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

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 nonzero 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  20130322 14:44:51 
Last modified on  20130322 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 