Proof of Basic Properties of the Trace

All matrices have real entries, products are matrix products, $(\cdot)^{\top}$ is the transpose, and $\operatorname{tr}$ is the trace, so that $\operatorname{tr}(M)=\sum_i M_{ii}$ for a square matrix $M$.

Claim 1. The diagonal entries of the entrywise combination $aM+bN$ are $aM_{ii}+bN_{ii}$, so

by the field axioms for the real numbers applied to finite sums.

Claim 2. By the definition of the transpose, $(M^{\top})_{ii}=M_{ii}$ for every $i$, so the two traces are sums of the same numbers.

Claim 3. By the definition of the matrix product, $(UV)_{ii}=\sum_{j=1}^{q}U_{ij}V_{ji}$, so

Likewise $(VU)_{jj}=\sum_{i=1}^{p}V_{ji}U_{ij}$, so $\operatorname{tr}(VU)=\sum_{j=1}^{q}\sum_{i=1}^{p}V_{ji}U_{ij}$. The two double sums have the same finitely many terms $U_{ij}V_{ji}$ and hence are equal, since finite sums of real numbers may be reordered by commutativity and associativity of addition.

Claim 4. By the definitions of transpose and product, $(U^{\top}V)_{jj}=\sum_{i=1}^{p}(U^{\top})_{ji}V_{ij}=\sum_{i=1}^{p}U_{ij}V_{ij}$, so

again by reordering a finite sum. Similarly $(UV^{\top})_{ii}=\sum_{j=1}^{q}U_{ij}(V^{\top})_{ji}=\sum_{j=1}^{q}U_{ij}V_{ij}$, so $\operatorname{tr}(UV^{\top})$ equals the same double sum. $\square$