Showing that a matrix equation is also satisfied by the transpose matrix

Question

I want to prove the following statement:

If a matrix L satisfies $L^T \eta L = \eta$ then the transpose matrix $L^T$ also satisfies the same equation.

When I try to solve this, I always end up with the equation $L \eta L^T=\eta$ which seems to be correct only if $L=L^T$ which I think is not the case in general. So my question is: How can I prove this statement in a proper way?

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NOTE: The matrix $\eta$ satisfies $\eta^T = \eta$ and its inverse is also equal to itself.

Answer 1

Here a slightly different solution and some comments:

Since $\eta$ is assume to be invertible, $L$ is also invertible. Thus, we have $$ \eta = \eta^{-1} = L^{-1}\eta^{-1} L^{-T} = L^{-1}\eta L^{-T}, $$ that is $$ L \eta L^T = \eta. $$

Notes:

1. $L$ is not uniquely determined by $\eta$ (nor vice versa $\eta$ is an eigen"vector" of $X\mapsto L^T XL$ w.r.t. eigenvalue $1$).

2. Also notice, $L$ needs not to be symmetric, not even normal. Consider the example $$L = \begin{bmatrix} \sqrt{2} & -1 \\ 1 & -\sqrt{2} \end{bmatrix} \qquad\text{and}\qquad \eta = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}. $$ We have $$ L^T \eta L = \eta $$ but $$ LL^T = \begin{bmatrix} * & 2\sqrt{2} \\ * & * \end{bmatrix} \ne \begin{bmatrix} * & -2\sqrt{2} \\ * & * \end{bmatrix} = L^T L. $$