$vv^TA-Avv^T$ where $A$ is skew-symmetric and $vv^T$ is rank one

Question

Suppose I have the following:

$$vv^TA-Avv^T$$

1. $v\in \mathbb{R}^{n}$ with $\|v\|_2=1$, so $vv^T$ is PSD, rank one and $\operatorname{tr}(vv^T)=1$
2. $A$ is skew-symmetric

Is there any nice properties or conditions I can use to simplify it?

---

I try to search some related problems:

1. Rank of skew-symmetric matrix

   • It is even.

Or

1. When is matrix multiplication commutative?

   • simultaneously diagonalizable

Is there any closer one to my question?

Answer 1

Let $u=-Av/\|Av\|_2$. Then your matrix becomes $S=\|Av\|_2(vu^T+uv^T)$. As $A$ is skew-symmetric, $u\perp v$. Therefore $u+v$ and $u-v$ are eigenvectors of $S$ corresponding to the eigenvalues $\|Av\|_2$ and $-\|Av\|_2$ respectively, and $S$ is orthogonally similar to $\|Av\|_2\operatorname{diag}(1,-1,0,\ldots,0)$.