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?


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}

  • $\begingroup$ Actually your example is wrong since -1+1+2=2 is an eigenvalue. What do you mean by eigenvalues of T are sum of pairs of eigenvalue of A? $\endgroup$ – deep12345 Dec 8 '18 at 16:18
  • $\begingroup$ I misundrstood the condition! I thought that it was “the sum of two eigenvalues is not an eigenvalue”. But then, yes, the statment is correct. I will edit my answer and provide a proof. $\endgroup$ – José Carlos Santos Dec 8 '18 at 16:26
  • $\begingroup$ I've edited my answer. What do you think now? $\endgroup$ – José Carlos Santos Dec 8 '18 at 16:37
  • $\begingroup$ Never thought of brute force to compute T directly, though the answer turns out to be surprisingly simple. Thanks! $\endgroup$ – deep12345 Dec 8 '18 at 16:44
  • $\begingroup$ What happens for dimensions other than $3\times 3$? For $2\times 2$ I get $$T(X) = \begin{bmatrix} 0 & \alpha x+\beta x \\ -\alpha x-\beta x & 0 \end{bmatrix}$$ $\endgroup$ – Jeppe Stig Nielsen Dec 8 '18 at 17:17

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