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If $B$ is a square symmetric matrix of order $n\times n$ then is true that $$ \det{\Big(A^T\cdot B\cdot A\Big)}=\det\Big(A^T\cdot A\Big)\det B $$ where $A$ is a matrix of order $n\times m$? Unfortunately I did not find a counterexample: in particular I tried to show that $$ A^T\cdot B\cdot A=A\cdot A^T\cdot B $$ so that the statement follows directely applying the Binet formula. So could someone help me, please?

Antonio Maria Di Mauro
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  • @Jujustum Okay, could you prove it? – Antonio Maria Di Mauro Jun 27 '21 at 20:49
  • It’s because for any matrices $A$ and $B$, $\det(A)\det(B)=\det(AB)$. This is not that easy to prove, though. But you should be able to find your answers on the internet. (or here: https://math.stackexchange.com/questions/60284/how-to-show-that-detab-deta-detb ) – Jujustum Jun 27 '21 at 20:51
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    @Jujustum it's only true for square matrices in general. Here, $A,A^T$ are not assumed to be square. – Vercassivelaunos Jun 27 '21 at 20:54
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    @Jujustum I know this result: however unfortunately it is true only for square matrices. Anyway for sake of completeness I point out that a more general result holds (Cauchy-Binet formula) but unfortunately it do not seems work here. – Antonio Maria Di Mauro Jun 27 '21 at 20:54
  • @Vercassivelaunos You say well. – Antonio Maria Di Mauro Jun 27 '21 at 20:55
  • Oh well, my bad. I read your message too quickly. – Jujustum Jun 27 '21 at 20:57
  • you should be able to eyeball this in terms of signature of $B$.

    For example, suppose $B$ has signature $(n-1,1)$ and $A$ is $n \times n-1$, with rows consisting of 3 orhornomal eigenvectors of $A$ associated with its positive eigenvalues. Then $ \det{\Big(A^T\cdot B\cdot A\Big)}\gt 0$ and $\det\Big(A^T\cdot A\Big)\cdot \det B = 1 \cdot \det B\lt 0$

     – user8675309 Jun 28 '21 at 17:35

1 Answers1

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It's not true. Take $A = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$. Then $$\det(A^\top B A) = \det(3) = 3,$$ but $$\det(A^\top A) \cdot \det(B) = \det(2) \cdot \det\begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix} = 4.$$

Theo Bendit
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