positivity of the trace of a matrix product

Question

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?

Answer 1

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 $X=0$.

Answer 2

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.

Answer 3

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.