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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?

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I've noticed you've posted several questions on linear algebra in the last day or two. If these are homework questions, it would be appropriate to add the homework tag. Cheers. :) –  cardinal Apr 22 '12 at 17:19
    
@cardinal: These are not homework problems. The next week is the week of finals, and I am preparing for my final in Linear Algebra coming next week. I am solving problems from a textbook that has only hints and partial solutions, and whenever I get stuck at some point, I post my question here. :)- –  M.Krov Apr 22 '12 at 17:39
    
Ok, fair enough. I do see that you've provided some of your own thoughts in the questions, which is, of course, appreciated. :) –  cardinal Apr 22 '12 at 17:40

1 Answer 1

up vote 4 down vote accepted

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*}

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

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