If $h, k$ be $(1,1)-$tensors on a Riemannian manifold $(M,g)$ then (similar to matrices) does the following property hold? $$\operatorname{tr}(h\circ k)=\operatorname{tr}(k\circ h).$$
More generally if $s$ be another $(1,1)-$tensor then does the following property hold? $$\operatorname{tr} (h\circ k \circ s)=\operatorname{tr}(s\circ h\circ k).$$
Every $(1,1)$ tensor can be written as a matrix accepting a vector and a covector as a column and row, respectively. Moreover, trace is coordinate-free, so it doesn't even matter how you choose to calculate the trace of these matrices.