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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.
in Mathematics by Diamond (88,200 points) | 197 views

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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

\(A B = B P\)


\(B A = P B\)

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 Diamond (88,200 points)

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