Let $A$, $B\in\mathbb{R}^{n\times n}$ with $\text{rank}(A)=\text{rank}(B)=n$ and define the Kronecker products $P=A\otimes B$ and $Q=B\otimes A$. Through example-based examination, I have found that

$\text{rank}(P-Q)=n^2-n$,

but I am struggling to formulate a proof of this. Thus far, I have focused on the fact that $P$ and $Q$ are permutation-similar such

$Q=ZPZ$,

where $Z=Z^T=Z^{-1}$. We know that $P$ and $Q$ have the same eigenvalues and that the eigenvectors, $\Phi_P$ and $\Phi_Q$, are related by $\Phi_P=Z\Phi_Q$. Furthermore,

$\Phi_P-Z\Phi_QZ=T=[T_1 \ldots T_{n^2}]$,

where $T_i=0$ for $i \in \{1:n+1:n^2\}$. I presume that this somehow shows the rank condition, but I cannot figure out the coupling. I would like to ask whether I am on the right track; and if yes, can you provide some pointers on how I proceed to show the rank condition?

Let $A$, $B\in\mathbb{R}^{n\times n}$ be full-rank random matrices and define the Kronecker products $P=A\otimes B$ and $Q=B\otimes A$. Through example-based examination, I have found that

$\text{rank}(P-Q)=n^2-n$,

but I am struggling to formulate a proof of this. Thus far, I have focused on the fact that $P$ and $Q$ are permutation-similar such

$Q=ZPZ$,

where $Z=Z^T=Z^{-1}$. We know that $P$ and $Q$ have the same eigenvalues and that the eigenvectors, $\Phi_P$ and $\Phi_Q$, are related by $\Phi_P=Z\Phi_Q$. Furthermore,

$\Phi_P-Z\Phi_QZ=T=[T_1 \ldots T_{n^2}]$,

where $T_i=0$ for $i \in \{1:n+1:n^2\}$. I presume that this somehow shows the rank condition, but I cannot figure out the coupling. I would like to ask whether I am on the right track; and if yes, can you provide some pointers on how I proceed to show the rank condition?

Let $A$, $B\in\mathbb{R}^{n\times n}$ and define the Kronecker products $P=A\otimes B$ and $Q=B\otimes A$. Through example-based examination, I have found that

$\text{rank}(P-Q)=n^2-n$,

but I am struggling to formulate a proof of this. Thus far, I have focused on the fact that $P$ and $Q$ are permutation-similar such

$Q=ZPZ$,

where $Z=Z^T=Z^{-1}$. We know that $P$ and $Q$ have the same eigenvalues and that the eigenvectors, $\Phi_P$ and $\Phi_Q$, are related by $\Phi_P=Z\Phi_Q$. Furthermore,

$\Phi_P-Z\Phi_QZ=T=[T_1 \ldots T_{n^2}]$,

where $T_i=0$ for $i \in \{1:n+1:n^2\}$. I presume that this somehow shows the rank condition, but I cannot figure out the coupling. I would like to ask whether I am on the right track; and if yes, can you provide some pointers on how I proceed to show the rank condition?

Let $A$, $B\in\mathbb{R}^{n\times n}$ with $\text{rank}(A)=\text{rank}(B)=n$ and define the Kronecker products $P=A\otimes B$ and $Q=B\otimes A$. Through example-based examination, I have found that

$\text{rank}(P-Q)=n^2-n$,

but I am struggling to formulate a proof of this. Thus far, I have focused on the fact that $P$ and $Q$ are permutation-similar such

$Q=ZPZ$,

where $Z=Z^T=Z^{-1}$. We know that $P$ and $Q$ have the same eigenvalues and that the eigenvectors, $\Phi_P$ and $\Phi_Q$, are related by $\Phi_P=Z\Phi_Q$. Furthermore,

$\Phi_P-Z\Phi_QZ=T=[T_1 \ldots T_{n^2}]$,

where $T_i=0$ for $i \in \{1:n+1:n^2\}$. I presume that this somehow shows the rank condition, but I cannot figure out the coupling. I would like to ask whether I am on the right track; and if yes, can you provide some pointers on how I proceed to show the rank condition?