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.

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