$\newcommand{\R}{\mathbb R}\newcommand{\tr}{\operatorname{tr}}$It follows from Proposition 7 and this recent answer that, for any positive-definite $n\times n$ symmetric real matrices $A$ and $B$,
$$\tr\big[A+B-2(A^{1/2}BA^{1/2})^{1/2}\big]\ge \Big(\frac{\|A-B\|}{\sqrt{\|A\|}+\sqrt{\|B\|}}\Big)^2, \tag{1}\label{1}$$ where $\tr$ denotes the trace and $\|M\|$ is the spectral norm of a matrix $M$.

That proof of \eqref{1} involves certain probabilistic arguments.

Since \eqref{1} is stated in purely matrix terms, the following question naturally arises:

Is there a proof of \eqref{1} involving (no probabilistic tools, but) only matrix analysis tools?

A related question:

Will inequality \eqref{1} hold (perhaps up to a universal positive real constant factor) if each instance of the spectral norm in \eqref{1} is replaced by that of the Frobenius one?

Any correct and complete answer to either one of these two questions will be considered a correct and complete answer to this entire post.

  • $\begingroup$ The inequality would be trivial if we had known that the expression in brackets is PSD. This seems rather natural but the formal proof escapes me for some reason. Can you enlighten me as to why that is (or is not necessarily) true? :-) $\endgroup$
    – fedja
    Jul 23 at 15:27

1 Answer 1


I'll try to answer both questions at once. First, let's find the reason why the LHS is positive at all. I claim that the spectrum $\mu_1\le\mu_2\le\dots\le\mu_n$ of $2\sqrt{A^{1/2}BA^{1/2}}$ is dominated by the spectrum $\lambda_1\le\lambda_2\le\dots\le\lambda_n$ of $A+B$ elementwise. Indeed, for each $k$, there is and $n-k+1$-dimensional subspace $V$ on which $$ \langle A^{1/2}BA^{1/2}x,x\rangle\ge \frac{\mu_k^2}{4}\|x\|^2 $$ for any vector $x\in V$, which is equivalent to stating that $\langle By,y\rangle\ge\frac{\mu_k^2}{4}\langle A^{-1}y,y\rangle$ for every $y\in W=A^{1/2}V$. But then on $W$, we have $$ \langle(A+B)y,y\rangle\ge \langle(A+\tfrac{\mu_k^2}4A^{-1})y,y\rangle\ge \mu_k\|y\|^2\, $$ which means that $\lambda_k\ge \mu_k$ (otherwise we could choose $y$ to be a linear combination of the first $k$ eigenvectors of $A+B$ and get the reverse inequality).

Now, with this subordination of eigenvalues, we can find a rotation $R$ such that $A+B-2R^*\sqrt{A^{1/2}BA^{1/2}}R=A+B-2X$ is positive semi-definite (and $A+B$ commutes with $X$, but that is not so important though once we have it, I'll shamelessly use it). Note that this rotation does not affect the trace in any way.

The next thing we need is the following trivial inequality. If $P,Q$ are positive semidefinite matrices, then $\operatorname{tr}(PQ)\le\|Q\|\operatorname{tr}P$. (just compute both traces in the orthogonal basis of the eigenvectors of $Q$). We shall use it with $P=A+B-2X$ and $Q=A+B+2X$. Then $\|Q\|\le(\sqrt{\|A\|}+\sqrt{\|B\|})^2$ while $$ \operatorname{tr}[PQ]=\operatorname{tr}[(A+B)^2-4X^2] \\ = \operatorname{tr}[A^2+AB+BA+B^2-4R^*A^{1/2}BA^{1/2}R] \\ =\operatorname{tr}[A^2-AB-BA+B^2]=\operatorname{tr}[(A-B)^2]=\|A-B\|_F^2\,. $$ and we are done.

This computation would be much simpler if, as I thought initially, the matrix on the LHS were PSD itself. Unfortunately, that is not always true, but the eigenvalue subordination and the extra rotation save the day nicely.

  • $\begingroup$ Very nice! Thank you for your answer. $\endgroup$ Jul 23 at 19:36

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