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[parent] $AB$ and $BA$ are almost isospectral (Corollary)

General case

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

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. $ \square$

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.



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Cross-references: algebras, operator, theory, algebra, unital, surjective, finite dimensional, linear mapping, spectrum, infinite dimensional, injective, implies, invertible, eigenvalues, isospectral, vector space, endomorphisms
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This is version 11 of $AB$ and $BA$ are almost isospectral, born on 2004-10-17, modified 2007-10-24.
Object id is 6386, canonical name is ABAndBAAreAlmostIsospectral.
Accessed 2248 times total.

Classification:
AMS MSC: 15A04 (Linear and multilinear algebra; matrix theory :: Linear transformations, semilinear transformations)
 16B99 (Associative rings and algebras :: General and miscellaneous :: Miscellaneous)
 47A10 (Operator theory :: General theory of linear operators :: Spectrum, resolvent)
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