Eigenvalues and eigenvectors of a linear transformation that transposes a matrix a) Consider the transformation $T$ in the space $M^{2\times2}$ of $2\times2$ matrices, $T(A)=A^{\top}$. Find all its eigenvalues and eigenvectors. Is it possible to diagonalize this transformation?
b) Can you do the same problem but in the space of $n\times n$ matrices?
Since all linear transformation can be represented as a matrix, there should be a matrix that multiplies a $2 \times 2$ matrix that turns it into its transpose. But based on what I've read so far, there is no such matrix that exist. I might be misunderstanding this question, but if we can't find this matrix, how are we suppose to find its eigenvalues and eigenvectors?
 A: For solving this problem, my suggestion is that you use isomorphism first. Since the domain is $M^{2\times2}$, the basis should be $\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}$, which I named as $\mathcal{B}=\{v_1,v_2,v_3,v_4\}$ orderly.
Then we hope $Tv_1=v_1,\;Tv_2=v_3,\;Tv_3=v_2,\;Tv_4=v_4$.
Thus $[T]_{\mathcal{B}} =\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\\end{bmatrix}$. Then $\det([T]_\mathcal{B}-\lambda I)=(\lambda-1)^3(\lambda+1)$.
Then compute the $\text{Ker}(T-\lambda I)$.
For $\lambda = 1$, $\text{Ker}(T-I)=\text{span}\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix} \right\}$.
For $\lambda = -1$, $\text{Ker}(T+I)=\text{span}\left\{ \begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix}\right\}$.
By written in this form, you can diagonalize it.
When problem goes to $M^{n\times n}$, you just write a basis in the same way and abstract it to a basis $\mathcal{B}={v_1,\dots,v_{n^2}}$, better in a relatively symmetric way (i.e. swap $v_i$ and $v_{n^2+1-i}$) so that your $[T]_{\mathcal{B}}$. In this way, you can compute the characteristic polynomial easier.
A: When you're treating matrices as objects in a vector space, instead of as representations of a transformation, then it's often useful to think of them as vectors with components.
A 2x2 matrix has four components, a 3x3 matrix has 9 components, a 4x4 matrix has 16 components, and so on.
Every 2x2 matrix can be written uniquely as:
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix} = a\begin{bmatrix}1 & 0 \\ 0 &0 \end{bmatrix} + b\begin{bmatrix}0 & 1 \\ 0 &0 \end{bmatrix} + c\begin{bmatrix}0 & 0 \\ 1 &0 \end{bmatrix}+d\begin{bmatrix}0 & 0 \\ 0 &1 \end{bmatrix}$$
In other words, these four matrices on the right hand side form a basis for the space.

Because we're treating these matrices as objects in a vector space, you are free to write them as vectors if you like.
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix} \leadsto \begin{bmatrix}a \\ b\\ c\\d\end{bmatrix}$$
Written in this form, transposition is the transformation:
$$T : \begin{bmatrix}a \\ b\\ c\\d\end{bmatrix} \mapsto \begin{bmatrix}a \\ c\\ b\\d\end{bmatrix}$$
which you can write in matrix form as:
$$\underbrace{\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 1& 0 & 0\\ 0&0&0&1\end{bmatrix}}_T\begin{bmatrix}a \\ b\\ c\\d\end{bmatrix} = \begin{bmatrix}a \\ c\\ b\\d\end{bmatrix}$$
and you can do the rest.

Of course, you can also get some intuition about transposition using intuition about eigenvectors, without relying on specific components.
If a linear transformation scales an object by a factor of $\lambda\neq 0$, then that object is an eigenvector with eigenvalue $\lambda$. (As a special case, if a linear transformation has no effect on an object, it's an eigenvector with eigenvalue $\lambda=1$.)
Let's exclude the zero matrix from consideration in everything that follows.  There are three kinds of matrices that are unaffected by transposition $T$:
$$\begin{bmatrix}a & 0\\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0\\ 0 & b\end{bmatrix}, 
\begin{bmatrix}0 & c\\ c & 0\end{bmatrix}$$
As a result, these are eigenvectors of $T$, and they all have eigenvalue 1.  The linear combinations of these matrices form a three dimensional space of matrices that  are all eigenvectors with eigenvalue 1:
$$\begin{bmatrix}a & c \\ c & d\end{bmatrix}\qquad \forall a, b, c$$
