(tI_{n}-A)\sum _{i=0}^{n-1}t^{i}B_{i}\\&=\sum _{i=0}^{n-1}tI_{n}\cdot t^{i}B_{i}-\sum _{i=0}^{n-1}A\cdot t^{i}B_{i}\\&=\sum _{i=0}^{n-1}t^{i+1}B_{i}-\sum _{i=0}^{n-1}t^{i}AB_{i}\\&=t^{n}B_{n-1}+\sum _{i=1}^{n-1}t^{i}(B_{i-1}-AB_{i})-AB_{0}~.\end{aligned}}}
Writing
p ( t ) I n = t n I n + t n − 1 c n − 1 I n + ⋯ + t c 1 I n + c 0 I n , {\displaystyle p(t)I_{n}=t^{n}I_{n}+t^{n-1}c_{n-1}I_{n}+\cdots +tc_{1}I_{n}+c_{0}I_{n}~,}
one obtains an equality of two matrices with polynomial entries, written as linear combinations of constant matrices with powers of t as coefficients.
Such an equality can hold only if in any matrix position the entry that is multiplied by a given power ti is the same on both sides; it follows that the constant matrices with coefficient ti in both expressions must be equal. Writing these equations then for i from n down to 0, one finds
B n − 1 = I n , B i − 1 − A B i = c i I n for 1 ≤ i ≤ n − 1 , − A B 0 = c 0 I n . {\displaystyle B_{n-1}=I_{n},\qquad B_{i-1}-AB_{i}=c_{i}I_{n}\quad {\text{for }}1\leq i\leq n-1,\qquad -AB_{0}=c_{0}I_{n}~.}
Finally, multiply the equation of the coefficients of ti from the left by Ai, and sum up:
A n B n − 1 + ∑ i = 1 n − 1 ( A i B i − 1 − A i + 1 B i ) − A B 0 = A n + c n − 1 A n − 1 + ⋯ + c 1 A + c 0 I n . {\textstyle A^{n}B_{n-1}+\sum \limits _{i=1}^{n-1}\left(A^{i}B_{i-1}-A^{i+1}B_{i}\right)-AB_{0}=A^{n}+c_{n-1}A^{n-1}+\cdots +c_{1}A+c_{0}I_{n}~.}
The left-hand sides form a telescoping sum and cancel completely; the right-hand sides add up to p ( A ) {\displaystyle p(A)} :
O = p ( A ) . {\displaystyle O=p(A)~.}
This completes the proof.
A proof using polynomials with matrix coefficients[edit]
This proof is similar to the first one, but tries to give meaning to the notion of polynomial with matrix coefficients that was suggested by the expressions occurring in that proof. This requires considerable care, since it is somewhat unusual to consider polynomials with coefficients in a non-commutative ring, and not all reasoning that is valid for commutative polynomials can be applied in this setting.
Notably, while arithmetic of polynomials over a commutative ring models the arithmetic of polynomial functions, this is not the case over a non-commutative ring (in fact there is no obvious notion of polynomial function in this case that is closed under multiplication). So when considering polynomials in t with matrix coefficients, the variable t must not be thought of as an "unknown", but as a formal symbol that is to be manipulated according to given rules; in particular one cannot just set t to a specific value.
( f + g ) ( x ) = ∑ i ( f i + g i ) x i = ∑ i f i x i + ∑ i g i x i = f ( x ) + g ( x ) {\displaystyle (f+g)(x)=\sum _{i}\left(f_{i}+g_{i}\right)x^{i}=\sum _{i}{f_{i}x^{i}}+\sum _{i}{g_{i}x^{i}}=f(x)+g(x)} .
Let M(n, R) be the ring of n×n matrices with entries in some ring R (such as the real or complex numbers) that has A as an element. Matrices with as coefficients polynomials in t, such as t I n − A {\displaystyle tI_{n}-A} or its adjugate B in the first proof, are elements of M(n, R[t]).
By collecting like powers of t, such matrices can be written as "polynomials" in t with constant matrices as coefficients; write M(n, R)[t] for the set of such polynomials. Since this set is in bijection with M(n, R[t]), one defines arithmetic operations on it correspondingly, in particular multiplication is given by
( ∑ i M i t i ) ( ∑ j N j t j ) = ∑ i , j ( M i N j ) t i + j , {\displaystyle \left(\sum _{i}M_{i}t^{i}\right)\left(\sum _{j}N_{j}t^{j}\right)=\sum _{i,j}(M_{i}N_{j})t^{i+j},}
respecting the order of the coefficient matrices from the two operands; obviously this gives a non-commutative multiplication.
Thus, the identity
( t I n − A ) B = p ( t ) I n . {\displaystyle (tI_{n}-A)B=p(t)I_{n}.}
from the first proof can be viewed as one involving a multiplication of elements in M(n, R)[t].
At this point, it is tempting to simply set t equal to the matrix A , which makes the first factor on the left equal to the null matrix, and the right hand side equal to p(A); however, this is not an allowed operation when coefficients do not commute. It is possible to define a "right-evaluation map" evA : M[t] → M, which replaces each ti by the matrix power Ai of A , where one stipulates that the power is always to be multiplied on the right to the corresponding coefficient.
But this map is not a ring homomorphism: the right-evaluation of a product differs in general from the product of the right-evaluations. This is so because multiplication of polynomials with matrix coefficients does not model multiplication of expressions containing unknowns: a product M t i N t j = ( M ⋅ N ) t i + j {\displaystyle Mt^{i}Nt^{j}=(M\cdot N)t^{i+j}} is defined assuming that t commutes with N, but this may fail if t is replaced by the matrix A.
