Exercise 4.3.3

Note, multiplying $P$ in front of a matrix permutates its rows, while multiplying $P$ on the back of a matrix permutates its columns.

Let the size of $A$ be $m\times n$, and the size of $B$ be $M\times N$.

Consider the $(i-1)M+k$th row in $A\otimes B$, this row comes from the product of the $i$th row from $A$, $a_{i,}$, with the $k$th row from $B$, i.e. $b_{k,}$. This row looks like

To have the same elements of $b_{\mathit{kq}}{a}_{\mathit{ip}}=a_{\mathit{ip}}{b}_{\mathit{kq}}$ in the matrix of $B\otimes A$, they are on the row of $(k-1)m+i$. This row looks like

So we should switch every $(i-1)M+k$th row in $A\otimes B$ with the $(k-1)m+i$th row.

We also need switch the columns in $B\otimes A$ as well, consider the entry of $b_{\mathit{kq}}{a}_{\mathit{ip}}$ of $B\otimes A$, its column position is $(q-1)n+p$. The element from $A\otimes B$ with the same value, i.e. $a_{\mathit{ip}}{b}_{\mathit{kq}}$ is at the position of $(p-1)N+q$ So we need swtich all columns between $(q-1)n+p$ and $(p-1)N+q$.

Combine the two conditions together, we see that $P$ should switch rows satisfying following condition:

$(i-1)M+k=(q-1)n+p$ and $(k-1)m+i=(p-1)N+q$

• For $A\otimes B$, its eigenvalues are always the products of eigenvalues from $A$ and $B$, this is also true for $B\otimes A$, so their eigenvalues are the same.