Positive semidefinite matrix and pseudoinverse 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 $Α \in R^{6 \times 6}$?
 A: 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.
