Let $J \in \mathbb{R}$ be a $m \times n$ matrix of rank $n$ and a $A \in \mathbb{R}$ be a $m \times m$ diagonal positive definite matrix. Denote by $J^+$ a pseudoinverse of $J$.

As was shown in this question $J^+AJ$ is not positive definite in general.

However, I was wondering whether one can prove or disprove that $J^+AJ$ has real and positive eigenvalues.


1 Answer 1


This question can be answered easily by performing singular value decomposition on $J$.

Since $J$ has full column rank, any singular value decomposition of $J$ must be in the form of $U\pmatrix{S\\ 0}V^T$ for some positive diagonal matrix $S$. Therefore $J^+AJ=VS^{-1}BSV^T$, where $B$ is the leading principal $n\times n$ submatrix of $U^TAU$. Since $A$ is positive definite, so is $B$ (Sylvester's criterion). Hence $J^+AJ$, being similar to $B$, must have a real positive spectrum.


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