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In his paper "On the existence of a connection with curvature zero", Milnor makes the following claim. Let $X$ and $Y$ be positive-definite $n \times n$ matrices. Then the trace of $X Y$ is positive. I can't figure out how to prove this. Can someone help me?

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1 Answer 1

up vote 7 down vote accepted

Let $A$ be the positive definite square root of $X$ and $B$ the positive definite square root of $Y$.

You have $$ \mbox{tr}(XY)=\mbox{tr}(AABB)=\mbox{tr}(BAAB)=\mbox{tr}((AB)^*AB)>0. $$

Indeed, the latter is the sum of all $c_{i,j}^2$ where $c_{i,j}=(AB)_{i,j}$.

So it is nonnegative.

And if it were zero, this would imply $AB=0$ hence $A=B=0$ since they are both invertible. Contradiction.

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Thanks for your help! –  Susan Feb 21 '13 at 23:46
    
@Susan You're welcome. Happy it helped. –  1015 Feb 21 '13 at 23:47

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