One can work around this difficulty in the particular situation at hand, since the above right-evaluation map does become a ring homomorphism if the matrix A is in the center of the ring of coefficients, so that it commutes with all the coefficients of the polynomials (the argument proving this is straightforward, exactly because commuting t with coefficients is now justified after evaluation).
Now, A is not always in the center of M, but we may replace M with a smaller ring provided it contains all the coefficients of the polynomials in question: I n {\displaystyle I_{n}} , A, and the coefficients B i {\displaystyle B_{i}} of the polynomial B. The obvious choice for such a subring is the centralizer Z of A, the subring of all matrices that commute with A; by definition A is in the center of Z.
This centralizer obviously contains I n {\displaystyle I_{n}} , and A, but one has to show that it contains the matrices B i {\displaystyle B_{i}} . To do this, one combines the two fundamental relations for adjugates, writing out the adjugate B as a polynomial:
( ∑ i = 0 m B i t i ) ( t I n − A ) = ( t I n − A ) ∑ i = 0 m B i t i ∑ i = 0 m B i t i + 1 − ∑ i = 0 m B i A t i = ∑ i = 0 m B i t i + 1 − ∑ i = 0 m A B i t i ∑ i = 0 m B i A t i = ∑ i = 0 m A B i t i . {\displaystyle {\begin{aligned}\left(\sum _{i=0}^{m}B_{i}t^{i}\right)(tI_{n}-A)&=(tI_{n}-A)\sum _{i=0}^{m}B_{i}t^{i}\\\sum _{i=0}^{m}B_{i}t^{i+1}-\sum _{i=0}^{m}B_{i}At^{i}&=\sum _{i=0}^{m}B_{i}t^{i+1}-\sum _{i=0}^{m}AB_{i}t^{i}\\\sum _{i=0}^{m}B_{i}At^{i}&=\sum _{i=0}^{m}AB_{i}t^{i}.\end{aligned}}}
Equating the coefficients shows that for each i, we have A Bi = Bi A as desired. Having found the proper setting in which evA is indeed a homomorphism of rings, one can complete the proof as suggested above:
ev A  ( p ( t ) I n ) = ev A  ( ( t I n − A ) B ) p ( A ) = ev A  ( t I n − A ) ⋅ ev A  ( B ) p ( A ) = ( A I n − A ) ⋅ ev A  ( B ) = O ⋅ ev A  ( B ) = O . {\displaystyle {\begin{aligned}\operatorname {ev} _{A}{\bigl (}p(t)I_{n}{\bigr )}&=\operatorname {ev} _{A}((tI_{n}-A)B)\\p(A)&=\operatorname {ev} _{A}(tI_{n}-A)\cdot \operatorname {ev} _{A}(B)\\p(A)&=(AI_{n}-A)\cdot \operatorname {ev} _{A}(B)=O\cdot \operatorname {ev} _{A}(B)=O.\end{aligned}}}
This completes the proof.
A synthesis of the first two proofs[edit]
In the first proof, one was able to determine the coefficients Bi of B based on the right-hand fundamental relation for the adjugate only. In fact the first n equations derived can be interpreted as determining the quotient B of the Euclidean division of the polynomial p(t)In on the left by the monic polynomial Int−A, while the final equation expresses the fact that the remainder is zero. This division is performed in the ring of polynomials with matrix coefficients. Indeed, even over a non-commutative ring, Euclidean division by a monic polynomial P is defined, and always produces a unique quotient and remainder with the same degree condition as in the commutative case, provided it is specified at which side one wishes P to be a factor (here that is to the left).
To see that quotient and remainder are unique (which is the important part of the statement here), it suffices to write P Q + r = P Q ′ + r ′ {\displaystyle PQ+r=PQ'+r'} as P ( Q − Q ′ ) = r ′ − r {\displaystyle P(Q-Q')=r'-r} and observe that since P is monic, P(Q−Q') cannot have a degree less than that of P, unless Q=Q' .
But the dividend p(t)In and divisor Int−A used here both lie in the subring (R[A])[t], where R[A] is the subring of the matrix ring M(n, R) generated by A: the R-linear span of all powers of A. Therefore, the Euclidean division can in fact be performed within that commutative polynomial ring, and of course it then gives the same quotient B and remainder 0 as in the larger ring; in particular this shows that B in fact lies in (R[A])[t].
But, in this commutative setting, it is valid to set t to A in the equation
p ( t ) I n = ( t I n − A ) B {\displaystyle p(t)I_{n}=(tI_{n}-A)B} ; in other words, to apply the evaluation map
ev A : ( R [ A ] ) [ t ] → R [ A ] {\displaystyle \operatorname {ev} _{A}:(R[A])[t]\to R[A]}
which is a ring homomorphism, giving
p ( A ) = 0 ⋅ ev A  ( B ) = 0 {\displaystyle p(A)=0\cdot \operatorname {ev} _{A}(B)=0}
just like in the second proof, as desired.
In addition to proving the theorem, the above argument tells us that the coefficients Bi of B are polynomials in A, while from the second proof we only knew that they lie in the centralizer Z of A; in general Z is a larger subring than R[A], and not necessarily commutative. In particular the constant term B0= adj(−A) lies in R[A]. Since A is an arbitrary square matrix, this proves that adj(A) can always be expressed as a polynomial in A (with coefficients that depend on A).
