# What is the determinant of a weighted orthogonal projection (based on the weighted pseudo-inverse)?

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.

• @Surb Sorry - I got that one wrong => J is square, I had omitted the $A$ at the beginning! Please re-check! Jun 25, 2019 at 8:47
• 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.
– Surb
Jun 25, 2019 at 8:53
• @Surb But $A$ is not square, which conflicts with Chris' derivation. Jun 25, 2019 at 8:54
• @JennyReininger right.
– Surb
Jun 25, 2019 at 8:54
• If $A$ isn't square, it must be a tall matrix and hence $\det(J)=0$. Jun 25, 2019 at 9:40

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 $$\det((A^T W A)^{-1} A^T W) = \frac{1}{\det A}$$
• Why assume that $A$ is square? Jun 25, 2019 at 8:29