Positive definitiveness of a product of matrices

Question

I have a matrix $J \in R^{8 \times 6}$ and I know that $rank(J)=6$, therefore $J^+ J=I_6$, thus positive definite. Is $J^+ AJ$ also positive definite, if $A$ is a diagonal positive definite matrix $Α \in R^{8 \times 8}$?

Answer 1

If the notation $J^+$ means the pseudoinverse of $J$, i.e. $J^\dagger$, the answer is no, and can be proved as follows.

Let $J=V\Sigma W^*$ be an SVD of $J$, and $J^\dagger=W\Sigma^\dagger V^*$. Then $M\triangleq J^\dagger AJ=W\Sigma^\dagger V^*AV\Sigma W^*$. First of all, note that $M$ may not be a Hermitian matrix at all, for which positive definiteness is usally defined. Even if we extend the definition of positive definiteness to the non-Hermitian $M$ (i.e. $M$ is p.d. if $x^*Mx>0$ for any $x\ne0$), $M$ still may not be p.d.

To see this, let $\hat A=V^*AV$, so $M=W\Sigma^\dagger\hat A\Sigma W^*$. Note that $M$ is p.d. if and only if $\Sigma^\dagger\hat A\Sigma$ is p.d. Let $\Sigma^\dagger=[D^{-1}\: 0]$, where $D$ is a $6\times6$ diagonal p.d. matrix, and let $\tilde A$ denote the upper-left $6\times 6$ submatrix of $\hat A$. Then $\Sigma^\dagger\hat A\Sigma=D^{-1}\tilde AD.$ Since $\hat A$ is clearly Hermitian and p.d., so is $\tilde A$. From this, we can easily find a counter-example. For instance, with $D=\textrm{diag}(10, 1, 1, 1, 1, 1) \textrm{, and}$

$$\tilde A=\left[ \begin{array}{ccc} 2 & 1 & 0_{1\times 4} \\ 1 & 1 & 0_{1\times 4} \\ 0_{4\times 1} & 0_{4\times 1} & I_4 \end{array} \right],$$

it's easy to see that

$$D^{-1}\tilde AD=\left[ \begin{array}{ccc} 2 & 0.1 & 0_{1\times 4} \\ 10 & 1 & 0_{1\times 4} \\ 0_{4\times 1} & 0_{4\times 1} & I_4 \end{array} \right],$$

which is not p.d., for if $x=[-1\:1\:0_{1\times 4}]^T$, then $x^*D^{-1}\tilde ADx=-7.1 < 0.$

Lastly, noting that we can certainly find an $A, V$ and $J$ that lead to the $\tilde A$ and $D$ above, we conclude that $J^\dagger AJ$ may not be p.d.