In fact, the equations found in the first proof allow successively expressing B n − 1 , … , B 1 , B 0 {\displaystyle B_{n-1},\ldots ,B_{1},B_{0}} as polynomials in A, which leads to the identity
adj  ( − A ) = ∑ i = 1 n c i A i − 1 , {\displaystyle \operatorname {adj} (-A)=\sum _{i=1}^{n}c_{i}A^{i-1},}
valid for all n×n matrices, where
p ( t ) = t n + c n − 1 t n − 1 + ⋯ + c 1 t + c 0 {\displaystyle p(t)=t^{n}+c_{n-1}t^{n-1}+\cdots +c_{1}t+c_{0}}
is the characteristic polynomial of A.
Note that this identity also implies the statement of the Cayley–Hamilton theorem: one may move adj(−A) to the right hand side, multiply the resulting equation (on the left or on the right) by A, and use the fact that
− A ⋅ adj  ( − A ) = adj  ( − A ) ⋅ ( − A ) = det ( − A ) I n = c 0 I n . {\displaystyle -A\cdot \operatorname {adj} (-A)=\operatorname {adj} (-A)\cdot (-A)=\det(-A)I_{n}=c_{0}I_{n}.}
See also: Faddeev–LeVerrier algorithm
A proof using matrices of endomorphisms[edit]
As was mentioned above, the matrix p(A) in statement of the theorem is obtained by first evaluating the determinant and then substituting the matrix A for t; doing that substitution into the matrix t I n − A {\displaystyle tI_{n}-A} before evaluating the determinant is not meaningful. Nevertheless, it is possible to give an interpretation where p(A) is obtained directly as the value of a certain determinant, but this requires a more complicated setting, one of matrices over a ring in which one can interpret both the entries A i , j {\displaystyle A_{i,j}} of A, and all of A itself. One could take for this the ring M(n, R) of n×n matrices over R, where the entry A i , j {\displaystyle A_{i,j}} is realised as A i , j I n {\displaystyle A_{i,j}I_{n}} , and A as itself. But considering matrices with matrices as entries might cause confusion with block matrices, which is not intended, as that gives the wrong notion of determinant (recall that the determinant of a matrix is defined as a sum of products of its entries, and in the case of a block matrix this is generally not the same as the corresponding sum of products of its blocks!). It is clearer to distinguish A from the endomorphism φ of an n-dimensional vector space V (or free R-module if R is not a field) defined by it in a basis e1, ..., en, and to take matrices over the ring End(V) of all such endomorphisms. Then φ ∈ End(V) is a possible matrix entry, while A designates the element of M(n, End(V)) whose i,j entry is endomorphism of scalar multiplication by A i , j {\displaystyle A_{i,j}} ; similarly In will be interpreted as element of M(n, End(V)). However, since End(V) is not a commutative ring, no determinant is defined on M(n, End(V)); this can only be done for matrices over a commutative subring of End(V). Now the entries of the matrix φ I n − A {\displaystyle \varphi I_{n}-A} all lie in the subring R[φ] generated by the identity and φ, which is commutative. Then a determinant map M(n, R[φ]) → R[φ] is defined, and det ( φ I n − A ) {\displaystyle \det(\varphi I_{n}-A)} evaluates to the value p(φ) of the characteristic polynomial of A at φ (this holds independently of the relation between A and φ); the Cayley–Hamilton theorem states that p(φ) is the null endomorphism.
In this form, the following proof can be obtained from that of (Atiyah & MacDonald 1969, Prop. 2.4) (which in fact is the more general statement related to the Nakayama lemma; one takes for the ideal in that proposition the whole ring R). The fact that A is the matrix of φ in the basis e1, ..., en means that
φ ( e i ) = ∑ j = 1 n A j , i e j for i = 1 , … , n . {\displaystyle \varphi (e_{i})=\sum _{j=1}^{n}A_{j,i}e_{j}\quad {\text{for }}i=1,\ldots ,n.}
One can interpret these as n components of one equation in Vn, whose members can be written using the matrix-vector product M(n, End(V)) × Vn → Vn that is defined as usual, but with individual entries ψ ∈ End(V) and v in V being "multiplied" by forming ψ ( v ) {\displaystyle \psi (v)} ; this gives:
φ I n ⋅ E = A tr ⋅ E , {\displaystyle \varphi I_{n}\cdot E=A^{\operatorname {tr} }\cdot E,}
where E ∈ V n {\displaystyle E\in V^{n}} is the element whose component i is ei (in other words it is the basis e1, ..., en of V written as a column of vectors). Writing this equation as
( φ I n − A tr ) ⋅ E = 0 ∈ V n {\displaystyle (\varphi I_{n}-A^{\operatorname {tr} })\cdot E=0\in V^{n}}
one recognizes the transpose of the matrix φ I n − A {\displaystyle \varphi I_{n}-A} considered above, and its determinant (as element of M(n, R[φ])) is also p(φ). To derive from this equation that p(φ) = 0 ∈ End(V), one left-multiplies by the adjugate matrix of φ I n − A tr {\displaystyle \varphi I_{n}-A^{\operatorname {tr} }} , which is defined in the matrix ring M(n, R[φ]), giving
0 = adj  ( φ I n − A tr ) ⋅ ( ( φ I n − A tr ) ⋅ E ) = ( adj  ( φ I n − A tr ) ⋅ ( φ I n − A tr ) ) ⋅ E = ( det ( φ I n − A tr ) I n ) ⋅ E = ( p ( φ ) I n ) ⋅ E ; {\displaystyle {\begin{aligned}0&=\operatorname {adj} (\varphi I_{n}-A^{\operatorname {tr} })\cdot ((\varphi I_{n}-A^{\operatorname {tr} })\cdot E)\\&=(\operatorname {adj} (\varphi I_{n}-A^{\operatorname {tr} })\cdot (\varphi I_{n}-A^{\operatorname {tr} }))\cdot E\\&=(\det(\varphi I_{n}-A^{\operatorname {tr} })I_{n})\cdot E\\&=(p(\varphi )I_{n})\cdot E;\end{aligned}}}
the associativity of matrix-matrix and matrix-vector multiplication used in the first step is a purely formal property of those operations, independent of the nature of the entries. Now component i of this equation says that p(φ)(ei) = 0 ∈ V; thus p(φ) vanishes on all ei, and since these elements generate V it follows that p(φ) = 0 ∈ End(V), completing the proof.
