Define $L(A) = A^T,$ for $A \in M_n(\mathbb{C}).$ Prove $L$ is diagonalizable and find eigenvalues Let $L:M_n(\mathbb{C}) \to M_n(\mathbb{C})$ be defined by $L(A) = A^T,$ where $A^T$ is the transpose of $A$ and $M_n(\mathbb{C})$ is the space of all $n \times n$ matrices with complex entries.
Prove that $L$ is diagonalizable and find the eignevalues of $L.$ 

I worked out a general $L$ that was $2 \times 2$ (might've found its inverse or something, see $L's$ representation below), it was bulkier than I thought it would be since its definition appears so simple.  The matrix I found is as follows,
Let $$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix},$$
and 
$$ L(A) = \begin{bmatrix} a & b \\ c & d \end{bmatrix}
\underbrace{\begin{bmatrix} ad-b^2 & cd-bd \\ ab-ac & ad-c^2 \end{bmatrix} \frac{1}{ad-bc}}_{L's \text{ representation}} = \begin{bmatrix} a & c \\ b & d \end{bmatrix} = A^T. $$
Now I suppose I could diagonalize this $L's$ representation, but that wouldn't accomplish much in the way of proving that $L$ can be diagonalized in general, as in the original question.  Any suggestions?
 A: Let's assume you have a finite dimensional vector space $V$ over a field with characteristic $\neq 2$ and an operator $T \colon V \rightarrow V$ that satisfies $T^2 = I$. The minimal polynomial of $T$ must divide $x^2 - 1 = (x - 1)(x + 1)$ and so $T$ must be diagonalizable and the only possible eigenvalues of $T$ are $\pm 1$. You can see this even without applying to the minimal polynomial argument by noting that
$$ V = V_{1} \oplus V_{-1} $$
where $V_{1} = \ker(T - I)$ and $V_{-1} = \ker(T + I)$. Every vector $v \in V$ can be written as
$$ v = \frac{v + Tv}{2} + \frac{v - Tv}{2} $$
where $\frac{v \pm Tv}{2} \in V_{\pm 1}$. The intersection is trivial since if $Tv = v$ and $Tv = -v$ then $2(Tv) = 0$ and so $v = Tv = 0$.
In your case, $V = M_{n}(\mathbb{C})$ and both $\pm 1$ are possible eigenvalues since if $A$ is a symmetric matrix then $L(A) = A$ and if $A$ is anti-symmetric then $L(A) = -A$. The decomposition
$$ A = \frac{A + A^T}{2} + \frac{A - A^T}{2} $$
writes every matrix $A \in M_n(\mathbb{C})$ uniquely as a sum of a symmetric and an anti-symmetric matrix.
A: Hint: In order to deduce that a linear operator is diagnolizable, it suffices to know the dimensionality of its eigenspaces. To this end we want to know the eigenvalues, which amounts to looking for $\lambda$ such that $L(A)=A^T = \lambda A$ for some $A$. But in that case, we also have $$A=(A^T)^T = (\lambda A)^T=\lambda A^T=\lambda^2 A.$$ From this we can deduce the eigenvalues, and from this characterize the eigenspaces. All that remains is to count the number of linearly-independent basis vectors in $M(\mathbb{C}_n)$, and confirm that this matches the dimensionality of $M(\mathbb{C}_n)$ itself.
A: To expand a bit on @David C. Ulrich comment about the $n^2 \times n^2$ representation.
If you allow yourself to vectorize the matrix : ${\bf a} = \text{vec}({\bf A})$ you can write the operator $L$ as a permutation matrix $\bf L$ so that:
$${\bf La} = \text{vec}({\bf A}^T)$$
Exactly how elements will be permuted depends on your choice of vectorization, but you can know for sure that ${\bf L}^2 = {\bf I}$ (why?). And the result is then easy to show for all potential eigenvalues must fulfil $\lambda^2 = 1$. But we can say that directly from construction of $\bf L$ as we know that the off diagonal element permutations can be paired in $2\times2$ blocks : $\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$ so the matrix must be permutation similar to a matrix which is a direct sum of such blocks and $1$ elements ( as the diagonal elements will of course map onto themselves ).
