Q:

Issues with a particular bilinear form and determining rank, signature, etc. of its restriction

Let $b: M_2(\mathbb{R}) \times M_2(\mathbb{R}) \to \mathbb{R}$ such that $b(X,Y)=trace(X^tAY)$, where $X^t$ is the transpose of $X$ and $A=\begin{pmatrix} 2 & 1\\1 & 0\end{pmatrix}$.
In my original exercise I had to determine the matrix associated to $b$ with respect to various bases, and I could do that without trouble. However, while doing this, I asked myself some additional questions (which unfortunately I wasn't able to answer). Here they are: 

How can I prove that this is indeed a bilinear symmetric form when $A$ is a generic matrx?
Considering $A=\begin{pmatrix} 2 & 1\\1 & 0\end{pmatrix}$, how can I determine the rank, kernel, image and signature of the restriction of b to the subspace of symmetric matrices of order $2$?

My issues are with the whole process of the proof in point (1) and with working with the restriction in point (2). Could you please show me with details how to proceed with these problems?

A:

For the first problem: note that
$$
b(X_1 + a X_2,Y) = \\
\operatorname{trace}[(X_1 + aX_2)^TAY] = \\
\operatorname{trace}[X_1^TAY + aX_2^TAY] =\\
\operatorname{trace}[X_1^TAY] + 
a\operatorname{trace}[X_2^TAY] =\\
b(X_1,Y) + ab(X_2,Y)
$$
this shows that $b$ is linear in the first argument.  We may do something similar for the other argument.

For the second: it suffices to find $b$ with respect to a basis for this space.  We can take
$$
v_1 = \pmatrix{1&0\\0&0}, \quad
v_2 = \pmatrix{0&1\\1&0}, \quad
v_3 = \pmatrix{0&0\\0&1}
$$
For ease of computation, note that
$$
\operatorname{trace} \pmatrix{u^T\\v^T} A \pmatrix{x&y} = u^TAx + v^TAy 
$$
Going through it all, you should find that the matrix of the bilinear form (over the restriction) is
$$
\pmatrix{2&1&0\\1&2&1\\0&1&0}
$$

