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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?

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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?

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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 $v=0$.

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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).

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