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$L(A)=A^T$ from $\Bbb R^{2\times 2}$ to $\Bbb R^{2\times 2}$

I know $\det(A^T) = \det(A)$ if $A$ is a square matrix. I also know that a basis for $\Bbb R^{2\times 2}$ is

$$\Bigg(\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}\Bigg)$$ but otherwise I'm not sure how to solve this problem. Any help would be appreciated.

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    $\begingroup$ Hint: Write down the matrix of $L$ w.r.t. that standard basis. $\endgroup$ – Mathematician 42 May 3 '16 at 23:49
  • $\begingroup$ $B=\begin{bmatrix}a&c\\b&d\end{bmatrix}$ like this? Then $det(B)=ad-bc$ $\endgroup$ – Lanous May 3 '16 at 23:53
  • $\begingroup$ No, if you have a linear map $T:V\rightarrow W$ with $\dim(V)=n$ and $\dim(W)=m$, then any matrix corresponding to this linear map will be a $m\times n$-matrix. $\endgroup$ – Mathematician 42 May 3 '16 at 23:56
  • $\begingroup$ Maybe it helps you to consider $\mathbb{R}^{2\times 2}\cong \mathbb{R}^4$, then $L$ becomes a linear transformation of $\mathbb{R}^4$. $\endgroup$ – Mathematician 42 May 3 '16 at 23:58
  • $\begingroup$ Would I end with something like this \begin{bmatrix}a&0&0&0\\0&b&0&0\\0&0&c&0\\0&0&0&c\end{bmatrix} $\endgroup$ – Lanous May 4 '16 at 0:17
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We have $L:\mathbb{R}^{2\times 2}\rightarrow\mathbb{R}^{2\times 2}:\begin{pmatrix} a&b\\c&d\end{pmatrix}\mapsto \begin{pmatrix}a&c\\b&d \end{pmatrix}$. Or after the identication $\mathbb{R}^{2\times 2}\cong \mathbb{R}^4$, we get the same linear map $L:\mathbb{R}^4\rightarrow \mathbb{R}^4:(a,b,c,d)\mapsto (a,c,b,d)$. Notice that $$L(a,b,c,d)=\begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1 \end{pmatrix}\begin{pmatrix} a\\b\\c\\d \end{pmatrix}.$$ Hence the matrix of $L$ w.r.t. the standard basis $\alpha$ is $$[L]_{\alpha}^{\alpha}=\begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1 \end{pmatrix}.$$ By definition $\det(L)=\det([L]_{\alpha}^{\alpha})$.

Now @JimmyK4542's answer is easier if you understand properly what's going on. But I feel you're struggling with determining the matrix of linear map w.r.t. some bases. So to make you understand it, try to find the matrix of $L$ w.r.t. the standard basis $\alpha$ and $\beta=\left\{\begin{pmatrix}1&1\\1&1\end{pmatrix},\begin{pmatrix}1&1\\1&0\end{pmatrix},\begin{pmatrix}1&1\\0&0\end{pmatrix},\begin{pmatrix}1&0\\0&0\end{pmatrix}\right\}$. Calculate the determinant of that matrix, it should be the same as th determinant of $[L]_{\alpha}^{\alpha}$.

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Hint: The matrices $$\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}0&-1\\1&0\end{bmatrix}\right\}$$

form a basis of $\mathbb{R}^{2 \times 2}$ and each of them is an eigenmatrix of $L$, i.e. $L(A) = \lambda A$ where $\lambda$ is a scalar.

What is the product of the eigenvalues of $L$?

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