Denote by $\sigma$ the spectral radius.

Is it true that $\sigma(AB) =\sigma(BA) $?

Edit: I am interested in the general case, i.e. $A$ is $n \times k$ and $B$ is $k \times n$.


If $A$ and $B$ are square matrices of the same size, then the products $AB$ and $BA$ have the same eigenvalues.

As was noted in comments by @Algebraic Pavel, the result still holds for rectangular matrices $A$ and $B$ (if the products $AB$ and $BA$ make sense). The non-zero eigenvalues of these products are the same.

  • $\begingroup$ Unfortunately I have to deal with the rectangular case. Does the result carry over? $\endgroup$ – ziutek Dec 3 '14 at 19:59
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    $\begingroup$ If $A$ and $B$ are such that $AB$ and $BA$ make sense then the nonzero eigenvalues of $AB$ are same as the nonzero eigenvalue of $BA$ and vice versa. That is, $vu^T$ has a nonzero eigenvalue if and only if $u^Tv\neq 0$. Therefore, the result carries over to the rectangular matrices. $\endgroup$ – Algebraic Pavel Dec 4 '14 at 10:47
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    $\begingroup$ @AlgebraicPavel thank you for this catch, I somehow mixed the norm and the spectral radius. $\endgroup$ – TZakrevskiy Dec 4 '14 at 11:07

$\sigma$ denotes here the set of eigenvalues! A $R^{m\times n}$ matrix is a linear operator $R^n\to R^m$; similarly for $C^{m\times n}$, so the following answers to you.

Theorem. Let $X,Y$ be vector spaces (with the same scalar field $R$ or $C$). Let $A:X\to Y$ and $B:Y\to X$ be linear.

(a) Then $\sigma(AB)\setminus\{0\} = \sigma(BA)\setminus\{0\}$.

(b) If both $X$ and $Y$ are $n$-dimensional, $n<\infty$, then $\sigma(AB) = \sigma(BA)$.

Proof: (a) Let $0\ne t\in \sigma(AB)$. Then $AB y=ty$ for some $y\in Y\setminus\{0\}$. Right-multiply by $B$ to get $BA(By)=t(By)$. Thus, $t\in\sigma(BA)$ (because $By\ne0$, as else $0=AB y=ty$). Therefore, $\sigma(AB)\setminus\{0\} \subset \sigma(BA)\setminus\{0\}$. Exchange $A$ with $B$ to get $\sigma(AB)\setminus\{0\} \supset \sigma(BA)\setminus\{0\}$.

(b) Assume $\dim X=n=\dim Y$. If $ABy=0$ for some $y\ne 0$, then $A$ or $B$ is singular (as $\det(AB)=\det(A)\det(B)$); hence so is then $BA$. Thus, $0\in\sigma(AB) \Rightarrow 0\in\sigma(BA)$. Exchange $A,B$. QED.

Remarks. (a) By the above proof, (a) holds for bounded and even unbounded operators (as long as they are defined on the whole space, as assumed in the theorem) over infinite-dimensional vector spaces.

(b1) Claim (b) is not true for non-square matrices. For example, $0\in\sigma(AB)\setminus\sigma(BA)$ if $A^T=(1,0)=B$, as then $BA=1$, $AB=(1,0; 0,0)$.

(b2) Claim (b) is not true for $n=\infty$. For example, let L, R be the left and right shifts on $R^N$, the set of sequences $(x_1,x_2,\cdots)$. Then $LR=I$ but $RL(1,0,0,0,\ldots)=R0=0$, so $0\in\sigma(RL)\setminus\sigma(LR)$.

(c) In the infinite-dimensional case, usually $\sigma_p$ denotes the eigenvalues and $\sigma$ is a bigger set, but here I define $\sigma:=\sigma_p$.


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