Question

Defining the right generalized inverse of a non-square Jacobian matrix $J$, $J^{\#}$, as

$J^{\#} = M^{-1} J^T \left(J M^{-1} J^T\right)^{-1}$

where the matrix $M \succ 0$ is positive definite and symmetric, can we infer that the following null space projection matrix

$\left(I - J^\# J \right)$

is non-negative definite?

For the engineering problem that I am tackling, I was able to show that

$M\left( I - J^\# J \right) \succeq 0$

Answer 1

No, consider the following counterexample: Take $$ M = \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix} , \quad J = \begin{pmatrix} 1 & 2\end{pmatrix},$$ then $J^# = \begin{pmatrix} 1 \\ 0\end{pmatrix}$ and your projection is given by $I - J^# J = \begin{pmatrix} 0 & -2\\ 0 & 1\end{pmatrix}$ and this is definitely not non-negative definite by your definition.

Edit: Can't seem to get the matrices right...

Answer 2

Modulo my comment above ($I-J^{\sharp}J$ is not Hermitian), here is what we can say: First, it is correct that $MJ^\sharp J=J^T(JM^{-1}J^T)^{-1}J$ is Hermitian. So is $M(I-J^\sharp J)$.

Once you know that $M(I-J^\sharp J)$ is semi-definite positive, there follows that $I-J^\sharp J$ is diagonalizable with non-negative real eigenvalues. More generally, if $G,H$ are Hermitian with $H$ positive definite (here $G=M(I-J^\sharp J)$ and $H=M^{-1}$), then $HG$ is diagonalizable with real eigenvalues, and the signs of the eigenvalues of $HG$ agree with those of $G$.