2
$\begingroup$

Suppose the matrix $A\in\mathbb{R}^{m\times n}$, $m\leq n$, and has full row rank $m$, $B\in\mathbb{R}^{n\times n}$ is a symmetric, $Z\in\mathbb{R}^{n\times(n-m)}$ is the matrix whose columns span $\ker A$, i.e., $AZ=0$. Suppose $Z^TBZ$ is positive definite. How to prove $$\left(\begin{array}{ccc} B & A^T \\ A & 0 \\ \end{array} \right)$$ is nonsingular?

Context

In the simple case of $m=n=1$ one can calculate the determinant to be $-A^2<0$, so the matrix is invertible. In general, the value of the determinant is not so easy to find; is there a better way?

$\endgroup$
4
$\begingroup$

Note that $\ker A = {\cal R}Z$.

Suppose $Bu + A^T v =0, Au = 0$. Then $u \in \ker A= {\cal R} Z$, hence $u = Zw$ for some $w$.

Then $Z^T B Zw + Z^T A^T v = Z^T B Zw + (AZ)^T v = Z^T B Zw = 0$. Hence $w=0$ since $Z^T BZ>0$.

Hence $u=0$ and since $A^T v = 0$ and $A$ has full row rank, we have $v=0$.

$\endgroup$
0
$\begingroup$

The matrix $Z^TBZ$ is positive definite and thus nonsingular. We can forget the positive definiteness as long as we keep the nonsingularity in mind. We can also forget the assumption that $A$ has rank $m$ and that $B$ is symmetric.

Let $v \in \ker \left( \begin{array}{ccc} B & A^T \\ A & 0 \\ \end{array} \right)$. We need to prove that $v = 0$.

Write $v$ in the block-matrix form $v = \left(\begin{matrix} x \\ y \end{matrix}\right)$, where $x$ is a length-$n$ vector and $y$ is a length-$m$ vector. Then, $\left(\begin{matrix} x \\ y \end{matrix}\right) = v \in \ker \left( \begin{array}{ccc} B & A^T \\ A & 0 \\ \end{array} \right)$, so that $0 = \left( \begin{array}{ccc} B & A^T \\ A & 0 \\ \end{array} \right) \left(\begin{matrix} x \\ y \end{matrix}\right) = \left(\begin{matrix} Bx+A^Ty \\ Ax \end{matrix}\right)$. Hence, $0 = Bx + A^Ty$ and $0 = Ax$.

We have $0 = Ax$, thus $x \in \ker A$. Hence, $x$ lies in the column space of $Z$ (since the columns of $Z$ span $\ker A$). That is, $x = Zz$ for some vector $z$. Fix such a $z$. We have $0 = B\underbrace{x}_{=Zz} + A^Ty = BZz + A^Ty$, so that $BZz = -A^Ty$. Multiplying this with $Z^T$ from the left, we obtain $Z^TBZz = -\underbrace{Z^TA^T}_{=\left(AZ\right)^T=0 \ \text{(since } AZ = 0 \text{)}}y = 0$. Hence, $z \in \ker \left(Z^TBZ\right)$, so that $z = 0$ (since $Z^TBZ$ is nonsingular).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.