Let me assume that $\mathbf{J} \in \mathbb{R}^{m \times n},~m<n$ is a full row rank matrix, $\mathbf{A} \in \mathbb{R}^{n \times n}$ is a symmetric positive definite matrix, and $\mathbf{J}^{-}$ is a weighted right pseudoinverse of the matrix $\mathbf{J}$ given as $\mathbf{J}^{-}=\mathbf{A}^{-1}\mathbf{J}^{T}(\mathbf{J}\mathbf{A}^{-1}\mathbf{J}^{T})^{-1}$.

It seems to me that $\mathbf{J}\mathbf{A}^{-1}\mathbf{J}^{-}$ is a regular matrix, but I have problems to prove it. I would be grateful if somebody has an idea.

  • $\begingroup$ So, the nullspace of $\mathbf A$ is empty, so $\forall \mathbf M:rank(\mathbf M \mathbf A)=rank(\mathbf M)$. $\endgroup$
    – davcha
    Oct 31, 2014 at 14:18
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    $\begingroup$ I agree with this, but this is not the proof that I want. $\endgroup$
    – Tarik
    Oct 31, 2014 at 14:22
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    $\begingroup$ Well, $JA^{-1}J^-=JA^{-2}J^T(JA^{-1}J^T)^{-1}$ is the product of two SPD (nonsingular) matrices. $\endgroup$ Oct 31, 2014 at 17:40

1 Answer 1


If you mean invertible by regular, then the answer is only if A is full-rank. The weighted pseudo-inverse formulation that you brought is only valid if A is full-rank. Otherwise, you can use a regularized pseudo-inverse (or damped least-square) $J^-=A^{-1}J^T(JA^{−1}J^T+\mu^2I)^{−1}$ where $\mu$ is a small number. Here is a more complete way of handling it: Weighted pseudo-inverse formulation


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