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Supposing $V$ is a finite dimensional vector space (over $\mathbb{R}$) of dimension $n$, and $A,B$ are symmetric positive definite linear mappings from $V$ to $V$, how can I show that in any orthonormal basis $\mathrm{tr}(AB) \geq 0$?

I noticed that since they are symmetric we have that $$\mathrm{tr}(AB) = \sum_{i=1}^n\sum_{j=1}^nA_{ij}B_{ji} = \sum_{i=1}^n\sum_{j=1}^nA_{ij}B_{ij}$$ which is the sum of the elements of the element-wise product of $A,B$. I don't know if this is helpful.

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I think you may as well assume $V = \mathbb{R}^n$ and that $A$ and $B$ are matrices. Symmetric positive-definite linear maps don't make much sense except in the presence of a chosen isomorphism $V \cong V^*$. – Zhen Lin Feb 25 '12 at 18:20
If this is homework, can you assume the spectral theorem? That is, can you assume the nice properties that hold about matrices' eigenvalues when they are symmetric & positive definite? – yep Feb 25 '12 at 18:20
Yes we may assume $V=\mathbb{R}^n$ and yes you may use the spectral theorem. But remember the result must hold for all orthonormal bases, not just the one where $A$ is diagonal.\ – nullUser Feb 25 '12 at 18:27
up vote 5 down vote accepted

Since this may be homework, I will only give hints.

  1. Without loss of generality you may assume that $V=R^n$.

  2. Trace is independent of the basis you use. Thus it suffices to show this in the basis where $A$ is diagonal.

  3. A positive semi-definite matrix has nonnegative diagonal. Why?

  4. Putting 1-3 together, one needs to show that the $tr(AB)\geq 0$ where $A$ is a nonnegative diagonal matrix and $B$ has nonnegative diagonal.

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As others have remarked, you might as well suppose that $A$ and $B$ are positive semidefinite matrices. We may write $A = X^{t}X$ and $B = Y^{t}Y$ where $X$ and $Y$ are $n \times n$ real matrices. Then ${\rm tr}(AB) = {\rm tr}(X^{t}X Y^{t}Y)$ = ${\rm tr}((YX^{t})(XY^{t})).$ The latter matrix has the form ${\rm tr}(UU^{t})$ for a real $n \times n$ matrix $U$, and such a trace is always non-negative.

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