Eigenvalues and Eigenvectors of the space of 2x2 matrices I wanted to find the eigenvalues and eigenvectors of the transformation of the space of $2\times 2$ matrices from $$T(A) = A^T$$ I initially thought representing the basis vectors in a $4 \times 4$ matrix like this would be helpful:
$$M = \left[\begin{array}{cccc}
    1 & 0 & 0 & 0  \\
    0 & 0 & 1 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 1
\end{array}\right]$$
But then finding the characteristic polynomial this way is really long and tedious. Is there an easier way to do this without having to represent the transformation as a matrix by perhaps using some property of Transpose?
 A: You don't need to find the characteristic polynomial of $M$ (or indeed the matrix $M$ at all) in order to find the eigenvalues and eigenvectors of $T$. You can work directly from the definition. If $\lambda$ is an eigenvalue of $T$ with associated eigenvector $A$, then by definition
$$A^T = \lambda A$$
Taking transposes of both sides gives
$$A = \lambda A^T$$
Substituting the previous equation into this one, we obtain
$$A = \lambda^2 A$$
Assuming $A$ is nonzero, which is required for any eigenvector, we conclude that the only possible eigenvalues are $\lambda = 1$ and $\lambda = -1$.
Now consider the two cases.
(1) $\lambda = 1$
In this case, the first equation becomes $A^T = A$, so $A$ is an associated eigenvector if and only if it is nonzero and symmetric, i.e. of the form
$$A = \begin{pmatrix}a & b \\ b & c\end{pmatrix}$$
Note that there are three degrees of freedom (the values of $a$, $b$, and $c$), so this eigenspace has geometric multiplicity $3$.
(2) $\lambda = -1$
In this case, the first equation becomes $A^T = -A$, so $A$ is an associated eigenvector if and only if it is nonzero and antisymmetric, i.e. of the form
$$A = \begin{pmatrix}0 & d \\ -d & 0\end{pmatrix}$$
Note that there is one degree of freedom (the value of $d$), so this eigenspace has geometric multiplicity $1$.
A: Since , each row sum of $M$ is $1$, $1$ is an eigenvalue of $M$.
You can also see by taking vector $v=(1,1,1,1)$ in $\mathbb{R}^{4}$ such that $Mv=1v$
Then , by definition of eigenvalue of a linear operator, $1$ is eigenvalue of $M$.
Now , we see that dimension of the kernel of $(M-1I_{4 \times 4})$ is $3$.
That mean , geometric multiplicity of eigenvalue $1$ is $3$.
Now ,as $M$ is real symmetric, so, M is diagnolisable.
Hence , algebraic multiplicity of eigenvalue $1 = $geometric multiplicity of $1$ $=3$
Now , $trace(M)=2$
So, if we take unknown eigenvalue as $\lambda$
then, $1+1+1+\lambda=2 \implies \lambda =-1 $
So, all eigenvalues are $1,1,1,-1$
