# Is trace invariant under cyclic permutation with rectangular matrices?

I'm working with trace of matrices. Trace is defined for square matrix and there are some useful rule, i.e. $$\text{tr}(AB) = \text{tr}(BA)$$, with $$A$$ and $$B$$ square, and more in general trace is invariant under cyclic permutation.

I was wondering if the formula $$\text{tr}(AB) = \text{tr}(BA)$$ holds even if $$A$$ and $$B$$ are rectangular, namely $$A$$ is $$n$$-by-$$m$$ and $$B$$ is $$m$$-by-$$n$$.

I figured out that if one completes the involved matrices to be square by adding zero entries in the right places, then the formula still works... but I want to be sure about this!

Yes, it holds true. Let $A$ be a $n\times m$ and $B$ be a $m \times n$ matrix over the commutative ring $R$, we have \begin{align*} \mathrm{tr}(AB) &= \sum_{i=1}^n (AB)_{ii}\\ &=\sum_{i=1}^n \sum_{j=1}^m A_{ij}B_{ji}\\ &= \sum_{j=1}^m \sum_{i=1}^n B_{ji}A_{ij}\\ &= \sum_{j=1}^m (BA)_{jj}\\ &= \mathrm{tr}(BA) \end{align*} So you can just prove the formula by computing.