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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

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

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The matrix $\left(I - J^\# J \right)$ is that of the projection onto $R(M^{-1}J^T)$, parallel to $\ker J$. It is not Hermitian, unless $R(J^T)$ (or equivalently $\ker J$) be stable under $M$. So, what do you mean by being non-negative definite ? –  Denis Serre Jun 20 '12 at 10:14
    
Non-negative definite is equivalent to semi-positive definite, i.e., I would like to know if for an arbitrary vector $q$, the following relation holds: $q^T(I-J^{\#} J)q \ge 0$ –  user24579 Jun 20 '12 at 12:26
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2 Answers

up vote 1 down vote accepted

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...

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It is really cool to discuss with mathematicians! Thanks Jan for your reply. You did show that for the general case, the projection matrix is not non-negative definite. –  user24579 Jun 20 '12 at 20:41
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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 $G$.

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Dear Denis, thanks for your reply! The properties of the projection matrix that you mentioned are very helpful to tackle my engineering problem. –  user24579 Jun 20 '12 at 20:38
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