Spectrum of Schur complement

Question

Good afternoon!

I have a question about the eigenvalues of a schur complement. Assume $\mu_1 \leq ... \leq \mu_n$ are eigenvalues of the schur-complement $S$ to a positive definite block-matrix $A$ and $0 \leq \lambda_1 \leq ...\leq \lambda_n$ eigenvalues from $A$. Now the relations $\mu_1 > 0$ and $\frac{\mu_n}{\mu_1} \leq \frac{\lambda_n}{\lambda_1}$. That $\mu_1 > 0$ one can see by showing that the schur complement matrix $S$ itself is positive definite and so all eigenvalues of $S$ have to be positive. But this second eigenvalue relation I cannot see. Has someone a hint/idea for me to start? Are there some usefull theorems to see that? Or can this be shown directly without special theorems?

Thanks a lot for helping me out!!!

Answer 1

Let $$ A=\begin{bmatrix}B&C\\C^T&D\end{bmatrix}>0. $$ We have $$\tag{1} \begin{bmatrix}B&C\\C^T&D\end{bmatrix} = \begin{bmatrix}I&0\\C^TB^{-1}&I\end{bmatrix} \begin{bmatrix}B&0\\0&S\end{bmatrix} \begin{bmatrix}I&B^{-1}C\\0&I\end{bmatrix}, $$ where $S=D-C^TB^{-1}C$. Let $$ \mathcal{V}:=\left\{ \begin{bmatrix}x\\y\end{bmatrix}:x+B^{-1}Cy=0 \right\}. $$ Note that if $z=[x^T,y^T]^T\in\mathcal{V}$, then $$ \begin{bmatrix}I&B^{-1}C\\0&I\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}0\\y\end{bmatrix}. $$ We have from (1) that $$ \lambda_\max(A)=\max_{z\neq 0}\frac{z^TAz}{z^Tz} \geq\max_{0\neq z\in\mathcal{V}}\frac{z^TAz}{z^Tz} =\max_{y\neq 0}\frac{y^TSy}{y^Ty}=\lambda_\max(S) $$ and $$ \lambda_\min(A)=\min_{z\neq 0}\frac{z^TAz}{z^Tz} \leq\min_{0\neq z\in\mathcal{V}}\frac{z^TAz}{z^Tz} \leq \min_{y\neq 0}\frac{y^TSy}{y^Ty}=\lambda_\min(S). $$

Answer 2

The derivation for $\lambda_{\max}$ in the other answer is incorrect. The fact that $\lambda_{\max}(S) \leq \lambda_{\max}(A)$ is obvious, because $S=D−C^T B^{-1}C \leq D$, and $\lambda_{\max}(D) \leq \lambda_{\max}(A)$.