Let $A\in M_n(\Bbb{R} )$ be a symmetric matrix. I want to find a general formula for the diagonals of the matrices of the form $g^{-1}Ag$, where $g\in O_n(\Bbb{R})$.
Here is what I did :
Since $A$ is symmetric, there is some $g\in O_n(\Bbb{R})$ such that $g^{-1}Ag$ is diagonal. Since similar matrices have the same trace, all the matrices in the orbit must have a trace equal to $\sum_{i=0}^k {n_i \lambda_i}$ where $\lambda_i$ are the eigenvalues of $A$ and $n_i$ their multiplicities.
My question is : Is every diagonal giving this trace in the orbit of $A$ or is there other restrictions ? And how to prove it?
$\mathbf{EDIT} :$ After the comments I got that helped me clarify a bit the situation, I want to continue a bit my thinking and formulate precisely what I need.
So far we have that the matrices in the orbit of $A$ must have the same trace $and $ the same eigenvalues. So it's a bit more complicated than thinking only about the trace. I'm still looking for a general formula for the diagonals of all the matrices in the orbit of $A$. Any idea would be very helpful.
Thank you
Best Answer
Let $A$ be a symmetric real $n\times n$ matrix with $spectrum(A)=(\lambda_i)$.
Then the orbit $\mathcal{O}_A$ of $A$, under the action $g\in O_n\mapsto g^{-1}Ag$, is the set of symmetric real matrices with the same spectrum as $A$.
Then the subset of $\mathcal{O}_A$ constituted by diagonal matrices has, up to a permutation -eventually with repetitions- of the entries of the diagonal, only one element: $diag(\lambda_1,\cdots,\lambda_n)$.
EDIT. Majorization is a complete description of the relationships between the eigenvalues and diagonal entries of real symmetric matrices.
$\textbf{Definition}$. Let $(\lambda)=\lambda_n\leq \cdots\leq \lambda_1$, $(a)=a_n\leq\cdots \leq a_1$; we say that $(\lambda)$ majorizes $(a)$ iff for every $k< n$, $\sum_{i=1}^k a_i\leq\sum_{i=1}^k \lambda_i$ and $\sum_{i=1}^n a_i=\sum_{i=1}^n \lambda_i$.
$\textbf{Theorem}$ (cf. Horn and Johnson, Matrix Analysis).
If $(\lambda)$ majorizes $(a)$, then there is a symmetic real matrix $A$ s.t. $spectrum(A)=(\lambda)$ and its diagonal is $(a_i)$.
The converse of the previous theorem is true; that gives the required result, up to a permutation of the entries of the diagonal.