One additional fact that follows from this proof is that the matrix A whose characteristic polynomial is taken need not be identical to the value φ substituted into that polynomial; it suffices that φ be an endomorphism of V satisfying the initial equations
φ ( e i ) = ∑ j A j , i e j {\displaystyle \varphi (e_{i})=\sum _{j}A_{j,i}e_{j}}
for some sequence of elements e1,...,en that generate V (which space might have smaller dimension than n, or in case the ring R is not a field it might not be a free module at all).
A bogus "proof": p(A) = det(AIn − A) = det(A − A) = 0[edit]
One persistent elementary but incorrect argument[18] for the theorem is to "simply" take the definition
p ( λ ) = det ( λ I n − A ) {\displaystyle p(\lambda )=\det(\lambda I_{n}-A)}
and substitute A for λ, obtaining
p ( A ) = det ( A I n − A ) = det ( A − A ) = 0 . {\displaystyle p(A)=\det(AI_{n}-A)=\det(A-A)=0~.}
There are many ways to see why this argument is wrong. First, in Cayley–Hamilton theorem, p(A) is an n×n matrix. However, the right hand side of the above equation is the value of a determinant, which is a scalar. So they cannot be equated unless n = 1 (i.e. A is just a scalar). Second, in the expression det ( λ I n − A ) {\displaystyle \det(\lambda I_{n}-A)} , the variable λ actually occurs at the diagonal entries of the matrix λ I n − A {\displaystyle \lambda I_{n}-A} . To illustrate, consider the characteristic polynomial in the previous example again:
det ( λ − 1 − 2 − 3 λ − 4 ) . {\displaystyle \det {\begin{pmatrix}\lambda -1&-2\\-3&\lambda -4\end{pmatrix}}.}
If one substitutes the entire matrix A for λ in those positions, one obtains
det ( ( 1 2 3 4 ) − 1 − 2 − 3 ( 1 2 3 4 ) − 4 ) , {\displaystyle \det {\begin{pmatrix}{\begin{pmatrix}1&2\\3&4\end{pmatrix}}-1&-2\\-3&{\begin{pmatrix}1&2\\3&4\end{pmatrix}}-4\end{pmatrix}},}
in which the "matrix" expression is simply not a valid one. Note, however, that if scalar multiples of identity matrices instead of scalars are subtracted in the above, i.e. if the substitution is performed as
det ( ( 1 2 3 4 ) − I 2 − 2 I 2 − 3 I 2 ( 1 2 3 4 ) − 4 I 2 ) , {\displaystyle \det {\begin{pmatrix}{\begin{pmatrix}1&2\\3&4\end{pmatrix}}-I_{2}&-2I_{2}\\-3I_{2}&{\begin{pmatrix}1&2\\3&4\end{pmatrix}}-4I_{2}\end{pmatrix}},}
then the determinant is indeed zero, but the expanded matrix in question does not evaluate to A I n − A {\displaystyle AI_{n}-A} ; nor can its determinant (a scalar) be compared to p(A) (a matrix). So the argument that p ( A ) = det ( A I n − A ) = 0 {\displaystyle p(A)=\det(AI_{n}-A)=0} still does not apply.
Actually, if such an argument holds, it should also hold when other multilinear forms instead of determinant is used. For instance, if we consider the permanent function and define q ( λ ) = perm  ( λ I n − A ) {\displaystyle q(\lambda )=\operatorname {perm} (\lambda I_{n}-A)} , then by the same argument, we should be able to "prove" that q(A) = 0. But this statement is demonstrably wrong. In the 2-dimensional case, for instance, the permanent of a matrix is given by
perm  ( a b c d ) = a d + b c . {\displaystyle \operatorname {perm} {\begin{pmatrix}a&b\\c&d\end{pmatrix}}=ad+bc.}
So, for the matrix A in the previous example,
q ( λ ) = perm  ( λ I 2 − A ) = perm  ( λ − 1 − 2 − 3 λ − 4 ) = ( λ − 1 ) ( λ − 4 ) + ( − 2 ) ( − 3 ) = λ 2 − 5 λ + 10. {\displaystyle {\begin{aligned}q(\lambda )&=\operatorname {perm} (\lambda I_{2}-A)=\operatorname {perm} {\begin{pmatrix}\lambda -1&-2\\-3&\lambda -4\end{pmatrix}}\\[6pt]&=(\lambda -1)(\lambda -4)+(-2)(-3)=\lambda ^{2}-5\lambda +10.\end{aligned}}}
Yet one can verify that
q ( A ) = A 2 − 5 A + 10 I 2 = 12 I 2 ≠ 0. {\displaystyle q(A)=A^{2}-5A+10I_{2}=12I_{2}\not =0.}
One of the proofs for Cayley–Hamilton theorem above bears some similarity to the argument that p ( A ) = det ( A I n − A ) = 0 {\displaystyle p(A)=\det(AI_{n}-A)=0} . By introducing a matrix with non-numeric coefficients, one can actually let A live inside a matrix entry, but then A I n {\displaystyle AI_{n}} is not equal to A, and the conclusion is reached differently.
