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Let $B\in \mathbb{R}^{n\times n}$ and $C\in\mathbb{R}^{m\times n}$ (where $m\leq n$) and let $v^TBv > 0$ $\forall v \neq 0$ with $Cv = 0$, where $C$ has full rank.

Proof that $A = \begin{pmatrix} B & C^T \\ C & 0 \end{pmatrix}$ is an invertible Matrix.

So far, I've concluded that $B$ is positive definite and therefore also $det(B)>0$, which could help prove that $det(A) \neq 0$. But other than that I had no idea for a solution.

Any help is appreciated, thanks in advance!

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2 Answers 2

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Take $B = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 0 \end{bmatrix}$ to see that $B$ does not have to be positive definite (replace the $-1$ with $0$ in $B$ to see non-invertibility of $B$).


To show that $A$ is invertible we will show $Av = 0 \implies v = 0$. We have

$$\begin{bmatrix} B & C^T \\ C & 0 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 $$

gives $Cv_1 = 0$ and $Bv_1 + C^Tv_2 = 0$. Left multiplying the second equation by $v_1^T$ yields

$$v_1^TBv_1 + v_1^TC^Tv_2 = 0 \iff v_1^TBv_1 + (Cv_1)^Tv_2 = 0$$

which gives $v_1^TBv_1 = 0$ since $Cv_1 = 0$. Thus $v_1 = 0$ (otherwise $v_1^TBv_1 > 0$). Plugging this back in gives $C^Tv_2 = 0$ which implies $v_2 = 0$ since $C^T$ is injective (full column rank).

$\square$

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This amounts to simply using the Schur complement.

In detail, note that $A$ is invertible if and only if the matrix $$ \pmatrix{I & 0\\-CB^{-1} & I} \pmatrix{B & C^T\\ C & 0} \pmatrix{I & 0\\-CB^{-1} & I}^T = \pmatrix{B&0\\0 & - CB^{-1}C^T} $$ is invertible. Thus, $A$ is invertible if and only if both $B$ is invertible and $CB^{-1}C^T$ is invertible. Using the fact that $C$ has full row-rank and $B^{-1}$ is positive definite, conclude that $CB^{-1}C^T$ is positive definite and therefore invertible.

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  • $\begingroup$ Hi Ben, I don't think $B$ has to be positive definite or even invertible (see counterexample in my answer). $\endgroup$
    – Coriolanus
    Commented Jan 29, 2021 at 16:01

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