Given a $4 \times 4$ symmetric matrix $A$, are there any matrices $B,C$ that: $BAC = I_{4}$ ?
I've thought of $B$ being a orthogonal matrix $P$ ($B=P$) and $ C = P^{T}$ so we get $PAP^{T} = \begin{bmatrix}\lambda_{1}&0&0&0\\0&\lambda_{2}&0&0\\0&0&\lambda_{3}&0\\0&0&0&\lambda_{4}\end{bmatrix} $
Where $\lambda_{i}$ with $i \in {1,2,3,4}$ are the eigenvalues of matrix $A$.
And then demanded that $\{\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}\} = \{1,1,1,1\}$ But I am not sure that this is correct.
Thank you for your help!