Taking positive part commutes with conjugating with $Y\geq 0$ on hermitian matrices?

Question

Let $X,Y\in\mathbb C^{n\times n}$ be hermitian and $Y$ positive semi-definite. Does $$ (YXY)^+=YX^+Y $$ hold, where $(\cdot)^+$ denotes the positive part of the respective hermitian matrix (i.e. if $A=U\operatorname{diag}(a_1,\ldots,a_n)U^\dagger$ hermitian with $U\in\mathbb C^{n\times n}$ unitary then $A^+=U\operatorname{diag}(\max\{a_1,0\},\ldots,\max\{a_n,0\})U^\dagger$)?

Answer 1

In some special cases it holds:

• if $X$ is positive (or negative)-semidefinite then $YXY$ is positive (or negative) semi-definite as well as is readily verified, so $(YXY)^\pm=(\pm)YXY=Y(\pm X)Y=YX^\pm Y$
• if $[X,Y]=0$ then there exists unitary $U\in\mathbb C^{n\times n}$ such that $UXU^\dagger=:D_X$, $UYU^\dagger=:D_Y$ are both diagonal. Then $YXY=U^\dagger D_YD_XD_Y U$ and thus $$(YXY)^+=U^\dagger (D_YD_XD_Y)^+ U=U^\dagger D_Y D_X^+D_YU=Y(U^\dagger D_X^+ U)Y=YX^+Y\,.$$

However, this is not true in general. For a counter-example consider $$ X=\begin{pmatrix} 0&2\\2&-3 \end{pmatrix}=\underbrace{\begin{pmatrix} \frac{-1}{\sqrt 5}&\frac{2}{\sqrt 5} \\\frac{2}{\sqrt 5}&\frac{1}{\sqrt 5}\end{pmatrix}}_{\text{unitary}} \begin{pmatrix} -4&0\\0&1 \end{pmatrix} \begin{pmatrix} \frac{-1}{\sqrt 5}&\frac{2}{\sqrt 5} \\\frac{2}{\sqrt 5}&\frac{1}{\sqrt 5}\end{pmatrix} $$ as well as $$ Y=\begin{pmatrix} \sqrt{\frac52}&0\\0&1 \end{pmatrix}\,. $$ Then $$ X^+=\begin{pmatrix} \frac{-1}{\sqrt 5}&\frac{2}{\sqrt 5} \\\frac{2}{\sqrt 5}&\frac{1}{\sqrt 5}\end{pmatrix}\begin{pmatrix} 0&0\\0&1 \end{pmatrix} \begin{pmatrix} \frac{-1}{\sqrt 5}&\frac{2}{\sqrt 5} \\\frac{2}{\sqrt 5}&\frac{1}{\sqrt 5}\end{pmatrix}=\frac15\begin{pmatrix}4&2\\2&1\end{pmatrix} $$ and $$ YX^+Y=\begin{pmatrix} 2&\sqrt{\frac25}\\\sqrt{\frac25}&\frac15 \end{pmatrix}\simeq \begin{pmatrix} \frac{11}5&0\\0&0 \end{pmatrix} $$ but $$ YXY=\begin{pmatrix} 0&\sqrt{10}\\\sqrt{10}&-3 \end{pmatrix}=\underbrace{\begin{pmatrix} -\sqrt{\frac{2}{7}}&\sqrt{\frac{5}{7}} \\\sqrt{\frac{5}{7}}&\sqrt{\frac{2}{7}}\end{pmatrix}}_{\text{unitary}} \begin{pmatrix} -5&0\\0&2 \end{pmatrix} \begin{pmatrix} -\sqrt{\frac{2}{7}}&\sqrt{\frac{5}{7}} \\\sqrt{\frac{5}{7}}&\sqrt{\frac{2}{7}}\end{pmatrix}\,. $$ Thus $YXY$ has largest eigenvalue $2$ but $YX^+Y$ has largest eigenvalue $\frac{11}5$ so $(YXY)^+\neq YX^+Y$. All that is written here naturally carries over to the negative part and the absolute value of a matrix.

I've never done this Q&A style-"answer your own question" thing before so if what I have written is already known or if anything can be improved here (e.g., tags, title, etc.) feel free to let me know!

Answer 2

The positive part of a Hermitian matrix $H$ can be expressed as $\frac12(|H|+H)$, where $|H|=\sqrt{H^2}$. Therefore $$ \begin{align} &(YXY)^+=YX^+Y\tag{1}\\ &\Leftrightarrow |YXY|+YXY=Y|X|Y+YXY\\ &\Leftrightarrow |YXY|=Y|X|Y\\ &\Leftrightarrow \sqrt{(YXY)^2}=\sqrt{(Y|X|Y)^2}\\ &\Leftrightarrow (YXY)^2=(Y|X|Y)^2\\ &\Leftrightarrow YXY^2XY=Y|X|Y^2|X|Y.\tag{2}\\ \end{align} $$ So, one sufficient condition for $(1)$ to hold is that $$ XY^2X=|X|Y^2|X|.\tag{3a} $$ In case $Y$ is positive definite, condition $(3a)$ is also necessary. By a change of orthonormal basis, we may assume that $X=D_1\oplus-D_2\oplus0$ where $D_1$ and $D_2$ are two positive diagonal matrices. If we partition $Y$ into a block $3\times3$ matrix accordingly, we see that $(3a)$ holds if and only if the $(1,2)$-th and the $(2,1)$-th subblocks of $Y$ must be zero. Thus $(3a)$ is equivalent to any one of $$ Y\mathscr P\perp Y\mathscr N,\quad Y^2\mathscr P\perp \mathscr N\quad\text{or}\quad \mathscr P\perp Y^2\mathscr N\tag{3b} $$ where $\mathscr P$ is the sum of all eigenspaces corresponding to positive eigenvalues of $X$ and $\mathscr N$ likewise is the sum of all negative eigenspaces of $X$.

In particular, $(1)$ holds when $X$ is positive/negative semidefinite, because one of $\mathscr P$ or $\mathscr N$ is zero. Condition $(1)$ also holds when $\mathscr P$ and $\mathscr N$ are invariant subspaces of $Y$. This occurs, e.g., when $X$ and $Y$ commute, but this may also occur in other circumstances, e.g., when $X$ takes the aforementioned form $D_1\oplus-D_2\oplus0$, the sufficient condition $(3b)$ holds when $Y$ is in the form of $P_1\oplus P_2\oplus P_3$ where $P_1$ does not commute with $D_1$ or $P_2$ does not commute with $D_2$.

There are also cases where $(1)$ holds when $X=D_1\oplus-D_2\oplus0$ but $Y$ is not block-diagonal. For instance, consider $$ X=\pmatrix{1\\ &-1\\ &&0}, \quad Y=\pmatrix{1&1&-3 \\ 1&2&1\\ -3&1&26} \quad\text{and}\quad Y^2=\pmatrix{11&0&-80 \\ 0&6&25\\ -80&25&686}. $$ One may readily verify that condition $(3a)$ is satisfied.