let $T:V\rightarrow V$ be a linear transformation defined by $T(X)=A(X)$. Compute $\det(T)$. Question: Let $V$ be a vector space of $2\times 2$ matrices over a field $F$.  Let $A=\begin{bmatrix} 
a & b \\
c & d 
\end{bmatrix}\in V$ and let $T:V\rightarrow V$ be a linear transformation defined by $T(X)=A(X)$.  Compute $\det(T)$.
My thoughts: So, $\det(A)=ad-bc$ (not sure if this will help).  Let's take a $2\times 2$ matrix $X=\begin{bmatrix} 
e & f \\
g & h 
\end{bmatrix}$, and compute $AX$.  So, $AX=\begin{bmatrix} 
ae+bg & af+bh \\
ce+dg & cf+dh 
\end{bmatrix}$.  So, since $T(x)=AX$, then wouldn't the determinant of $T(x)$ just be $(cf+dh)(ae+bg)-(ce+dg)(af+bh)$?
 A: Given a matrix $X \in V$, you've computed $T(X)$. Now to represent matrices in $V$ as vectors, you can just reshape the $2 \times 2$ matrices as vectors in $\mathbb{R}^4$. Then the matrix representation of $T$ is the $4\times 4$ matrix $M$ such that
$$M \begin{bmatrix} e \\ f \\ g \\ h\end{bmatrix} = \begin{bmatrix} ae+bg \\ af+bh \\ ce+dg \\ cf+dh\end{bmatrix}$$
and compute $\det M$.
(Note that different "reshapings" of a $2 \times 2$ matrix into a vector will lead to slightly different matrices $M$, but the determinant will always be the same because the determinant does not change under changes of basis.)
A: We can start by the case that $A$ is diagonalisable and let $\lambda$ and $\mu$ be its eigen values and respectively $x$ and $y$ its eigen vectors. Since $(x,y)$ is a basis of $F^2$ then you can easily prove that $\left(xx^T, xy^T, yx^T, yy^T\right)$ is a basis of $V$. In this basis the matrix of $T(A)$ is : \begin{align}\begin{bmatrix}\lambda & 0 & 0 & 0\\ 0 & \lambda & 0 & 0\\ 0 & 0 & \mu & 0\\ 0 & 0 & 0 & \mu\end{bmatrix}\end{align} So $\det(T(A)) = \lambda^2 \mu^2 = \det(A)^2$.
This will hold also for every matrix $A$ since every matrix is a limit of diagonalisable matrices.
This method also can be easily generalized to the $n\times n$ case :
$$\det (T(A)) = \det(A)^n$$
