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Homework question, have to prove the following:

Let $A$ be $m\times n$, rank $m$ and $L$ be an $n\times n$ symmetric and positive definite on the subspace $M = \{\mathbf{x} : A\mathbf{x}=\mathbf{0}\}$. Show that the following $(n+m)\times (n+m)$ matrix

$$ \begin{pmatrix} L& A^T \\ A&0 \end{pmatrix} $$

is non-singular.

My attempt: The given matrix is symmetric and hence, has an eigenvector base characterization, with real eigenvectors. I suppose I have to use this, but dont know how to carry on. Any hints? Thanks

user191919
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1 Answers1

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Write a vector $v\in \mathbb R^{n+m}$ as $(v_1,v_2)$, $v_1\in \Bbb R^n$, $v_2\in \Bbb R^m$. Denote your big matrix $B$ and suppose that $Bv=0$. It immediately follows that $Av_1=0$ (consider last $m$ lines of $Bv$).

Now consider scalar product $0=(Bv,v)$. It obviously writes $v_1^TLv_1 + v_1^TA^Tv_2$. The last term is zero, because $Av_1=0$. So we arrive to $v_1^TLv_1=0$, which gives us $v_1=0$.

Therefore, by considering the first $n$ lines of $Bv$ we obtain that $A^Tv_2=0$, but since $rank (A)=m$, we obtain that $v_2=0$. We immediately conclude that $B$ is non-singular.

TZakrevskiy
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