3
$\begingroup$

Let $A$ and $B$ be positive semidefinite matrices of the same size. If the largest eigenvalues of $A$ and $B$ are less than equal to $1$. probe that: $$AB+BA\geq- \frac{1}{4}I$$

In the hint, it says use the fact: $0\leq (A+B-\frac{1}{2}I)^{2}$. In order for $ (A+B-\frac{1}{2}I)^{2}$ to be positive semidefinite, the matrix $ (A+B-\frac{1}{2}I)$ has to be positive semidefinite. The hint given above confused me because I know that $A+B$ is definitely positive semidefinite since it is the sum of two positive semidefinite matrices, but we can't be sure that $A+B-\frac{1}{2}I$ is positive semidefinite. Does anyone know how to prove that: $0\leq (A+B-\frac{1}{2}I)^{2}$?

Now assume the given hint is true, then: $0\leq A\leq I$ and $0\leq B\leq I$. It follows that:$$0\leq (A+B-\frac{1}{2}I)^{2}=A^{2}+B^{2}+\frac{1}{4}I+A(B-I)+B(A-I)\leq A+B+\frac{1}{4}I+A(B-I)+B(A-I)$$

I need to prove that:$A(B-I)+B(A-I)\leq 0$. This will be true if the product of a positive semidefinite matrix and a negative semidefinite matrix is a negative semidefinite matrix. Is this true?

$\endgroup$
4
$\begingroup$

you have $0 \le (A + B - \frac 12 I)^2$ as the left hand side is a square: \[ \textstyle \langle (A + B - \frac 12 I)^2x, x\rangle = \langle (A+B-\frac 12I)x, (A+B-\frac 12I)^*x\rangle =\langle (A+B-\frac 12I)x, (A+B-\frac 12I)x\rangle = \|(A + B -\frac 12I)x\|^2 \ge 0 \] And for the second part we have $A^2 \le A$ as you wrote. I think you perhaps didn't remember you wanted to have $AB + BA + \frac 14I \ge 0$, for \begin{align*} \left(A + B - \frac 12\right)^2 &= A^2 + B^2 + \frac 14 I - A - B + AB + BA\\\ &\le A + B - A - B + A B + BA + \frac 14I\\\ &= AB + BA + \frac 14I. \end{align*}

$\endgroup$
  • $\begingroup$ Thanks a lot for this nice and neat proof:)- $\endgroup$ – M.Krov Apr 22 '12 at 17:43
  • $\begingroup$ @M.Krov Why no upvote. The proof is nice indeed. $\endgroup$ – Sasha Apr 22 '12 at 18:03
  • $\begingroup$ @Sasha: Good idea. I have already upvoted the answer $\endgroup$ – M.Krov Apr 22 '12 at 18:12

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.