Abstraction and generalizations[edit]
The above proofs show that the Cayley–Hamilton theorem holds for matrices with entries in any commutative ring R, and that p(φ) = 0 will hold whenever φ is an endomorphism of an R module generated by elements e1,...,en that satisfies
φ ( e j ) = ∑ a i j e i , j = 1 , … , n . {\displaystyle \varphi (e_{j})=\sum a_{ij}e_{i},\qquad j=1,\ldots ,n.}
This more general version of the theorem is the source of the celebrated Nakayama lemma in commutative algebra and algebraic geometry.
See also[edit]
Companion matrix
Remarks[edit]
Jump up ^ Due to the non-commutative nature of the multiplication operation for quaternions and related constructions, care needs to be taken with definitions, most notably in this context, for the determinant. The theorem holds as well for the slightly less well-behaved split-quaternions, see Alagös, Oral & Yüce (2012). The rings of quaternions and split-quaternions can both be represented by certain 2 × 2 complex matrices. (When restricted to unit norm, these are the groups SU(2) and SU(1, 1) respectively.) Therefore it is not surprising that the theorem holds.

There is no such matrix representation for the octonions, since the multiplication operation is not associative in this case. However, a modified Cayley-Hamilton theorem still holds for the octonions, see Tian (2000).
Jump up ^ An explicit expression for these coefficients is
c i = ∑ k 1 , k 2 , … , k n ∏ l = 1 n ( − 1 ) k l + 1 l k l k l ! tr  ( A l ) k l , {\displaystyle c_{i}=\sum _{k_{1},k_{2},\ldots ,k_{n}}\prod _{l=1}^{n}{\frac {(-1)^{k_{l}+1}}{l^{k_{l}}k_{l}!}}\operatorname {tr} (A^{l})^{k_{l}},}
where the sum is taken over the sets of all integer partitions kl ≥ 0 satisfying the equation
∑ l = 1 n l k l = n − i . {\displaystyle \sum _{l=1}^{n}lk_{l}=n-i.}
Jump up ^ See, e.g., p. 54 of Brown 1994, which solves Jacobi's formula,
∂ p ( λ ) / ∂ λ = p ( λ ) ∑ m = 0 ∞ λ − ( m + 1 ) tr  A m = p ( λ ) tr  I λ I − A ≡ tr  B , {\displaystyle \partial p(\lambda )/\partial \lambda =p(\lambda )\sum _{m=0}^{\infty }\lambda ^{-(m+1)}\operatorname {tr} A^{m}=p(\lambda )~\operatorname {tr} {\frac {I}{\lambda I-A}}\equiv \operatorname {tr} B~,}
where B is the adjugate matrix of the next section. There also exists an equivalent, related recursive algorithm introduced by Urbain Le Verrier and Dmitry Konstantinovich Faddeev—the Faddeev–LeVerrier algorithm, which reads
M 0 ≡ O c n = 1 ( k = 0 ) M k ≡ A M k − 1 − 1 k − 1 ( tr  ( A M k − 1 ) ) I c n − k = − 1 k tr  ( A M k ) k = 1 , … , n . {\displaystyle {\begin{aligned}M_{0}&\equiv O&c_{n}&=1\qquad &(k=0)\\M_{k}&\equiv AM_{k-1}-{\frac {1}{k-1}}(\operatorname {tr} (AM_{k-1}))I\qquad \qquad &c_{n-k}&=-{\frac {1}{k}}\operatorname {tr} (AM_{k})\qquad &k=1,\ldots ,n~.\end{aligned}}}
(see, e.g., p 88 of Gantmacher 1960.) Observe A−1 = −Mn /c0 as the recursion terminates. See the algebraic proof in the following section, which relies on the modes of the adjugate, Bk ≡ Mn−k . Specifically, ( λ I − A ) B = I p ( λ ) {\displaystyle (\lambda I-A)B=Ip(\lambda )} and the above derivative of p when one traces it yields
λ p ′ − n p = tr  ( A B ) , {\displaystyle \lambda p'-np=\operatorname {tr} (AB)~,} (Hou 1998), and the above recursions, in turn.
Notes[edit]
^ Jump up to: a b Crilly 1998
^ Jump up to: a b Cayley 1858, pp. 17–37
Jump up ^ Cayley 1889, pp. 475–496
^ Jump up to: a b Hamilton 1864a
^ Jump up to: a b Hamilton 1864b
^ Jump up to: a b Hamilton 1862
Jump up ^ Atiyah & MacDonald 1969
Jump up ^ Hamilton 1853, p. 562
Jump up ^ Zhang 1997
^ Jump up to: a b Frobenius 1878
Jump up ^ Zeni & Rodrigues 1992
Jump up ^ Barut, Zeni & Laufer 1994a
Jump up ^ Barut, Zeni & Laufer 1994b
Jump up ^ Laufer 1997
Jump up ^ Curtright, Fairlie & Zachos 2014
Jump up ^ Stein, William. Algebraic Number Theory, a Computational Approach (PDF). p. 29.
