In my problem, I came across numerical ranges of rank-one positive semidefinite matrices. Through Toeplitz-Hausdorff theorem and some other extensions, I know if there are at most three matrices, then the numerical range is convex for any given set of hermitian matrices. Is there any results on rank-one positive semi-definite matrices?

## 1 Answer

I don't expect that you can get results that say too much beyond what is true in the general Hermitian case. You mentioned convexity of the (joint) numerical range in your question, so I'll address that in this answer.

**Short version:** restricting to positive semi-definite matrices of rank 1 doesn't change the convexity results for the numerical range at all.

**Long version:** The numerical range is still convex if you have 3 or fewer positive semi-definite matrices of rank 1 (and the dimension is at least 3) --- this follows immediately from the result about Hermitian matrices.

Conversely, the numerical range can still fail to be convex if you have 4 or more positive semi-definite matrices of rank $1$. To see this, consider the following 4 matrices:

$$ \begin{bmatrix}1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}, \begin{bmatrix}1 & 1 & 0\\ 1 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}, \begin{bmatrix}1 & i & 0\\ -i & 1 & 0\\ 0 & 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1\end{bmatrix} $$

The joint numerical range of these 4 matrices is the set of all 4-tuples of the following form, where $x,y,z \in \mathbb{C}$ are such that $|x|^2 + |y|^2 + |z|^2 = 1$:

$$ (|x|^2, 2|x+y|^2, 2|x+iy|^2, |z|^2). $$

This set is not convex because it contains the two points $(1,2,2,0)$ (take $x = 1, y = z = 0$) and $(0,2,2,0)$ (take $y = 1, x = z = 0$), but it does not contain the point $(\tfrac{1}{2},2,2,0)$ (checking this last fact is just a calculation -- notice that we would need $|x| = |y| = \tfrac{1}{\sqrt{2}}$ and go from there).

You can also embed this example in larger dimensions to see that 4 matrices can still lead to non-convexity in any dimension $\geq 3$. For dimension 2, we can similarly see that 3 matrices can lead to non-convexity (just like in the general Hermitian case) by taking the top-left 2-by-2 block of the first 3 matrices above.