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For square $n \times n$ matrices, if either $A$ or $B$ are invertible, show that $A B$ and $B A$ are similar.
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Here is a proof that, for square $n \times n$ matrices $A$ and $B$, if either $A$ or $B$ is invertible, then $A B$ and $B A$ are similar:

Let $P$ be the matrix $P = B^{-1} A$. Then we have

$A B = B P$

and

$B A = P B$

Since A B and B A are both square $n \times n$ matrices, they are similar if there exists a non-singular matrix $S$ such that

$A B = S B A$

Substituting the expressions for $A B$ and $B A$, we get

$B P = S P B$

Since $B$ is invertible, we can multiply both sides by $B^{-1}$ to get

$P = S P$

Since $P$ is a non-singular matrix, we must have $S = I$, the identity matrix. This shows that $A B$ and $B A$ are similar.

Alternatively, we can show that $B A$ and $A B$ are similar in a similar way, using the matrix $Q = A^{-1} B$ instead of $P$.

This proves that, for square $n \times n$ matrices $A$ and $B$, if either $A$ or $B$ is invertible, then $A B$ and $B A$ are similar.
by Platinum (101k points)

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