## The trace of a Kronecker product is the product of traces

Let $A$ and $B$ be square matrices over a ring $R$. Recall that the trace of a square matrix is the sum of its diagonal entries. Let $A\ \mathsf{kr}\ B$ denote the Kronecker product of $A$ and $B$. Prove that $\mathsf{tr}(A \mathsf{kr}\ B) = \mathsf{tr}(A) \mathsf{tr}(B)$.

First we will give a concrete recursive characterization of the Kronecker product.

Let $A$ and $B$ be matrices. If $A = [a]$, then $A\ \mathsf{kr}\ B = aB$. If $A = \left[ \begin{array}{c|c} a & A_{1,2} \\ \hline A_{2,1} & A_{2,2} \end{array} \right]$, then $A\ \mathsf{kr}\ B = \left[ \begin{array}{c|c} aB & A_{1,2}\ \mathsf{kr}\ B \\ \hline A_{2,1}\ \mathsf{kr}\ B & A_{2,2}\ \mathsf{kr}\ B \end{array} \right]$.

Now to the main result; we will proceed by induction on the dimensions of $A$. If $A = [a]$, then certainly $\mathsf{tr}(A) = a$, and $\mathsf{tr}(A\ \mathsf{kr}\ B) = \mathsf{tr}(A)\mathsf{tr}(B)$.

For the inductive step, suppose the result holds when $A$ has dimension $n$, and let $A$ be a matrix with dimension $n+1$. Say $A = \left[ \begin{array}{c|c} a & A_{1,2} \\ \hline A_{2,1} & A_{2,2} \end{array} \right]$. Then $\mathsf{tr}(A\ \mathsf{kr}\ B) = \mathsf{tr}\left( \left[ \begin{array}{c|c} aB & A_{1,2}\ \mathsf{kr}\ B \\ \hline A_{2,1}\ \mathsf{kr}\ B & A_{2,2}\ \mathsf{kr}\ B \end{array} \right] \right)$ $= \mathsf{tr}(aB) + \mathsf{tr}(A_{22} \mathsf{kr}\ B)$ $= \mathsf{tr}(A)\mathsf{tr}(B)$ as desired.