Jump up ^ Bhatia 1997, p. 7
Jump up ^ Garrett, 2007 & p. 381
References[edit]
Alagös, Y.; Oral, K.; Yüce, S. (2012). "Split Quaternion Matrices". Miskolc Mathematical Notes. 13 (2): 223&ndash, 232. ISSN 1787-2405 (open access)
Atiyah, M. F.; MacDonald, I. G. (1969), Introduction to Commutative Algebra, Westview Press, ISBN 978-0-201-40751-8
Barut, A. O.; Zeni, J. R.; Laufer, A. (1994a). "The exponential map for the conformal group O(2,4)". J. Phys. A: Math. Gen. 27 (15): 5239&ndash, 5250. arXiv:hep-th/9408105. Bibcode:1994JPhA...27.5239B. doi:10.1088/0305-4470/27/15/022.
Barut, A. O.; Zeni, J. R.; Laufer, A. (1994b). "The exponential map for the unitary group SU(2,2)". J. Phys. A: Math. Gen. 27 (20): 6799&ndash, 6806. arXiv:hep-th/9408145. Bibcode:1994JPhA...27.6799B. doi:10.1088/0305-4470/27/20/017.
Bhatia, R. (1997). Matrix Analysis. Graduate texts in mathematics. 169. Springer. ISBN 978-0387948461.
Brown, Lowell S. (1994). Quantum Field Theory. Cambridge University Press. ISBN 978-0-521-46946-3.
Cayley, A. (1858). "A Memoir on the Theory of Matrices". Philos. Trans. 148.
Cayley, A. (1889). The Collected Mathematical Papers of Arthur Cayley. (Classic Reprint). 2. Forgotten books. ASIN B008HUED9O.
Crilly, T. (1998). "The young Arthur Cayley". Notes Rec. R. Soc. Lond. 52 (2): 267–282. doi:10.1098/rsnr.1998.0050.
Curtright, T L; Fairlie, D B; Zachos, C K (2014). "A compact formula for rotations as spin matrix polynomials". SIGMA. 10 (2014): 084. arXiv:1402.3541. Bibcode:2014SIGMA..10..084C. doi:10.3842/SIGMA.2014.084.
Frobenius, G. (1878). "Ueber lineare Substutionen und bilineare Formen". J. Reine Angew. Math. 84: 1–63.
Gantmacher, F.R. (1960). The Theory of Matrices. NY: Chelsea Publishing. ISBN 978-0-8218-1376-8.
Garrett, Paul B. (2007). Abstract Algebra. NY: Chapman and Hall/CRC. ISBN 978-1584886891.
Hamilton, W. R. (1853). Lectures on Quaternions. Dublin.
Hamilton, W. R. (1864a). "On a New and General Method of Inverting a Linear and Quaternion Function of a Quaternion". Proceedings of the Royal Irish Academy. viii: 182&ndash, 183. (communicated on June 9, 1862)
Hamilton, W. R. (1864b). "On the Existence of a Symbolic and Biquadratic Equation, which is satisfied by the Symbol of Linear Operation in Quaternions". Proceedings of the Royal Irish Academy. viii: 190&ndash, 101. (communicated on June 23, 1862)
Hou, S. H. (1998). "Classroom Note: A Simple Proof of the Leverrier--Faddeev Characteristic Polynomial Algorithm". SIAM Review. 40 (3): 706–709. Bibcode:1998SIAMR..40..706H. doi:10.1137/S003614459732076X. "Classroom Note: A Simple Proof of the Leverrier--Faddeev Characteristic Polynomial Algorithm"
Hamilton, W. R. (1862). "On the Existence of a Symbolic and Biquadratic Equation which is satisfied by the Symbol of Linear or Distributive Operation on a Quaternion". The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. series iv. 24: 127&ndash, 128. ISSN 1478-6435. Retrieved 2015-02-14.
Householder, Alston S. (2006). The Theory of Matrices in Numerical Analysis. Dover Books on Mathematics. ISBN 978-0486449722.
Laufer, A. (1997). "The exponential map of GL(N)". J. Phys. A: Math. Gen. 30 (15): 5455&ndash, 5470. arXiv:hep-th/9604049. Bibcode:1997JPhA...30.5455L. doi:10.1088/0305-4470/30/15/029.
Tian, Y. (2000). "Matrix representations of octonions and their application". Advances in Applied Clifford Algebras. 10 (1): 61&ndash, 90. arXiv:math/0003166v2. CiteSeerX 10.1.1.237.2217. doi:10.1007/BF03042010. ISSN 0188-7009.
Zeni, J. R.; Rodrigues, W.A. (1992). "A thoughful study of Lorentz transformations by Clifford algebras". Int. J. Mod. Phys. A. 7 (8): 1793 pp. Bibcode:1992IJMPA...7.1793Z. doi:10.1142/S0217751X92000776.
Zhang, F. (1997). "Quaternions and matrices of quaternions". Linear Algebra and its Applications. 251: 21&ndash, 57. doi:10.1016/0024-3795(95)00543-9. ISSN 0024-3795 – via ScienceDirect (open archive).
External links[edit]
Hazewinkel, Michiel, ed. (2001) [1994], "Cayley–Hamilton theorem", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4
A proof from PlanetMath.
The Cayley–Hamilton theorem at MathPages
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values: {\displaystyle \mathbb {Q} [\alpha _{1},\ldots ,\alpha _{k}]}

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values: \operatorname {adj} (M)\cdot M=\det(M)I_{n}=M\cdot \operatorname {adj} (M)~.

