show that symmetric and anti-symmetric matrices are eigenvectors for linear map

Solution 1:

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.

Solution 2:

Remark: You have to investigate the linear operator $L$ !

1. Let $M \ne 0$ be symmetric, then

$$L(M)=4M-7M=-3M,$$

hence $-3$ is an eigenvalue of $L$ and $M$ an corresponding eigenvector.

2. 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.

Solution 3:

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 $$X^2-8X-33=(X-11)(X+3).$$