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

Question

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)?

Answer 1

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.

Answer 2

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.

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