# Theorem: $AB$ and $BA$ have the same non-zero eigenvalues [duplicate]

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.

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.
• 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. – Tamshin Dion Dec 31 '19 at 8:33