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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.

VN7
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    If $A$ and $B$ are not square matrices, the statement is not true: all you can say is that $AB$ and $BA$ have the same nonzero eigenvalues. For example, try $A =\pmatrix{1 & 0 & 0\cr 0 & 1 & 0\cr}$ and $B = \pmatrix{1 & 0\cr 0 & 1\cr 0 & 0\cr}$. – Robert Israel Dec 30 '19 at 18:28
  • What happens if $AB$ has indeed $0$ as an eigenvalue – marwalix Dec 30 '19 at 18:29
  • List of older posts within an $\epsilon$-neighbourhood:
    • https://math.stackexchange.com/questions/124888/are-the-eigenvalues-of-ab-equal-to-the-eigenvalues-of-ba-citation-needed
    • https://math.stackexchange.com/questions/402281/eigenvalues-and-eigenvectors-of-ab-and-ba-proof
    • https://math.stackexchange.com/questions/1173614/eigenvalues-of-ab-and-ba-qquad (itself marked as duplicate)
    • https://math.stackexchange.com/questions/1648096/a-variation-on-the-ab-vs-ba-nonzero-eigenvalues-question
     – Hanno Jan 02 '20 at 19:52
  • List continued
    • https://math.stackexchange.com/questions/2029376/do-ab-and-ba-have-the-same-eigenvalues (duplicate)
    • https://math.stackexchange.com/questions/2061035/eigenvalues-of-ab-and-ba-the-same
    • https://math.stackexchange.com/questions/2494974/prove-that-if-%ce%bb-is-an-eigenvalue-of-ab-then-%ce%bb-is-also-an-eigenvalue-of-ba
    • https://math.stackexchange.com/questions/2806072/does-ab-and-ba-have-the-same-number-of-each-non-zero-eigenvalue (duplicate)
    • https://math.stackexchange.com/questions/3313167/eigenvalues-of-ab-vs-eigenvalues-of-ba-incl-infinite-dimensional-case
     – Hanno Jan 02 '20 at 19:53
  • Yeah thanks Hanno, I wanted people to comment on my proof, which they very helpfully did; I've edited the proof so time to move on. – VN7 Jan 02 '20 at 19:55

1 Answers1

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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.

Jacky Chong
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