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Let $\mathbf{A}\in\mathbb{R}^{N \times N}$ be symmetric positive definite. For some $1\leq k<N$, partition $$\mathbf{A}=\begin{pmatrix}\mathbf{A}_{11} & \mathbf{A}_{12} \\ \mathbf{A}_{12}^T & \mathbf{A}_{22}\end{pmatrix},$$ where $\mathbf{A}_{11}$ is $k\times k$ and $\mathbf{A}_{22}$ is $(N-k)\times (N-k)$.

Show that $$\|\mathbf{A}_{22}-\mathbf{A}_{12}^T\,\mathbf{A}_{11}^{-1}\,\mathbf{A}_{12}\|_2\leq \|\mathbf{A}\|_2.$$

I know that $\|\mathbf{A}_{ij}\|_2 \le \|\mathbf{A}\|_2$ for $1 \le i,j \le 2$, however I am unsure of how to use that fact to prove this inequality is true.

Does anyone have any ideas?

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Hint: See the Schur Complement.

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Using the block inversion formula, we have $$ A^{-1}=\pmatrix{*&*\\*&S^{-1}}, \quad S:=A_{22}-A_{12}^TA_{11}^{-1}A_{12}. $$ The Cauchy interlacing theorem implies that $$ \lambda_{\min}(A^{-1})\leq\lambda_{\min}(S^{-1})\leq\lambda_{\max}(S^{-1})\leq\lambda_{\max}(A^{-1}) $$ and hence, since for SPD $X$, $\lambda_{\max}(X^{-1})=1/\lambda_{\min}(X)$ and $\lambda_{\min}(X^{-1})=1/\lambda_{\max}(X)$, $$ \lambda_{\min}(A)\leq\lambda_{\min}(S)\leq\lambda_{\max}(S)\leq\lambda_{\max}(A). $$ The statement then follows from the last inequality and the fact that, for SPD matrices, $\lambda_{\max}$ and $\|\cdot\|_2$ are equal.

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