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I have a matrix $J \in R^{8 \times 7}$ and I know that $rank(J)=6$. Is $J^+ J$ a positive semi-definite matrix? If yes, is $J^+ AJ$ also positive semi-definite, where $A$ is a diagonal positive definite matrix $Α \in R^{7 \times 7}$? What if $J \in R^{8 \times 6}$ and $rank(J)=6$ and $Α \in R^{6 \times 6}$?

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  • $\begingroup$ see pseudoinverse#Limit_relations : $\displaystyle J^+ J = \lim_{\delta \searrow 0} (J^* J + \delta I)^{-1} J^* J$ $\endgroup$ – reuns Mar 15 '16 at 11:58
  • $\begingroup$ $A$ is also a real matrix. Does this change anything? $\endgroup$ – Controller Mar 15 '16 at 12:02
  • $\begingroup$ no $A = A^* $ hence if $J^+ \ne J^*$ then $(J^+ A J)^* \ne J^+ A J$ hence it is not self-adjoint, so it can't be positive semi-definite... and with the limit definition, $J J^+$ is positive-semi definite. and because $J J^+ J = J$, we get that if $J^+ J$ was not positive-semi definite, $J (J^+ J) J^* = J J^* $ would not be positive-semi definite ? hence, $J^+ J$ is also positive-semi definite (I think) $\endgroup$ – reuns Mar 15 '16 at 12:03
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The answer to your first question is yes.

To prove this, you can use the limit representation suggested in the comments: $$ J^+=\lim_{\delta\to{0+}}J^+_\delta, J^+_\delta:=(J^TJ+\delta I)^{-1}J^TJ,\\ J^+J=\lim_{\delta\to{0+}}J^+_\delta J. $$ Now, let $J=U^T\Sigma V$ be the singular value decomposition of $J$. Here, matrices $U, V$ are orthogonal ($UU^T=VV^T=I$) and $\Sigma$ is diagonal positive semidefinite. Then we can write $$ J^+_\delta J=(J^TJ+\delta I)^{-1}J^TJ=(V^T\Sigma UU^T\Sigma V+\delta V^TV)^{-1}V^T[\Sigma UU^T\Sigma]V=(V^T[\Sigma^2+\delta I]V)^{-1}V^T[\Sigma^2]V=\underbrace{V^{-1}}_{V^T}[\Sigma^2+\delta I]^{-1}VV^T[\Sigma^2]V=V^T([\Sigma^2+\delta I]^{-1}\Sigma^2)V. $$

Now, let's look at the associated quadratic form: $$ x^TJ_\delta^+Jx=\underbrace{(Vx)^T}_{y^T}([\Sigma^2+\delta I]^{-1}\Sigma^2)Vx $$ so the matrix in question is positive semidefinite if the matrix $M_\delta:=[\Sigma^2+\delta I]^{-1}\Sigma^2$ is also positive semidefinite.

However, it's easy to see that $M_\delta$ is indeed positive semidefinite: it is diagonal (as a product of diagonal matrices) with diagonal elements $$ M_{\delta ii}=\frac{\Sigma_{ii}^2}{\Sigma_{ii}^2+\delta}. $$

Thus, taking the limit, we see that $M=\lim_{\delta\to{0+}}M_\delta$ is positive semidefinite (because it is again diagonal with nonnegative entries) and thus $J^+J$ is positive semidefinite as well.

Concerning your second question, I don't see a reason yet why it should be positive semidefinite.

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  • $\begingroup$ Sorry I forgot to mention that $A$ is a diagonal matrix... $\endgroup$ – Controller Mar 15 '16 at 16:59

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