values: {\displaystyle B=\operatorname {adj} (tI_{n}-A).}

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values: {\displaystyle \operatorname {ev} _{A}:(R[A])[t]\to R[A]}

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values: \operatorname {ev} _{A}:(R[A])[t]\to R[A]

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values: \operatorname {adj} (-A)=\sum _{i=1}^{n}c_{i}A^{i-1},

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values: {\displaystyle q(\lambda )=\operatorname {perm} (\lambda I_{n}-A)}

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values: q(\lambda )=\operatorname {perm} (\lambda I_{n}-A)

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The Truth About Jane
From Wikipedia, the free encyclopedia
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The Truth About Jane
The Truth About Jane
Directed by
Lee Rose
Produced by
Orly Adelson
Written by
Lee Rose
Starring
Ellen Muth
Stockard Channing
Kelly Rowan
Music by
Terence Blanchard
Cinematography
Eric Van Haren Noman
Edited by
Peter V. White
Production
company
Orly Adelson Productions
Hearst Entertainment Productions
Distributed by
Lifetime Television
Release date
7 August 2000
Running time
87 minutes
Language
English
The Truth About Jane is a Lifetime Original Movie, directed by Lee Rose, that first aired on TV on August 7, 2000 and stars Stockard Channing, Ellen Muth, Kelly Rowan Jenny O Hara and RuPaul (credited under RuPaul Charles). It's about a teenage girl named Jane (Muth) who struggles not only with her sexuality, but with a mother (Channing) who refuses to accept her.
The film was nominated for several awards, including Outstanding TV Movie by GLAAD, Outstanding Performance by a Female Actor in a Television Movie or Miniseries (Channing) Screen Actors Guild, and Original Long Form by WGA.
The film ends with a dedication to Matthew Shepard and to "all the men & women who love differently".
Plot[edit]
The movie begins with Jane narrating about her life from her birth, to her first birthday and even her first day of kindergarten. Throughout the events, Jane talks about how overprotective her parents have been of her. The film jumps to Jane at age 15, starting her first day of high school. Although Jane appears to be happy and a normal teenage girl to her friends and family, inside she's feeling alone and different, particularly due to her lack of interest in boys which her friends take note of. Her feelings grow more confusing when a new girl named Taylor arrives in class (played by Alicia Lagano) who Jane sees (and says in a voice over) as "different, smarter, wiser". She then continues to say in the voice over that "maybe because she wasn't from here or maybe it was just her. I'm going to go with the last one, because she ended up changing my life."
The two initially become innocent friends by Jane helping her get caught up in class due to delayed paperwork for her transfer. Eventually Jane's feelings start to become gradually clearer and she realizes that she has a crush on Taylor. Because of Taylor's abusive home life, she misses school for several days, causing Jane to worry about her and to visit her at her house. When Taylor's mother starts yelling at her for being at the door, Taylor tells Jane to leave and that she'll be fine. Worried about her friend and struggling with figuring herself out, Jane starts to become withdrawn from her parents, who are starting to grow concerned.
When Jane's mother, Janice, finally tries to talk about what's been going on, they're interrupted by Jane's father coming in while announcing that Taylor has come for a visit. After they're left alone by Jane's parents, Taylor confesses about her home life and how scared she is sometimes. Jane, wanting to comfort her, brushes Taylor's hair away and begins stroking it. Taylor stares at her for a moment, before she and Jane share their first kiss. In a voice over, Jane wonders to herself if kissing Taylor made her gay and that, at the time, she convinced herself that "it was just a phase". She then goes on to say that "it was the first time [she] ever felt connected to someone". After that, the two become an official couple.
Janice does not like Taylor, believing she's a bad influence on Jane, since she is no longer hanging out with her old friends and doesn't spend as much time with the family like she previously has. A while later, Taylor invites Jane over to her house since her mother is going to be out of town. Despite being terrified of her still confused feelings, Jane and Taylor have sex for the first time. The next day at school, Jane tells Taylor that "it was a mistake" and that she's not gay. Taylor says it's not about being gay, it's about being with the person you want to be with. Hurt, Taylor breaks up with Jane.
With Taylor gone, Jane is more withdrawn and heartbroken. When she finally gets over her confusion, she asks Taylor for a second chance, but Taylor turns her down. Distraught, Jane begins to cry and is then confronted by her English teacher/guidance counselor Ms. Walcott (Rowan). Ms. Walcott takes Jane into her office and, with some gentle encouraging, Jane confesses that she had sex with someone for the first time and about her situation with Taylor. However, she's careful to not specifically say the person is another girl and instead uses such euphemisms as "they're" and "this person".
Ms. Walcott then suggests that Jane write Taylor a note to express how she feels and why she acted the way she did. Jane does and a few days later, Taylor shows up at her house, while Jane is babysitting her brother while her parents are out. Thinking her brother is asleep on the couch, Jane takes Taylor up to her room and, after reconciling, the two share a kiss. While they do so, they are unaware of Jane's brother watching them through her partially open bedroom door.
Jane's brother tells his close classmate in class about Jane and Taylor, who happens to be the younger brother of one of Jane's former friends whom she stopped seeing in order to see Taylor more. As a result, one night after dinner, Janice gets an anonymous phone call from someone asking if she knows her daughter is a lesbian. Not knowing what to do, Jane's parents confront her about it and ask if it's true. Fearing that her parents will hate her, Jane lies and claims that she and Taylor were just practicing, which is what her father already assumed they were doing.
Confused and consumed with guilt for lying to her parents about her true relationship with Taylor, Jane seeks advice from her mother's gay friend Jimmy (RuPaul) and comes out to him. Jimmy explains that it will get easier and that when the time is right, she'll tell her parents the truth. Eventually the harassment and name calling gets too much for Jane at school and, during dinner, Jane says how she wants to see Taylor only for her younger brother to call Taylor a dyke. Snapping, Jane attacks him by pulling him across the dining room table and hitting him. When her parents yell at her and say that she should rise above gossip, Jane confesses that it's not just gossip and comes out to her parents as a lesbian.
