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 (and there exists an orthonormal eigenbasis for both matrices). 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?

  • 2
    $\begingroup$ $\lambda_k(AB) = \lambda_k(\sqrt{B}A\sqrt{B})$. $\endgroup$ Mar 11 '13 at 0:09
  • 1
    $\begingroup$ It's "positive definite," not "positive defined." $\endgroup$ Mar 11 '13 at 7:59

\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*}

  • 2
    $\begingroup$ @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. $\endgroup$
    – Lukas
    Mar 11 '13 at 16:11

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.