How does the restriction of a function change the coordinate matrix?

Question

I have the quadratic form $$q: \mathbb R^{2\times 2} \rightarrow \mathbb R: A \mapsto \det(A)$$ and its corresponding bilinear form: $$\sigma: (A,B) \mapsto \frac12tr(A*B^\#)$$ whereas $tr$ denotes the trace of a matrix and $B^\#$ denotes the cofactor matrix of $B$. I have an excercise that wants me to find the coordinate matrix of the restriction of $\sigma$ in various subspaces. For the subspace $S_1=\mathbb R^{2\times 2} $, I just found the coordinate matrix as usual and I end up with: $$\begin{pmatrix} 0 &0&0&0.5\\0&0&-0.5&0\\0&-0.5&0&0\\0.5&0&0&0\end{pmatrix}$$ Now I am asked to find the coordinate matrix for $\sigma|_{{S_2}\times{S_2}}$ with $S_2$ being the subspace of $\mathbb R^{2\times 2}$ of all matrices with $tr(A) = 0$. What confuses me is that I do not understand how the restriction of this function would change my coordinate matrix in any way. I can just put a matrix of $S_2$ into my old matrix and would end up with the same result. What am I missing here?

Answer 1

The subspace $S_2$ is a three-dimensional subspace of $\mathbb{R}^{2 \times 2}$ so you should get a $3 \times 3$ matrix. Let us choose some basis for $S_2$ such as

$$ e = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, f = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. $$

Then the matrix representing $\sigma|_{S_2}$ with respect to the basis $(e,f,g)$ will be

$$ \begin{pmatrix} \sigma(e,e) & \sigma(e,f) & \sigma(e,h) \\ \sigma(f,e) & \sigma(f,f) & \sigma(f,h) \\ \sigma(h,e) & \sigma(h,f) & \sigma(h,h) \end{pmatrix} $$

which you can compute explicitly. Since this matrix is symmetric and you probably used the standard basis of $\mathbb{R}^{2 \times 2}$ for the matrix of $\sigma$ on $\mathbb{R}^{2 \times 2}$, you already computed the terms which involve only $e,f$ so you only need to compute $\sigma(e,h), \sigma(f,h), \sigma(h,h)$.