Invertibility of a Kronecker Product

Solution 1:

The previous answers assume that $A$ and $B$ are square. This is fair because it is not generally possible for $A\otimes B$ to be invertible when $A$ and $B$ are rectangular.

In the more general setting where $A$ is $m\times n$ and $B$ is $r\times s$ (and $mr = ns$) it is still easy to see that $A\otimes B$ is equivalent to $B \otimes A$ up to permuting the rows and columns.

Both $A \otimes B$ and $B \otimes A$ contain exactly the same entries $a_{ij} b_{kl}$. More explicitly, one places $a_{ij} b_{kl}$ in row $(i-1)r+k$ and column $(j-1)s+l$, and the other places it in row $(k-1)m+i$ and column $(l-1)n+j$. Note that the row numbers are uniquely determined by $i,k$ and the column numbers determined independently by $j,l$.

Solution 2:

Hint: If $A$ is a $m\times m$ matrix and $B$ a $n\times n$ matrix then $$\det(A\otimes B)=(\det A)^n\cdot (\det B)^m.$$

Solution 3:

Do you know that if two matrices $A$ and $B$ represent two linear maps $a : V \to V'$ and $b : W \to W'$ (with respect to some chosen bases of the vector spaces $V$, $V'$, $W$ and $W'$), then the Kronecker product $A \otimes B$ represents the linear map $a \otimes b : V \otimes W \to V' \otimes W'$ (with respect to the appropriately defined bases: e.g., if our bases of $V$ and $W$ are $\left(e_i\right)_{i \in I}$ and $\left(f_j\right)_{j \in J}$, then for $V \otimes W$ we use the basis $\left(e_i \otimes f_j\right)_{\left(i,j\right)\in I\times J}$) ? If so, then you will realize that your question is equivalent to the following question: If the tensor product $a \otimes b$ of two linear maps $a$ and $b$ is invertible, then so is $b \otimes a$. And this follows from the following commutative diagram:

$ \begin{array}{ccc} V \otimes W & \overset{\cong}{\longrightarrow} & W \otimes V \\ \downarrow a \otimes b & & \downarrow b \otimes a \\ V' \otimes W' & \overset{\cong}{\longrightarrow} & W' \otimes V' \end{array} $

where the horizontal arrows are canonical isomorphisms.

This is essentially an algebraic version of Erick Wong's proof.

Solution 4:

Added: Assuming that $A$ and $B$ are square matrices and neither is $0\times 0$:

You can show that $A\otimes B$ is invertible if and only if $A$ and $B$ are invertible, from which this result follows.

If $A$ and $B$ are invertible, do you know how to find $(A\otimes B)^{-1}$?

If $A$ is not invertible, then there exists $C\neq 0$ such that $CA=0$. It follows that $(C\otimes I)(A\otimes B)=0$, which implies that $A\otimes B$ is not invertible.