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Let $A,B\in M_n(\mathbb{R})$ be two symmetric positive definite matrices, i.e.: $$\forall x\in\mathbb{R}^n, x\neq 0, (Ax,x)>0, (Bx,x)>0,$$ where $(\cdot,\cdot)$ is the usual scalar product in $\mathbb{R}^n$.

It is equivalent to saying that eigenvalues of A and B are strictly positive. We order these eigenvalues which are not necessarily distinct: $$\lambda_1(A)\leq ...\leq\lambda_n(A)$$ and $$\lambda_1(B)\leq ...\leq\lambda_n(B).$$ It is not hard to prove the minimax principle for these eigenvalues: $$\lambda_k(A)=\min_{\substack{F\subset \mathbb{R}^n \\ \dim(F)=k}} \left( \max_{x\in F\backslash \{0\}} \frac{(Ax,x)}{(x,x)}\right).$$

For $\lambda_1(A)$ and $\lambda_n(A)$ we have simpler expressions: $$\lambda_1(A)=\min_{x\in\mathbb{R}^n \backslash \{0\}} \frac{(Ax,x)}{(x,x)}\hspace{1cm}\text{and}\hspace{1cm}\lambda_n(A)=\max_{x\in\mathbb{R}^n \backslash \{0\}} \frac{(Ax,x)}{(x,x)}.$$

We now consider the matrix $AB$. It is possible to prove that the eigenvalues of $AB$ are the same as for the matrix $\sqrt{B}\cdot A\cdot \sqrt{B}$, where $\sqrt{A}$ denotes the unique square root matrix of $A$: $\sqrt{A}$ is real, symmetric, positive definite and $\sqrt{A}\cdot \sqrt{A}=A$. The eigenvalues of $AB$ are hence real and strictly positive. We order them similarly like we did for A and B. Prove that, $\forall 1\leq k\leq n$: $$\lambda_k(A)\lambda_1(B)\leq \lambda_k(AB)\leq \lambda_k(A)\lambda_n(B).$$

Note that we cannot use the minimax principle for the $\lambda_k(AB)$ since $AB$ is not necessarily symmetric. However, the hint of the exercise suggests we use the the minimax principle at some point.

References to some books? Any thoughts how I should attack the problem? Maybe I am missing some obvious observations?

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$\lambda_k(AB) = \lambda_k(\sqrt{B}A\sqrt{B})$. –  achille hui Mar 11 '13 at 0:09
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It's "positive definite," not "positive defined." –  John Moeller Mar 11 '13 at 7:59

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up vote 3 down vote accepted

\begin{align*} \lambda_k(AB)=\lambda_k(\sqrt{B}A\sqrt{B}) &=\min_{\substack{F\subset \mathbb{R}^n \\ \dim(F)=k}} \left( \max_{x\in F\backslash \{0\}} \frac{(\sqrt{B}A\sqrt{B}x,x)}{(x,x)}\right)\\ &=\min_{\substack{F\subset \mathbb{R}^n \\ \dim(F)=k}} \left( \max_{x\in F\backslash \{0\}} \frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)} \frac{(Bx,x)}{(x,x)}\right). \end{align*}

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@usser1551 . Yes. You forgot to mention that $\sqrt{B}A\sqrt{B}$ is always symmetric if $\sqrt{B}$ and $A$ are symmetric which is the case. Then we can finish the proof by a simple evaluation taking $F=\operatorname{span}((\sqrt{B})^{-1}a_1,...,(\sqrt{B})^{-1}a_k)$ where $a_i$ are the orthonormal eigenvectors of A. Thanks a lot. I was indeed missing an obvious observation. –  Lukas Mar 11 '13 at 16:11

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