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

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 $k \times n$.

• – Ivan Di Liberti Dec 3 '14 at 19:54
• @Ivan. Unfortunately the link you gave isn't really applicable, since I am not trying to bound the spectral radii of $A$ and $B$ separately. – ziutek Dec 3 '14 at 20:03
• See this, this or this. – user1551 Dec 3 '14 at 21:13

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.

• Unfortunately I have to deal with the rectangular case. Does the result carry over? – ziutek Dec 3 '14 at 19:59
• 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. – Algebraic Pavel Dec 4 '14 at 10:47
• @AlgebraicPavel thank you for this catch, I somehow mixed the norm and the spectral radius. – 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 $$\sigma:=\sigma_p$$.