Linear Operator Matrix Problem

Question

I was doing some problems on my book and i stuck to the question below. Here's the question:

$T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} c & a \\ d & b \end{bmatrix}$ for all $\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \mathbb{R}^{2\times 2}$. If $A$ is eigenvector $T$ for eigenvalue $-1$, then $\det(A) = ... $

The first thing which came to my head was trying to figure out the operation of $T$, I didn't think all my work lead me anywhere. Any thought would be appreciated. Thanks

Answer 1

If $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} $ and $A$ is an eigenvector for the eigenvalue $-1$, then $T(A)=-A$.

Hence $A=-T(A)=\begin{bmatrix} -c & -a \\ -d & -b \end{bmatrix}$. Therefore

$ \det(A)=ad-bc= \det(\begin{bmatrix} -c & -a \\ -d & -b \end{bmatrix})=cb-ad= - \det(A)$, thus $\det(A)=0$.

Answer 2

If $A = \pmatrix{ a & b \\ c & d}$ is an eigenvector for eigenvalue $-1$ then $$\pmatrix{ c & a \\ d & b} = T(A) = -A = -\pmatrix{ a & b \\ c & d}$$

so $c = -a, a = -b, d = -c, b = -d$.

We get $A = \pmatrix{ a & -a \\ -a & a}$ for arbitrary $a \in \mathbb{R}$ so

$$\det A = a^2 - a^2 = 0$$