Assume $A$ is a real symmetric positive semidefinite $k\times k$ matrix. Denote by $A^{(a:b,c:d)}$ the submatrix of $A$ obtained by keeping only the rows between $a$ and $b$, and columns between $c$ and $d$. Now define the $(k-1) \times (k-1)$ matrix defined by
$ \tilde{A} = A^{(1:k-1,1:k-1)} – \bigg(A^{(1:k-1,k)}(A^{(k,k)})^{-1}\bigg)A^{(k,k)}\bigg(A^{(1:k-1,k)}(A^{(k,k)})^{-1}\bigg)^{\prime} $.
I'm pretty sure that this matrix is positive semidefinite, but I need some to help to verify that claim rigorously.
Best Answer
Writing $$ A = \pmatrix{ A_1 & a \\ a^T & c} $$ we have $$ \tilde A = A_1 - aa^T c^{-1}. $$ Let $x\in \mathbb R^{k-1}$. Then $$ x^T\tilde Ax = x^TA_1x - (x^Ta)(a^Tx) c^{-1}. $$ We want to use the matrix $A$, so let me augment $x$ by a real number $b$ to obtain $$ 0\le \pmatrix{x \\ b }^TA\pmatrix{x \\ b } = x^TA_1x + 2b a^Tx + b ^2c. $$ Setting $b=-c^{-1} a^Tx$ yields $$ 0\le \pmatrix{x \\ -c^{-1} a^Tx }^TA\pmatrix{x \\ -c^{-1} a^Tx }=x^TA_1x - (x^Ta)(a^Tx) c^{-1}, $$ which shows positive semidefinitness.