My question stems from quantum physics but is essentially a linear algebra problem, so I hope this is the right place to ask it.
I have a matrix $A$ and I want to maximize the trace of the product of two matrices (assuming the maximum exists, so no infinities...), Tr $A B$ by varying $B$. However I only want to consider matrices $B$, with Tr$B = 1$ and non-negative eigenvalues ($B$ is basically a density matrix).
Unfortunately I am not so versatile when it comes to mathematically rigorous proofs, but here are my thoughts: The trace should be independent of the used basis, so I choose the one, in which $A$ is diagonal (hence the eigenbasis with eigenvectors $v_n$ ). In this basis, the diagonal of $A$ consists of all the eigenvalues $a_n$. Then the trace reads Tr $AB = \sum_n a_n p_n$, where $p_n = v_n^T B v_n$. With the properties of $B$, the maximum of this expression is (**) simply $a_{n_0} \cdot 1$, where $a_{n_0}$ denotes the largest eigenvalue of $A$. Hence the matrix B has then a "$1$" on the diagonal at row/column $n_0$, and the rest are zeros, the non-diagonal entries of $B$ do not matter and lead to the same maximum value for the trace.
To me the step (**) seems somehow "clear" to me but failed to proof it, which usually means that its either trivial or wrong.
A side question which is related to this: I know that the trace of a matrix (sum of diagonal terms) is equal to the sum of the eigenvalues. Is it also possible to say that the maximum diagonal term (under change within normalized basis) is the maximum eigenvalue? I have the impression that if the maximum eigenvalue can be exceeded, my argumentation from before fails. (or at least becomes more difficult)
Best Thanks already :)
Edit: Maybe there will be more comments if I try to reduce my thoughts to the essentials: I guess everyone agrees that the trace can be written as
$Tr (AB) = \sum_n a_n p_n$
with $a_n$ being the eigenvalues of $A$ and
$1 \geq p_n = v_n^T B v_n \geq 0$
as well as
$ \sum_n p_n =1$.
There are now two "unclear" points to me:
(i) Given said properties, is the following statement correct? $\max_{p_n} \sum_n a_n p_n = a_{n_0} \cdot 1$, where $a_{n_0} = \max_n a_n$ is the largest eigenvalue of $A$
(ii) I write that the trace is independent of the used basis, but does this hold true when calculating the maximum of the trace?
Matrix multiplication is linear in a larger space in the sense we can express "multiplication by A from left" with a matrix using Kronecker products.
In this space the trace operator will be scalar product with a row vector consisting of ones on the positions of the diagonal and 0 in all other positions.
One approach could then be to try and solve:
$$\min_{B}\{\|Tr(B) - 1\|_2^2 + \epsilon\|Tr(AB) - b\|_2^2\}$$ where b is some big scalar number.