Trace

For a square matrix $A$, its trace is defined as the sum of its diagonal entries

$$\mathrm{Tr}A=\sum _{k=1}^n{a}_{k,k}.$$

Theorem 1(Theorem).

Let $A$ and $B$ be matrices of size $m\times n$ and $n\times m$ respectively. Then, $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$.

Proof

$$\mathrm{Tr}(AB)=\sum _{j=1}^m\sum _{i=1}^n{a}_{j,i}{b}_{i,j}=\sum _{i=1}^n\sum _{j=1}^m{b}_{i,j}{a}_{j,i}=\mathrm{Tr}(BA)$$

Another way to prove this theorem would be to consider two linear transformations $T,T_1:M_{n\times m}\to {\mathbb{F}}^1$(from the space of all $n\times m$ matrices to a field $\mathbb{F}$) defined by

$$T(X)=\mathrm{Tr}(AX),T_1(X)=\mathrm{Tr}(XA).$$

To prove the theorem it is sufficient to prove $T=T_1$.
Recall that a linear transformation is completely defined by its values on a generating system. An example of a simple generating system for $M_{n\times m}$ would be

$$S:=\{X_{i,j}|1<i\le n,1<j\le m\}$$

where $X_{i,j}$ is a matrix where $x_{i,j}=1$, and all other cells are zero. Thus, we only have to show that

$$T(X)=T_1(X)\forall X\in S,$$

which is trivial.