I noticed that, in $\mathbb{R}^2$, holds the following identity:

$$ \mathrm{det} \begin{pmatrix} <LX|X> & <LX|Y> \\ <LY|X> & <LY|Y>\\ \end{pmatrix} = \mathrm{det}(L)\mathrm{det}\begin{pmatrix} <X|X> & <X|Y> \\ <Y|Y> & <Y|Y>\\ \end{pmatrix} $$

where $L$ is a real and symmetric operator, $<X|Y>$ is the scalar product and $X$, $Y$ are independent vectors. I prove the equality writing explicitly the determinant of both sides, and it was very tedious. The question is:

Does this equality hold also in $\mathbb{R}^n$? And how can I prove this equality in a better, quickly way?


P.S. I think it is not true that $A = LS$ where $A$ is the first matrix and $S=\begin{pmatrix} <X|X> & <X|Y> \\ <Y|Y> & <Y|Y>\\ \end{pmatrix}$.


Yes. Since $\langle\cdot,\cdot\rangle$ is an inner product, there exists a positive definite matrix $P$ such that $\langle x,y\rangle=x^TPy$ for every pair of vectors $x$ and $y$. It follows that $$ S_L:=\pmatrix{\langle Lx_1,x_1\rangle&\cdots& \langle Lx_1,x_n\rangle\\ \vdots&\cdots&\vdots\\ \langle Lx_n,x_1\rangle&\cdots& \langle Lx_n,x_n\rangle} =\pmatrix{x_1^TL^TPx_1&\cdots&x_1^TL^TPx_n\\ \vdots&\cdots&\vdots\\ x_n^TL^TPx_1&\cdots&x_n^TL^TPx_n} =(LX)^TPX. $$ Therefore $\det(S_L)=\det(L)\det(X^TPX)=\det(L)\det(S_I)$.

  • $\begingroup$ Perfect:) Thanks $\endgroup$ – Alex Mar 28 '18 at 9:51

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