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Theorem: $AB$ and $BA$ have the same eigenvalues, where $A_{n\times n}$ and $B_{n\times n}$, and $\alpha\neq0$

Step 1: Let $v$ be an eigenvector corresponding to the eigenvalue $\alpha\neq0$ of $AB$. $ABv=\alpha v$ and by definition $v\neq0$. We are looking for a vector $w$ such that $BAw=\alpha w$

2. If we apply $B$ to both sides of $ABv=\alpha v$, we have $BABv=B\alpha v=\alpha Bv$. Then, $BA(Bv)=\alpha (Bv)$ and $w=Bv$. If we can show that $Bv\neq0$ then $w$ is an eigenvector, $\alpha\neq0$ is an eigenvalue of $BA$, and $AB$ and $BA$ have the same eigenvalues when $\alpha\neq0$.

3. $\alpha\neq0$ and take $Bv=0$. If $Bv=0$, $ABv=0=\alpha v$. As $v\neq0$, $\alpha=0$, however this is a contradiction and therefore $Bv\neq0$ and $w$ is an eigenvector. As $w$ is an eigenvector, $\alpha\neq0$ is an eigenvalue of $BA$, and $AB$ and $BA$ have the same eigenvalues when $\alpha\neq0$.

Q.E.D.

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An alternative way to prove the statement is the use the identity \begin{align} \det(z I_n+AB) = \det(z I_n+BA) \end{align} where $A, B \in \mathcal{M}_{n\times n}(\mathbb{R})$.

To prove the identity, observe \begin{align} \begin{pmatrix} zI_n+AB & A\\ 0 & I_n \end{pmatrix} \begin{pmatrix} I_n & 0\\ -B & zI_n \end{pmatrix} = M_z := \begin{pmatrix} zI_n & zA\\ -B & zI_n \end{pmatrix} = \begin{pmatrix} zI_n & 0\\ -B & I_n \end{pmatrix} \begin{pmatrix} I_n & A\\ 0 & zI_n+BA \end{pmatrix} \end{align} then it follows \begin{align} \det(M_z) = z^n\det(zI_n+AB)=z^n\det(zI_n+BA). \end{align}

This approach will also give you information when $A, B$ are not square matrices.

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  • $\begingroup$ You are right. More generally, $\det(\lambda I_n+AB)=\lambda^{n-m}\det(\lambda I_m +BA)$ Therefore we deduce that $AB$ and $BA$ has the same nonzero eigenvalues. $\endgroup$ – Tamshin Dion Dec 31 '19 at 8:33
  • $\begingroup$ @TamshinDion Exactly $\endgroup$ – Jacky Chong Dec 31 '19 at 8:40

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