3
$\begingroup$

Let $A,B\geq0$ be positive semidefinite matrices of arbitrary size $n\times n$. Denote by $\alpha$ and $\beta$ their largest eigenvalues.

It is well-known that the eigenvalues of the expression $AB + BA$ are bounded by [F. Zhang, "Matrix Theory", Sec. 7.2]

$$ -\frac{1}{4}\alpha\beta I \,\,\leq\,\, AB + BA \,\,\leq\,\, 2\alpha\beta I. $$

I can show that $$ - \rm{tr}(AB)I - \rm{tr}(A)B - \rm{tr}(B)A \,\leq\, AB + BA \,\leq\, \rm{tr}(AB)I + \rm{tr}(B)A + \rm{tr}(A)B \quad (\dagger) $$

Now normalize to $\rm{tr}(A) = \rm{tr}(B) = 1$. When $\rm{tr}(AB) = 0$, that is $A\perp B$ in terms of the Hilbert-Schmidt inner product, the expression reduces to

$$ -(A+B) \,\,\leq\, AB + BA \,\leq\,\, (A + B) \quad (\ddagger) $$

These inequalities $(\dagger)$ and $(\ddagger)$ look rather simple and are for $n\geq 3$ even tight. However I failed to find them in the literature, including in Bernstein's book on "Matrix Facts" or in the books by Bhatia. Am I missing something, are these known or can they straightforwardly be derived from other known expressions?

edit: they are weaker than the simple sum-of-squares, $(A-B)^2 \geq 0$; see answer below.

$\endgroup$
3
$\begingroup$

It seems both bounds are much weaker than what can be obtained from a simple sum-of-squares: namely, for all Hermitian matrices $A$ and $B$, one has $(A-B)^2 \geq 0$. Expand

$$ AB + BA \leq A^2 + B^2 $$

Replacing $A\rightarrow -A$ yields also the lower bound

$$ -(A^2 + B^2) \leq AB + BA $$

Because of $\rm{tr}(A) = \rm{tr}(B)=1$, these are stronger than both $(\dagger)$ and $(\ddagger)$ in the question.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.