Let $A$ be a matrix whose only non-zero entries are on the diagonal, those entries are all non-negative and at least one is positive; $B$ a matrix whose only non-zero elements are on the diagonal and those are all positive, $R$ and $S$ rotation matrices. I can see that $$ \mathrm{tr}(\mathrm{adj}(RAR^T)SBS^T) \geq 0 $$ since by the invariance of the trace to cyclic permutations of the argument product one can write it as $\mathrm{tr}(X^TX)$. But is it necessarily positive?
3 Answers
If $X=(x_{ij})$ then $$(X^TX)_{ii}=\sum_{j=1}^n x_{ij}^2 \\ \mathrm{tr}(X^TX)=\sum_{i=1}^m \sum_{j=1}^n x_{i,j}^2$$
This is always $\geq 0$ and is equal to $0$ if and only if $X=0$.
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$\begingroup$ Oh of course, how could I have missed that. Many (slightly embarrassed) thanks. $\endgroup$– n00bJan 30, 2017 at 23:44
Edited to give a complete solution.
First, note that because $R$ and $R^{T}$ are orthogonal, the adjugates of $R$ and $R^{T}$ are orthogonal matrices.
$\mbox{tr}(\mbox{adj}(RAR^{T})SBS^{T})=\mbox{tr}(\mbox{adj}(R)\mbox{adj}(A)\mbox{adj}(R)^{T}SBS^{T})$
$\mbox{tr}(\mbox{adj}(RAR^{T})SBS^{T})=\mbox{tr}(\mbox{adj}(R)\mbox{adj}(A^{1/2})\mbox{adj}(A^{1/2})\mbox{adj}(R)^{T}SB^{1/2}B^{1/2}S^{T})$
$\mbox{tr}(\mbox{adj}(RAR^{T})SBS^{T})=\mbox{tr}(B^{1/2}S^{T}\mbox{adj}(R)\mbox{adj}(A^{1/2})\mbox{adj}(A^{1/2})\mbox{adj}(R)^{T}SB^{1/2})$
$\mbox{tr}(\mbox{adj}(RAR^{T})SBS^{T})=\mbox{tr}(X^{T}X)$
where $X=\mbox{adj}(A^{1/2})\mbox{adj}(R)^{T}SB^{1/2}$
$\mbox{tr}(X^{T}X)$ is always nonnegative, because the matrix $X^{T}X$ is positive semidefinite, and the trace of $X^{T}X$ is the sum of its eigenvalues which are all non-negative.
$\mbox{tr}(X^{T}X)=0$ if and only if $X^{T}X=0$.
Consider the range of $X^{T}X$. Each of the matrices in the product is invertible (the adjugate of an orthogonal matrix is orthognal) except for $\mbox{adj}(A)$, which has non-zero rank. Thus the range of $X^{T}X$ is non-trivial, and $X^{T}X$ is not 0.
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$\begingroup$ But this does not preclude the trace being zero, right? $\endgroup$– n00bJan 30, 2017 at 23:38
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$\begingroup$ Correct, but that can only happen if all of the eigenvalues are 0, or equivalently if $X^{T}X$ is the 0 matrix. $\endgroup$ Jan 31, 2017 at 0:17
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The product of the diagonal matrix with terms $d_1,d_2,\dots,d_n$ and the diagonal matrix with terms $e_1,e_2,\dots,e_n$ is the diagonal matrix with entries $e_1d_1,e2d2,\dots,e_nd_n$.
Since the $e_i$'s are non-negative and so are the $d_i's$ we have that $e_id_i\geq 0$ for all $i$. We also have that at least one $d_k$ is positive, so for that one we have $d_ke_k>0$.
we conclude that $e_1d_1+\dots+e_nd_n\geq e_kd_k>0$. So the trace is positive.