# Is there any interesting relationship between a Hermitian matrix and its corresponding entrywise absolute?

In general, a Hermitian matrix can have complex off-diagonal terms. Given any Hermitian matrix $[A]_{n,m}$, I can construct another matrix $[\vert A\vert ]_{n,m} =\vert A_{n,m} \vert$. I would like to know if there are any known relationships between two such matrices. Some more specific questions that I was thinking about but don't have an answer to are the following:

1. Is it possible to transform using a unitary $UAU^{\dagger}$ or using a convex combination of unitary matrices $\sum_i p_i U_iAU_i^{\dagger}$ where $\sum_i p_i =1$? It seems to be possible for $2 \times 2$ positive semi-definite matrices.

2. Are there any inequalities relative their trace norms $\operatorname{tr}(\sqrt{A^\dagger A})$ and $\operatorname{tr}(\sqrt{\vert A \vert^\dagger \vert A \vert})$? Such as if one is always bigger than the other.

Any information will be helpful for general Hermitian matrices will be helpful, but if you would like to know, I am looking at two special cases matrices right now:

a) $A$ is a positive definite matrix.

b) $A$ is a hollow (zero main diagonal), banded, Toeplitz and Hermitian matrix that looks like the following:

\begin{bmatrix} 0 & a_1 & a_2 & \ldots & a_n \\ a_1^* & 0 & a_1 &\ldots & \vdots \\ a_2^* &a_1^* & 0 & \ldots &\vdots \\ \vdots & & & \ddots & \vdots\\ a_n^* & a_{n-1}^* & a_{n-2}^* &\ldots &0 \end{bmatrix}

Thanks for all your help.

## 1 Answer

1. The condition $$\tag{1} B=\sum_jp_jU_jAU_j^*,\ \ \ \sum p_j=1,\ \ p_j\geq0$$ implies that $B$ has the same trace as $A$. If $$A=\begin{bmatrix}1&1\\1&-1\end{bmatrix},$$ neither $A$ is a convex combination of unitary conjugates of $|A|$, nor $|A|$ is a convex combination of unitary conjugates of $A$. $$\$$ It does work, as you say, for $2\times2$ positive-semidefinite matrices. In this case the main diagonal of $A$ consists of nonnegative entries, and so if $A_{12}=e^{it}\,|A_{12}|$, $$UAU^*=\begin{bmatrix}1&0\\0&e^{it}\end{bmatrix}\, \begin{bmatrix}A_{11}&e^{it}|A_{12}|\\ e^{-it}|A_{21}|&A_{22}\end{bmatrix}\, \begin{bmatrix}1&0\\0&e^{-it}\end{bmatrix}\, = \begin{bmatrix}A_{11}&|A_{12}|\\ |A_{21}|&A_{22}\end{bmatrix}\,=|A|.$$ $$\$$ Note that $(1)$ is equivalent to the majorization $\lambda(B)\prec\lambda(A)$ (i.e. majorization between the vectors of eigenvalues). For the matrix $$A=\begin{bmatrix}0&-2&-2\\ -2&0&-3\\-2&-3&0\end{bmatrix},$$ we have $\lambda(|A|)=-\lambda(A)$. So to be able to obtain $|A|$ in the form $(1)$, we would need $-\lambda(A)\prec\lambda(A)$, which one can check explicitly it fails.$$\$$ Similarly, it also fails when $A$ is positive-semidefinite: for $$A=\begin{bmatrix}4&-2&-2\\ -2&5&-3\\-2&-3&6\end{bmatrix},$$ again $\lambda(A)$ and $\lambda(|A|)$ do not majorize each other, so we cannot obtain one from the other with an expression of the form $(1)$.

2. Since $\text{tr}(A^*A)=\text{tr}(|A|^*|A|)$ always, , $$\text{tr}(\sqrt{A^*A})\leq n^{1/2}\,\text{tr}(A^*A)^{1/2}= n^{1/2}\,\text{tr}(|A|^*|A|)^{1/2}\leq n^{1/2}\,\text{tr}(\sqrt{|A|^*|A|})$$