Shocked, hurt and angry, both her parents initially react badly to it, assuming that they had done something wrong as parents and decide that they will do whatever it takes to make their daughter's life "better" for her. They forbid Jane from seeing Taylor and send her to conversion therapy since they feel it's the only way to "fix" the situation. Therapy proves useless and Jane and Taylor still continue to see each other, even going as far as sneaking out to a gay bar, which Jane gets grounded for when she comes home at 4 AM. Janice, meanwhile, goes to her two close friends to talk about the situation. Despite both her friends saying that Janice should accept Jane for who she is, as it is a mother's duty to love her child, Janice refuses to as she believes Jane is ruining her life. Soon, the sneaking around and drama becomes too much for Taylor and she breaks up with Jane, breaking her heart.
Ms. Walcott, who happened to have been by, stops to comfort Jane when seeing her crying. She then explains to Jane that it will get better and that she understands what she's going through. At first Jane doesn't believe her, but Ms. Walcott comes out to her as a lesbian and tells her the story of her first time falling in love with a girl named Barbara when she was sixteen. She explains how she naively thought that she and Barbara would be together forever, but was eventually dumped by Barbara for a male football star at their school. As Ms. Walcott says she thought it was "the end of the world" for her at the time, but eventually emotionally healed and moved forward, Jane starts to feel better. However, during lunch at school, she's confronted by her old friends who begin taunting her about being a lesbian. Jane ends up attacking one of them, causing her to get suspended.
After Janice picks up Jane from school following the suspension and they walk out of the high school, a group of boys make homophobic remarks about Jane in front of her mother, who is hurt by the comments. Jane sarcastically says she doesn't know why her mother should care as she is "one of them". At home, Janice tries to confront her daughter about her remark and claims that what Jane is doing isn't normal, Jane angrily begins to rant that gay people are perfectly normal, unintentionally outing Ms. Walcott in the process.
Believing that Ms. Walcott is influencing her daughter to be a lesbian, Janice declares that such a person shouldn't be a teacher. Distressed, Jane collapses into her father's arms and begins to sob for how her mother is reacting and treating her. Janice angrily confronts Ms. Walcott at the school, demanding that she stay away from her daughter and threatening to go to the school board if she doesn't.
Later on at home, Jane's parents announce that they feel sending her away to boarding school will be the best way to handle the situation, explaining how they're at a loss for what to do. Jane tells them that sending her away isn't going to change who she is and storms off. Having nowhere else to go, she runs away to Ms. Walcott's house, meeting her partner. Jane apologizes to her for outing her and tells her that she is considering suicide as she can't stand having her parents hate her so much.
Knowing exactly how she feels and wanting to help Jane before it is too late, Ms. Walcott goes to Jane's parents and tells them her own personal experience growing up as a lesbian with parents who refused to accept her and, when Janice claims to know Jane better than she does, tells them about Jane considering suicide. Stunned into realization that they need to learn to accept Jane before losing her forever, they go along with Ms. Walcott back to her home. Janice and Jane talk and the two reconcile, though their relationship is still strained. In an attempt to make her mother more comfortable with things, Jane takes her to a PFLAG (Parents, Families and Friends of Lesbians and Gays) meeting, where they hear a story from another mother whose situation mirrored their own. Janice continues to go to the meetings, despite still being uncomfortable with things and still being unable to admit that her daughter is a lesbian.
When it comes time for a pride rally, Janice announces at dinner that she's not going to go because she's still not ready, which obviously hurts Jane. Later that night, Janice apologizes and says she's trying. But Jane says that accepting her in private isn't enough as it still proves a part of her mother is still ashamed of her. Jane later attends the rally with her father, brother, Jimmy and another friend of her mother's and are eventually joined by Ms. Walcott and her partner.
In the ending of the film, when it's time for the PFLAG parents from Janice's group to speak, Jane looks around at the crowd and sees her mother walking through the crowd towards her. Realizing her mother has finally learned to accept her, the two happily smile at each other as Janice joins Jane and their family and friends by their side.
Cast[edit]
Stockard Channing as Janice
Ellen Muth as Jane
Kelly Rowan as Ms. Lynn Walcott
Jenny O'Hara as Beth
RuPaul as Jimmy
Noah Fleiss as Ned
James Naughton as Robert
Alicia Lagano as Taylor
Kacy Clark as Brad
Mark Siegel as Doctor
Erika Moore as Jane, age 9
Tiffany Westlie as Kate
Kenny McKeon as Grant
Michelle Lane as Betsy
Amy Davidson as Elizabeth
Melissa Wolfklain as Claire
Lillie Richardson as Dr. Sanders
John T. Huggins Jr. as Boy
Makyle Kaplan-Dailey as Gay Boy
Robert Harvey as Mr. Anderson
Victor Kravetz as Boy
Molly Schaffer as Margaret
Lucky Hayes as Dorothy
Terey Summers as Lorraine
Maedell Dixon as Susan
Will Bribiescas as Raver / Club Scene
Krysten Jackson as Student
Chuck Rahill as Ambulance Patient
John Thomas as Boy
Bob Huff as Doctor with Briefcase
George M. Kostuch as Teacher
Jose Rosete as Diner Patron
Cheryl Schmidt as Student
External links[edit]
The Truth About Jane on IMDB.com
The Truth About Jane on AfterEllen.com
Retrieved from "https://en.wikipedia.org/w/index.php?title=The_Truth_About_Jane&oldid=794446550"
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