Let $A$ be a real $n \times n$ matrix with positive determinant. Suppose that $A$ is diagonal, and that
$ A=S-\lambda (S^T)^{-1}$, where $S$ is another invertible real $n \times n$ matrix, and $\lambda$ is some real number.
Must $S$ be diagonal?
Edit:
It suffices to prove that $S$ is symmetric, at least in the generic case where $A$ has distinct diagonal entries.
Indeed, the assumption is equivalent to $S^T(S-A)=\lambda I$, hence
$$S^T(S-A)=\lambda I=(\lambda I)^T=(S^T(S-A))^T=(S^T-A^T)S,$$ i.e.
$ S^TA=A^TS$. Since $A$ is symmetric, we obtain $ S^TA=AS$. Thus, if we assume $S^T=S$, then $S$ commutes with $A$, hence it must be diagonal.
Best Answer
There's exceptions, at least in 2 dimensions: if $$ S = \begin{pmatrix} a & b \\ c & d \end{pmatrix} , $$ the equation is equivalent to $$ A = \begin{pmatrix} a - \lambda d/\Delta & b + \lambda c/\Delta \\ c + \lambda b/\Delta & d - \lambda a/\Delta \end{pmatrix} , $$ where $\Delta = ad-bc$. Asking for the off-diagonal entries to be $0$ forces $ c = \pm b $ and $b = \sqrt{ \lambda \pm ad} $, provided that they are not $0$ to start with, and then $\Delta = \pm \lambda $, so either $$ A = (a+d)I \quad \text{or} \quad A = \operatorname{diag}(a-d,-a+d) , $$ and the second is excluded since $\det{A}>0$. This will also extend to $n$ dimensions any time $A$ has two equal eigenvalues, of course. In particular, the second one will be allowed for matrices such as $\operatorname{diag}(\mu,-\mu,-1)$, so for dimensions larger than $2$, both types of solution are possible.
This leaves the general case of nonequal roots. We can improve on sufficiency of $S$ being symmetric to $S$ being (real-)normal:
In general, we have $$ S^T A = S^T S - \lambda I \quad \text{and} \quad AS^T = SS^T - \lambda I , $$ and taking the transpose of the first gives $$ AS = A^T S = S^T S - \lambda I $$ so $S^T-S = A^{-1} [S^T,S] $: if $S$ is normal, it is symmetric.
$ (S^T)^{-1} = \lambda^{-1} (A - S) $, so $ S^T = \lambda (A-S)^{-1} $, and I'm now also stuck.