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Let $X$ be an $n\times p$ matrix and $A$ be a $n\times n$ matrix. When is it true that

$$\det (X^{\top}AX) = \det(A)\det(X^{\top}X)?$$

Dexter
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    You would also be stating (as a corollary) that $$x^TAx=\det(A)\times |x|_2^2$$ If you pick a non-positive definite matrix with positive determinant, you can find a counterexample. For instance, $$A=\begin{pmatrix}-2&1\-5&1\end{pmatrix}\x=\binom{1}{0}$$ $$\det(A)=3\ x^TAx=-2\\det(A)\times|x||_2^2 =3$$ – Inquest Aug 27 '13 at 15:39

2 Answers2

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(Edited to reflect change in problem) In general not true:

Let $X = e_1$ and $A$ be any matrix with $A_{11} = 0$. $X^T A X = 0$ whereas the the other quantity will be in general nonzero.

Now to think on it... Intuitively I suspect $X$ has to be full dimensional (ruling out silly cases like $0$ matrices), and at least if $\det(X) = 1$, then it's clear that $\det(X^T A X) = \det((X^T)^{-1} X^T A X X^{-1}) = \det(A)$.

Evan
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  • What if we restrict $A$ to be a diagonal matrix with positive diagonal? – Dexter Aug 27 '13 at 15:31
  • Good question. Let's see... but still not in general, if we take symmetric $A = UDU^T$ and let $X$ be some columns of $U$ corresponding to the null space, same thing happens. Perhaps should say $A$ is non-singular? Even then it's still not true for symmetric $A$ and $X$ subset of eigenvectors as $X^TAX$ in general takes a subset of the eigenvalues. Then $X^T$ is always $1$ but $\det(X^TAX) \neq \det(A)$ – Evan Aug 27 '13 at 15:36
  • I am going to accept this answer since I have to sign out now and this was helpful, but the general question remains open. Thanks. – Dexter Aug 27 '13 at 16:41
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In the case that $A$ is a positive diagonal matrix, (i.e. all non-diagonal entries are zero) we can set $$ X=\sqrt\lambda_2e_1-\sqrt\lambda_1e_2 $$ where $\lambda_1,\lambda_2$ are the first two diagonal entries of $A$.

The result is $X^TAX=0$

Ben Grossmann
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