Prove that a non-positive definite matrix has real and positive eigenvalues I have a $2\times2$ matrix $J$ of rank $2$ and a $2\times2$ diagonal positive definite matrix $Α$. Denote by $J^+$ a pseudoinverse of $J$. 
I can find many counterexamples for which $J^+AJ$ is not positive definite (e.g. $J=\left(\begin{smallmatrix}2&1\\1&1\end{smallmatrix}\right)$ and $A=\left(\begin{smallmatrix}1&0\\0&2\end{smallmatrix}\right)$), but for all of them $J^+AJ$ has real and positive eigenvalues. So, I was wondering whether I can prove or disprove that indeed this is the case for this form.
What would happen if $J$ was still of rank $2$, but $3\times2$ this time? Of course, $A$ would be correspondingly a $3\times3$ real positive definite matrix.
 A: $J$ is $2 \times 2$ and has rank $2$, so $J$ is invertible.  Consequently, the Moore-Penrose pseudoinverse of $J$, $J^+ = J^{-1}$.  Then \begin{align*}
    J^+ A J = J^{-1} A J
\end{align*} is similar to $A$ and has the same eigenvalues as $A$.  $A$ is a diagonal matrix, so is its own symmetric part.  Therefore, since $A$ is a real positive definite matrix, $A$ and $J^+ A J$ have all positive eigenvalues.

Edit:
Assuming your $3 \times 3$ version of $A$ is also diagonal...  Take \begin{align*}
    J &= \begin{pmatrix} 1 & 4 \\ 0 & 1 \\ 0 & 1 \end{pmatrix}  \text{,}  \\
    A &= \begin{pmatrix} 1/2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}  \text{.}
\end{align*}  Then the rank of $J$ is $2$.  (To see this, subtract row 3 from row 2 to get row-echelon form and note that we have two nonzero rows.  Alternatively, the two columns are linearly independent.)  We find \begin{align*}
J^+ &= \begin{pmatrix} 1 & -2 & -2 \\ 0 & \frac{1}{2} & \frac{1}{2} \end{pmatrix}  \text{,}  \\
B = J^+ A J &= \begin{pmatrix} \frac{1}{2} & -2 \\ 0 & 1 \end{pmatrix}  \text{,}
\end{align*} and the symmetric part of $B$ is \begin{align*}
B_S &= \begin{pmatrix} \frac{1}{2} & -1 \\ -1 & 1 \end{pmatrix}  \text{.}
\end{align*}  The eigenvalues of $B_S$ are $\frac{1}{4}(3 \pm \sqrt{17})$ and the smaller one is negative.  Thus, we have found choices for $J$ and $A$ yielding a $J^+ A J$ which is not positive definite.
(Note, these values aren't "special".  They're more or less the first thing I tried after using a CAS to find the expressions for the eigenvalues.  Two more: let the right column of $J$ and $A$ be $(85, -21, -42)$ and $\mathrm{diag}(15, 2173/29, 58)$, respectively, or $(-271, 0,-18)$ and $\mathrm{diag}(93, 70, 148)$, respectively.  These were found with that same CAS using a random number function for several numbers and messing around with the remaining ones.)
