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Let

$$M =\left[\begin{array}{cc} A & B\\ B^{T} & D\end{array}\right]$$

where blocks $A$ and $D$ are not invertible, but both are positive semidefinite. Are there conditions such that $M$ is positive semi-definite?

For example, consider the case where

$$A=\left[\begin{array}{cc} a & -a\\-a & a\end{array}\right], \qquad B=\left[\begin{array}{cc} b_{1} & b_{2}\\ b_{2} & b_{1}\end{array}\right], \qquad D=\left[\begin{array}{cc}0 & 0\\0 & 0\end{array}\right]$$

where $a, b_1, b_2 \in \Bbb R$.

I read that if $D$ is invertible and $A-BD^{-1}B^T$ is positive semidefinite then $M$ is positive semidefinite (using Schur complements).

I am wondering if there is a way to show positive semidefiniteness when neither $A$ nor $D$ is invertible; especially, when the matrices $A$, $B$, and $D$ could be written in the form given in the example.

Rodrigo de Azevedo
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zola
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1 Answers1

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The quadratic form represented by the block matrix is $q\left((x,y)\right)=x^TAx+2x^TBy+y^TDy$. In order that it is positive semidefinite, we must have $\ker(A)\subseteq\ker(B^T)$ and $\ker(D)\subseteq\ker(B)$, i.e. $B^T(I-A^+A)=0$ and $B(I-D^+D)=0$.

When these two conditions are satisfied, $q$ is positive semidefinite if and only if it is positive semidefinite on $\ker(A)^\perp\times\ker(D)^\perp$. Since the inverse of $D$ on $\ker(D)^\perp$ is $D^+$, $q$ is positive semidefinite if and only if $A-BD^+B^T\ge0$.

In summary, $q\ge0$ if and only if $B^T(I-A^+A)=0,\,B(I-D^+D)=0$ and $A-BD^+B^T\ge0$.

user1551
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