What is the determinant of a weighted orthogonal projection (based on the weighted pseudo-inverse)? E.g. I have

$$ J = A \left( A^\intercal W A \right)^{-1} A^\intercal W $$

and would like to know $\det (J)$. Note that $A$ is not square while $W$, $J$, and $A^\intercal W A$ are square.

  • $\begingroup$ @Surb Sorry - I got that one wrong => J is square, I had omitted the $A$ at the beginning! Please re-check! $\endgroup$ Jun 25, 2019 at 8:47
  • $\begingroup$ Well then, the properties of the determinant listed in @Chris answer should give you what you want, namely $\det(J)=1$ if I'm not wrong. $\endgroup$
    – Surb
    Jun 25, 2019 at 8:53
  • $\begingroup$ @Surb But $A$ is not square, which conflicts with Chris' derivation. $\endgroup$ Jun 25, 2019 at 8:54
  • $\begingroup$ @JennyReininger right. $\endgroup$
    – Surb
    Jun 25, 2019 at 8:54
  • 1
    $\begingroup$ If $A$ isn't square, it must be a tall matrix and hence $\det(J)=0$. $\endgroup$
    – user1551
    Jun 25, 2019 at 9:40

2 Answers 2


The kernel of any projection onto a proper subspace is nontrivial. If $A$ is not square, then its columns don’t span the entire ambient space, therefore $J$ is rank-deficient and $\det(J)=0$.


Given that $$ \det A^{-1} = \det(A)^{-1} $$ and $$ \det(AB) = \det(A)\det(B) $$ I would say $$ \det((A^T W A)^{-1} A^T W) = \frac{1}{\det A} $$

  • 1
    $\begingroup$ Why assume that $A$ is square? $\endgroup$ Jun 25, 2019 at 8:29
  • $\begingroup$ Thank you for helping me debug the question! Now, I just need to find the answer... $\endgroup$ Jun 25, 2019 at 8:57

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