Determinant of Kronecker product $D = B⊗A$ is equeal $D=A^kB^n$

Question

Here determinants: $$ A = \begin{vmatrix} a_{11} & ... & a_{1n}\\ ... & ... & ...\\ a_{n1} & ... & a_{nn}\\ \end{vmatrix} B = \begin{vmatrix} b_{11} & ... & b_{1k}\\ ... & ... & ...\\ b_{k1} & ... & b_{kk}\\ \end{vmatrix} $$ And Kronecker product $D = B⊗A$: $$ D = \begin{vmatrix} a_{11}b_{11} & ... & a_{1n}b_{11} & a_{11}b_{12} & ... & a_{1n}b_{12} && ... && a_{11}b_{1} & ... & a_{1n}b_{1k}\\ ... & ... & ... & ... & ... & ... & ... & ... & ... & ... & ... & ... \\ a_{n1}b_{11} & ... & a_{nn}b_{11} & a_{n1}b_{12} & ... & a_{nn}b_{12} && ... && a_{n1}b_{1k} & ... & a_{nn}b_{1k}\\ ... & ... & ... & ... & ... & ... & ... & ... & ... & ... & ... & ... \\ a_{11}b_{k1} & ... & a_{1n}b_{k1} & a_{11}b_{k2} & ... & a_{1n}b_{k2} && ... && a_{11}b_{kk} & ... & a_{1n}b_{kk}\\ ... & ... & ... & ... & ... & ... & ... & ... & ... & ... & ... & ... \\ a_{n1}b_{k1} & ... & a_{nn}b_{k1} & a_{n1}b_{k2} & ... & a_{nn}b_{k2} && ... && a_{n1}b_{kk} & ... & a_{nn}b_{kk}\\ \end{vmatrix} $$ Need to proof that: $$ D = A^kB^n $$ I tried to separate this by groups with size $(n,n)$ and took out the $A$ from each row of $D$, then using properties of determinant and got: $\det(AD')=\det(A)^k\det(D')$. I can't understand what I shall do next, and not sure that way is right, because this properties can be not right for groups.

Answer 1

The hints have given you an efficient way to do this.

But you can get a proof by continuing along the line you have started on.

You've got, as you have observed, that $$ D= \det\begin{pmatrix} Ab_{11}& &\dots & Ab_{1n}\\ \vdots &\ddots &\vdots\\ Ab_{n1}& &\dots & Ab_{nn}\\ \end{pmatrix} = \det\begin{pmatrix} A& \\ &\ddots &\\ & & A\\ \end{pmatrix} \det\begin{pmatrix} Ib_{11}& &\dots & Ib_{1n}\\ \vdots &\ddots &\vdots\\ Ib_{n1}& &\dots & Ib_{nn}\\ \end{pmatrix} $$ If we now permute the rows and columns of the second determinant we'll get $$ D=\pm \det\begin{pmatrix} A& \\ &\ddots &\\ & & A\\ \end{pmatrix} \det\begin{pmatrix} B& \\ &\ddots &\\ & & B\\ \end{pmatrix} $$ and we can check that the sign is $+$ by considering the leading term, or some special case.