2013/05/9

eigenvalues under commutation

May 09th 2013 Posted to Uncategorized

\(A\) and \(B\) are two square matrices. The eigenvalues of \(AB\) and \(BA\) are the same.

Proof: Suppose \(\lambda\) and \(v\) are an eigenvalue and the corresponding eigenvector for \(AB\), so that \(ABv = \lambda v\). Let \(q = Bv\). Then \(Aq = \lambda v\), so \(BAq = \lambda Bv = \lambda q\). So \(\lambda\) is also an eigenvalue of \(BA\), and its associated eigenvector is \(q = Bv\).

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