Matrix Representation of Bilinear Form/Map in Matrix Space

Question

I am trying to understand bilinear forms and have a related problem, but unfortunately all of the examples I've been able to find deal with $V = \mathbb{R}^n$. I am dealing with $V = M_{n \times n}^{\mathbb{R}}$.

This question probably stems from my incomplete understanding of linear subspaces, so I apologize in advance if it's silly, but please give me a direction.

The matrix representation of a bilinear form is defined thus:

If $f : V\times V \rightarrow \mathbb{R}$ and $B = \{b_1, b_2, ..., b_n\}$ a basis of $V$ then:

$[f]_B = A$ so that $A_{ij} = f(b_i, b_j)$.

However, as far as I know, in the matrix space the basis has two parameters (for example, $b_{ij}$), so the above definition seems meaningless. How to reconcile this?

Right now I'm working with:

$f(A, B) = tr(A^tMB)$, where $M = \begin{pmatrix} 1 & 2 \\ 3 & 5\end{pmatrix}$

How to determine $[f]_B$? An example with B as the standard basis would be appreciated.

Answer 1

Your bilinear form is \begin{array}{ccc} M_{2\times 2}(\Bbb R)\times M_{2\times 2}(\Bbb R) & \xrightarrow{f} & \Bbb R \\ (A,B) & \mapsto & \DeclareMathOperator{trace}{trace}\trace(A^\top M B) \end{array} where $$ M=\begin{bmatrix}1&2\\3&5\end{bmatrix} $$ Note that $$ \beta=\left\{ e_1= \begin{bmatrix}1&0\\0&0\end{bmatrix}, e_2= \begin{bmatrix}0&1\\0&0\end{bmatrix}, e_3= \begin{bmatrix}0&0\\1&0\end{bmatrix}, e_4= \begin{bmatrix}0&0\\0&1\end{bmatrix} \right\} $$ is a basis for $M_{2\times 2}(\Bbb R)$. This implies that $[f]_\beta$ is an $4\times 4$ matrix whose $(i,j)$th entry is $f(e_i,e_j)$. For example, $$ f(e_2,e_4)=\trace\begin{bmatrix}0&0\\0&2\end{bmatrix}=2 $$ so $$ [f]_\beta= \begin{bmatrix} *&*&*&*\\*&*&*&2\\*&*&*&*\\*&*&*&* \end{bmatrix} $$ Can you fill in the rest of the matrix?

Answer 2

The standard basis for $M_2(\mathbb{R})$ is
$$ B =\left\{m_{1},m_{2},m_{3},m_{4} \right\} =\left\{ \left[\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right], \left[\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right] , \left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right] ,\left[\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right] \right\}. $$ Then $$ [f]_B =\left[ \begin{array}{cccc} f(m_1,m_1) & f(m_1,m_1) & f(m_1,m_3) & f(m_1,m_4) \\ f(m_2,m_1) & f(m_2,m_1) & f(m_2,m_3) & f(m_2,m_4) \\ f(m_3,m_1) & f(m_3,m_1) & f(m_3,m_3) & f(m_3,m_4) \\ f(m_4,m_1) & f(m_4,m_1) & f(m_4,m_3) & f(m_4,m_4) \\ \end{array} \right] $$ Now compute $$ f(m_1,m_1)=\mathrm{tr} \left( \left[\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]^t \left[\begin{array}{cc} 1 & 2 \\ 3 & 5 \end{array} \right] \left[\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right] \right) = \mathrm{tr} \left(\left[\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right] \right)=1. $$ Similarly, $$ f(m_1,m_2)=\mathrm{tr} \left( \left[\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]^t \left[\begin{array}{cc} 1 & 2 \\ 3 & 5 \end{array} \right] \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right] \right) = \mathrm{tr} \left(\left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right] \right)= 0 $$ Continuing in this fashion, we get $$ [f]_B=\left[ \begin {array}{cccc} 1&0&2&0\\0&1&0&2 \\ 3&0&5&0\\ 0&3&0&5\end {array} \right] $$