$AB$ and $BA$ are almost isospectral

0.1 General case

Let $A$ and $B$ be endomorphisms of a vector space $V$. Let $\sigma (AB)$ and $\sigma (BA)$ denote, respectively, the spectra (http://planetmath.org/spectrum) of $AB$ and $BA$.

The next result shows that $AB$ and $BA$ are “almost” isospectral, in the sense that their spectra is the same except possibly the value $0$.

Theorem - Let $A$ and $B$ be as above. We have

1. $\sigma (AB)\cup \{0\}=\sigma (BA)\cup \{0\}$, and moreover

2. $AB$ and $BA$ have the same eigenvalues, except possibly the zero eigenvalue.

Proof : Let $\lambda \ne 0$.

1. If $\lambda \in \sigma (AB)$ then ${\lambda }^{-1}AB-I$ is not invertible. By the result in the parent entry, this implies that ${\lambda }^{-1}BA-I$ is not invertible either, hence $\lambda \in \sigma (BA)$.

   A similar argument proves that every non-zero element of $\sigma (BA)$ also belongs to $\sigma (AB)$. Hence $\sigma (AB)\cup \{0\}=\sigma (BA)\cup \{0\}$.

2. If $\lambda$ is an eigenvalue of $AB$, then $I-{\lambda }^{-1}AB$ is not injective. By the result in the parent entry, this implies that $I-{\lambda }^{-1}BA$ is also not injective, hence $\lambda$ is an eigenvalue of $BA$.

   A similar argument proves that non-zero eigenvalues of $BA$ are also eigenvalues of $AB$. $\mathrm{\square }$

Remark : Note that for infinite dimensional vector spaces the spectrum of a linear mapping does not consist solely of its eigenvalues. Hence, 1 and 2 above are two different statements.

0.2 Finite dimensional case

When the vector space $V$ is finite dimensional we can strengthen the above result.

Theroem - $AB$ and $BA$ are isospectral, i.e. they have the same spectrum. Since $V$ is finite dimensional, this means that $AB$ and $BA$ have the same eigenvalues.

Proof : By the above result we only need to prove that: $AB$ is invertible if and only if $BA$ is invertible.

Suppose $AB$ is not invertible. Hence, $A$ is not invertible or $B$ is not invertible.

For finite dimensional vector spaces invertibility, injectivity and surjectivity are the same thing. Thus, the above statement can be rewritten as: $A$ is not injective or $B$ is not surjective.

Either way $BA$ is not invertible.

A similar argument shows that if $BA$ is not invertible, then $AB$ is also not invertible, which concludes the proof. $\mathrm{\square }$

0.3 Comments

The first theorem can be proven in a more general context : If $A$ and $B$ are elements of an arbitrary unital algebra, then

$\sigma (AB)\cup \{0\}=\sigma (BA)\cup \{0\}$.

This humble result plays an important role in the spectral theory of operator algebras.

Title	$AB$ and $BA$ are almost isospectral
Canonical name	ABAndBAAreAlmostIsospectral
Date of creation	2013-03-22 14:44:51
Last modified on	2013-03-22 14:44:51
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	14
Author	asteroid (17536)
Entry type	Corollary
Classification	msc 15A04
Classification	msc 47A10
Classification	msc 16B99