A⁢B and B⁢A are almost isospectral

A

⁢

B

and

B

⁢

A

are almost isospectral

0.1

General case

Let

A

and

B

be

endomorphisms

of a

vector space

V

. Let

σ

⁢

(

A

⁢

B

)

and

σ

⁢

(

B

⁢

A

)

denote, respectively, the spectra (

http://planetmath.org/

spectrum

) of

A

⁢

B

and

B

⁢

A

.

The next result shows that

A

⁢

B

and

B

⁢

A

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.

σ

⁢

(

A

⁢

B

)

∪

{

0

}

=

σ

⁢

(

B

⁢

A

)

∪

{

0

}

, and moreover

2.

A

⁢

B

and

B

⁢

A

have the same

eigenvalues

, except possibly the zero eigenvalue.

Proof :

Let

λ

≠

0

.

1.

If

λ

∈

σ

⁢

(

A

⁢

B

)

then

λ

-

1

⁢

A

⁢

B

-

I

is not

invertible

. By the result in the parent entry, this implies that

λ

-

1

⁢

B

⁢

A

-

I

is not invertible either, hence

λ

∈

σ

⁢

(

B

⁢

A

)

.

A

similar

argument

proves that every non-zero element of

σ

⁢

(

B

⁢

A

)

also belongs to

σ

⁢

(

A

⁢

B

)

. Hence

σ

⁢

(

A

⁢

B

)

∪

{

0

}

=

σ

⁢

(

B

⁢

A

)

∪

{

0

}

.

2.

If

λ

is an eigenvalue of

A

⁢

B

, then

I

-

λ

-

1

⁢

A

⁢

B

is not

injective

. By the result in the parent entry, this implies that

I

-

λ

-

1

⁢

B

⁢

A

is also not injective, hence

λ

is an eigenvalue of

B

⁢

A

.

A similar argument proves that non-zero eigenvalues of

B

⁢

A

are also eigenvalues of

A

⁢

B

.

□

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 -

A

⁢

B

and

B

⁢

A

are isospectral, i.e. they have the same spectrum. Since

V

is finite dimensional, this means that

A

⁢

B

and

B

⁢

A

have the same eigenvalues.

Proof :

By the above result we only need to prove that:

A

⁢

B

is invertible if and only if

B

⁢

A

is invertible.

Suppose

A

⁢

B

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

B

⁢

A

is not invertible.

A similar argument shows that if

B

⁢

A

is not invertible, then

A

⁢

B

is also not invertible, which concludes the proof.

□

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

σ

⁢

(

A

⁢

B

)

∪

{

0

}

=

σ

⁢

(

B

⁢

A

)

∪

{

0

}

.

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

Title

A

⁢

B

and

B

⁢

A

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

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