How to show this matrix is positive definite?

Given: $A,A_0 \in \mathbb{R}^{n \times n};A,A_0$ symmetric and pos. def.; $A>A_0(\Leftrightarrow A-A_0$ pos. def. or $v^TAv>v^TA_0v,\forall v);B \in \mathbb{R}^{m \times n}$ of row full rank $m, m<n$

I want to show that $M:=\left( \begin{array}{cc} AA_0^{-1}A-A && (AA_0^{-1}-I)B^T \\ B(A_0^{-1}A-I) && BA_0^{-1}B^T \end{array} \right)$ is pos. def..

I started with $(v,w)^TM\left( \begin{array}{c} v \\ w \end{array} \right)=v^T(AA_0^{-1}A-A)v + v^T(AA_0^{-1}-I)B^Tw+w^TB(A_0^{-1}A-I)v +w^TBA_0^{-1}B^Tw$

I know that both $AA_0^{-1}A-A$ and $BA_0^{-1}B^T$ are pos. def.

Now I was told to use Cauchy-Schwarz and Young's inequality in the following way to get an estimate for the other terms: $a:=w^TB, b:=(A-A_0)v$

$w^TB(A_0^{-1}A-I)v=aA_0^{-1}b \leq \sqrt{aA_0^{-1}a}\sqrt{bA_0^{-1}b} \leq \frac{\epsilon}{2}aA_0^{-1}a+\frac{1}{2\epsilon}bA_0^{-1}b,\epsilon \in \mathbb{R}^+$ but I’m not quite sure how that gives me an useful estimate. Would appreciate any help.

Edit: With back substitution I get

$|w^TB(A_0^{-1}A-I)v| \leq \frac{1}{2}aA_0^{-1}a+\frac{1}{2}bA_0^{-1}b = \frac{1}{2}w^TBA_0^{-1}B^Tw + \frac{1}{2}v^T(A-A_0)^TA_0^{-1}(A-A_0)v = \frac{1}{2}w^TBA_0^{-1}B^Tw + \frac{1}{2}v^T(AA_0^{-1}A-A)v-\frac{1}{2}v^T(A-A_0)v$

and

$|w^TB(A_0^{-1}A-I)v| + |v^T(AA_0^{-1}-I)B^Tw| \leq w^TBA_0^{-1}B^Tw + v^T(AA_0^{-1}A-A)v-v^T(A-A_0)v$

which should give me what I need since $v^T(A-A_0)v >0$, or am I wrong?

Solution 1:

You don't need those inequalities. As $A>A_0>0$, we have $A_0^{-1}>A^{-1}>0$. Therefore $P=(A_0^{-1}-A^{-1})^{1/2}$ exists and $$ M =\pmatrix{ AA_0^{-1}A-A & (AA_0^{-1}-I)B^T\\ B(A_0^{-1}A-I) & BA_0^{-1}B^T} =\underbrace{\pmatrix{ AP & 0\\ BP & BA^{-1/2}}}_X \pmatrix{ PA & PB^T\\ 0 & A^{-1/2}B^T} =XX^T $$ is a Gram matrix. Since $B$ has full row rank, so does $X$. Hence $M>0$.

Solution 2:

From the Cauchy-Schwarz inequality and the Young's inequality, you have

$\begin{align} & \left| v^T (A A_0^{-1} - I) B^T w \right| \\ &= \left| v^T A (A_0^{-1} - A^{-1}) B^T w \right| \\ &\le \sqrt{v^T A (A_0^{-1} - A^{-1}) A^T v} \sqrt{w^T B (A_0^{-1} - A^{-1}) B^T w} && \text{(by Cauchy-Schwarz inequality)} \\ &\le \frac{1}{2} v^T A (A_0^{-1} - A^{-1}) A^T v + \frac{1}{2} w^T B (A_0^{-1} - A^{-1}) B^T w && \text{(by Young's inequality)} . \end{align}$

Hence,

$\begin{align} \left| v^T (A A_0^{-1} - I) B^T w + w^T B (A_0^{-1} A - I) v \right| &\le \left| v^T (A A_0^{-1} - I) B^T w \right| + \left| v^T (A A_0^{-1} - I) B^T w \right| \\ &\le v^T A (A_0^{-1} - A^{-1}) A^T v + w^T B (A_0^{-1} - A^{-1}) B^T w . \end{align}$

It follows that

$\begin{align} & v^T (A A_0^{-1} A - A) v + v^T (A A_0^{-1} - I) B^T w + w^T B (A_0^{-1} A - I) v + w^T B A_0^{-1} B^T w \\ & \ge \left( v^T (A A_0^{-1} A - A) v + w^T B A_0^{-1} B^T w \right) - \left( v^T A (A_0^{-1} - A^{-1}) A^T v + w^T B (A_0^{-1} - A^{-1}) B^T w \right) \\ &= w^T B A_0^{-1} B^T w - w^T B (A_0^{-1} - A^{-1}) B^T w \\ &= w^T B A^{-1} B^T w > 0 \end{align}$

for $ w \neq \mathbf{0} $, which implies that $ M $ is positive definite.