show that symmetric and anti-symmetric matrices are eigenvectors for linear map Consider the linear map
$$L:Mat_{nxn}(\mathbb{R})\rightarrow Mat_{nxn}(\mathbb{R}),\\
M\mapsto 4M-7M^T$$
a) Prove that any symmetric or anti-symmetric matrix is an eigenvector for $L$.
b) Is $L$ diagonalizable? Justify.
Solution:
Let $M$ be symmetric ($M=M^T$), then $L(M)=4M-7M^T=4M-7M=-3M=(-3)M$ where $-3$ is an eigenvalue an $M$ the eigenvector.
The same for $M$ anti-symmetric ($M=-M^T)$. Does this prove a)?
 A: An operator $T: V \to V$ is diagonalizable if $V$ is spanned by the eigenvectors of $T$.
Once you have the all symmetric and anti-symmetric matrices are eigenvectors, the next step is to see if all of $\text{Mat}_{n\times n}$ is spanned by those matrices. Now observe that any matrix $A$ can be written as the sum of a symmetric and anti-symmetric matrix as follows:
$$A = \frac{A - A^T}{2} + \frac{A+A^T}{2}  $$
Therefore, your operator is diagonalizable.
A: Remark: You have to investigate the linear operator $L$ !

*

*Let $M \ne 0$ be symmetric, then

$$L(M)=4M-7M=-3M,$$
hence $-3$ is an eigenvalue of $L$ and $M$ an corresponding eigenvector.


*Let $M \ne 0$ be anti-symmetric, then

$$L(M)=4M+7M=11M,$$
hence $11$ is an eigenvalue of $L$ and $M$ an corresponding eigenvector.
A: As a quirky alternative for part b, you could show that for every $n\times n$-matrix $M$ you have
\begin{eqnarray*}
L^2(M)&=&L(4M-7M^{\top})\\
&=&4(4M-7M^{\top})-7(4M-7M^{\top})^{\top}\\
&=&65M-56M^{\top},
\end{eqnarray*}
from which it follows that
$$L^2(M)-8L(M)-33M=0,$$
and so the minimal polynomial of $L$ divides
$$X^2-8X-33=(X-11)(X+3).$$
