Clarification about finding a bilinear form's matrix

Question

Say $V:=M_{2\times 3}(\mathbb{R})$ and let $f:V\times V \to \mathbb{R}$ be defined $f(X,Y)=Tr(X^TAY)$ for $A=\begin{pmatrix} 1 & 2 \\ 3 & 4\\ \end{pmatrix} \, $.
So I want to find the the matrix of $f$ for the standart basis of $M_{2\times 3}(\mathbb{R})$. If this matrix is $B$ so I know $(B)_{ij}=f(v_i,v_j)$, but I'm kind of getting confused about which $v_i$ I should calculate, and what is the order of $B$ anyway? If someone coukd focus me and explain a bit it'd be great. Thanks.

Answer 1

Take $v_1=\begin{pmatrix} 1 & 0 & 0\\ 0 & 0& 0 \end{pmatrix}$, $v_2=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0& 0 \end{pmatrix}$, $v_3=\begin{pmatrix} 0 & 0 & 1\\ 0 & 0& 0 \end{pmatrix}$, similarly define $v_4,v_5,v_6$. Using your definition of $B_{ij}$, you will get a $6\times 6$ matrix for $B$.