Let \(A\) and \(B\) be \(n\times n\) matrices. Show that \(AB\) and \(BA\) have the same eigenvalues.
Solution
1. For arbitrary \(n\times n\) matrices \(X,Y\),\[ \det(I-XY) = \det(I-YX). \]
Consider the block matrix \[ M = \begin{pmatrix} I&X\\ Y&I \end{pmatrix}. \] Left multiplication by \[ \begin{pmatrix} I&-X\\ 0&I \end{pmatrix} \] has determinant \(1\) and gives \[ \begin{pmatrix} I-XY&0\\ Y&I \end{pmatrix}. \] Therefore \[ \det M = \det(I-XY). \]
On the other hand, left multiplication by \[ \begin{pmatrix} I&0\\ -Y&I \end{pmatrix} \] also has determinant \(1\) and gives \[ \begin{pmatrix} I&X\\ 0&I-YX \end{pmatrix}. \] Hence \[ \det M = \det(I-YX). \] Equating the two expressions for \(\det M\) proves the identity.
2. For every nonzero scalar \(\lambda\),\[ \det(\lambda I-AB) = \det(\lambda I-BA). \]
Factor out \(\lambda\): \[ \begin{aligned} \det(\lambda I-AB) &= \lambda^n \det\left( I-\lambda^{-1}AB \right)\\ &= \lambda^n \det\left( I-B\lambda^{-1}A \right) \end{aligned} \] by step 1, with \[ X=\lambda^{-1}A, \qquad Y=B. \] Since scalars commute with matrices, \[ B\lambda^{-1}A = \lambda^{-1}BA. \] Factoring \(\lambda\) back in gives the desired equality.
3. The characteristic polynomials of \(AB\) and \(BA\) are equal.
Both \[ \det(\lambda I-AB) \] and \[ \det(\lambda I-BA) \] are polynomials in \(\lambda\) of degree \(n\). Step 2 shows that they agree for every nonzero scalar \(\lambda\). Over \(\RR\) or \(\CC\), this is an infinite set, so the two polynomials are identical.
4. The matrices \(AB\) and \(BA\) have the same eigenvalues, with the samealgebraic multiplicities.
The eigenvalues, counted with algebraic multiplicity, are precisely the roots of the characteristic polynomial. Apply step 3.
5. Q.E.D.
Step 4 proves the requested statement.