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Prove that A Kronecker Product B is invertible if and only if B Kronecker Product A is invertible. I dont have a clue where to start to be honest. I am not very familiar yet to the Kronecker Product so could you please help me with providing an easyily understandable proof. Thanks in advance. |
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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 $j,l$. |
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Hint: If $A$ is a $m\times m$ matrix and $B$ a $n\times n$ matrix then $$\det(A\otimes B)=(\det A)^n\cdot (\det B)^m.$$ |
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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. |
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