$$ M=\left[\begin{array}{cc} A & B\\ B^{T} & D \end{array}\right] $$ where $A$ and $D$ are not invertible, but both are positive semi-definite, e.g., consider the case where $A=\left[\begin{array}{cc} a & -a\\ -a & a \end{array}\right]$, $B=\left[\begin{array}{cc} b_{1} & b_{2}\\ b_{2} & b_{1} \end{array}\right]$, and $D=\left[\begin{array}{cc} 0 & 0\\ 0 & 0 \end{array}\right]$; $a,b_1$ and $b_2$ are real numbers.

Are there conditions such that $M$ is positive semi-definite? I read that if $D$ is invertible and $A-BD^{-1}B^T$ is positive semi-definite then $M$ is positive semi-definite (using Schur complements).

I am wondering if there is a way to show positive semi-definiteness 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.


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 $A-BD^+B^T\ge0$.


It seems the example given cannot be positive semi-definite unless $a \ge 0$ and $b_1=b_2=0$. Solving for the eigenvalues of $M$ using symbolic MATLAB we get
$$\begin{align*} \lambda_{1} & =b_{1}+b_{2}\\ \lambda_{2} & =a-\sqrt{a^{2}+b_{1}^{2}-2b_{1}b_{2}+b_{2}^{2}}\\ \lambda_{3} & =-b_{1}-b_{2}\\ \lambda_{3} & =a+\sqrt{a^{2}+b_{1}^{2}-2b_{1}b_{2}+b_{2}^{2}} \end{align*}$$ and thus, the conditions $a \ge 0$ and $b_1=b_2=0$ should hold for all the eigenvalues to be non-negative. If at all possible, I will be happy to learn a way of showing this without solving for the eigenvalues.


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