Questions on symmetric matrices and skew-symmetric matrices

Question

Let $A$ be a $3\times 3\;$ symmetric matrix. Let $U$ be the set of all $3\times 3\;$ skew-symmetric matrices. Let $T : U\to U$ be defined as $T(B)=AB+BA.$
Prove that $T$ is bijective iff the sum of eigenvalues of $A$ is not an eigenvalue of $A.$

This question makes me stuck for days, since I cannot relate the eigenvalues with the map. Can anyone help me?

Answer 1

I am assuming that you are working over the real field. So, $A$ is diagonalizable and you can assume without loss of generality that $A$ is diagonal:$$A=\begin{bmatrix}\alpha&0&0\\0&\beta&0\\0&0&\gamma\end{bmatrix}$$and $\alpha$, $\beta$, and $\gamma$ are the eigenvalues of $A$. So, if$$X=\begin{bmatrix}0&x&-z\\-x&0&y\\z&-y&0\end{bmatrix},$$then$$T(X)=\begin{bmatrix}0 & \alpha x+\beta x & -\alpha z-\gamma z \\ -\alpha x-\beta x & 0 & \beta y+\gamma y \\ \alpha z+\gamma z & -\beta y-\gamma y & 0\end{bmatrix}$$and therefore the eigenvalues are $\alpha+\beta$, $\alpha+\gamma$, and $\beta+\gamma$. Therefore\begin{align}T\text{ is bijective}&\iff T\text{ is injective}\\&\iff0\text{ is not an eigenvalue of }T\\&\iff\alpha+\beta,\alpha+\gamma,\beta+\gamma\neq0\\&\iff\alpha+\beta+\gamma\neq\gamma\text{, }\alpha+\beta+\gamma\neq\beta\text{, and }\alpha+\beta+\gamma\neq\alpha\\&\iff\alpha+\beta+\gamma\text{ is not an eigenvalue of }A.\end{align}