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Actually i'm interested in the answer of this question and precisely i like to know how they achieved the following inequality: $$\lambda_k(A)\lambda_1(B)\leq \lambda_k(AB)$$

based on this trick: \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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Bob
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1 Answers1

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So, the assumption is that $A,B$ are symmetric positive-definite. Then each has asymmetric positive-definite square root. Since the eigenvalues of the product do not depend on order, $$\lambda_k(AB)=\lambda_k(\sqrt B\,A\,\sqrt B).$$ The next two equalities are the min-max theorem.

We have, since $B$ is symmetric, $$ \lambda_1(B)\leq\frac{\langle Bx,x\rangle}{\langle x,x\rangle}\leq\lambda_n(B) $$ for all $x\ne0$.

So $$ \max_{x\in F\backslash \{0\}} \frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)} \frac{(Bx,x)}{(x,x)} \geq\lambda_1(B)\, \max_{x\in F\backslash \{0\}} \frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)} $$ Now as you take the minimum over all $F$ of dimension $k$, since $\sqrt B$ is invertible you get \begin{align} \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) &\geq\lambda_1(B)\,\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)}\right)\\ \ \\ &=\lambda_1(B)\,\min_{\substack{F\subset \mathbb{R}^n \\ \dim(F)=k}} \left( \max_{x\in F\backslash \{0\}} \frac{(Ax,x)}{(x,x)} \right)\\ \ \\ &=\lambda_1(B)\lambda_k(A). \end{align}

Martin Argerami
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