I-AB is invertible if and only if I-BA is invertible

I-AB is invertible if and only if I-BA is invertible

In this entry

A

and

B

are

endomorphisms

of a

vector space

V

. If

V

is finite dimensional, we may choose a basis and regard

A

and

B

as

square matrices

of equal

dimension

.

Theorem

-

Let

A

and

B

be endomorphisms of a vector space

V

. We have that

1.

I

-

A

⁢

B

is

invertible

(

http://planetmath.org/LinearIsomorphism

) if and only if

I

-

B

⁢

A

is invertible, and moreover

2.

I

-

A

⁢

B

is

injective

if and only if

I

-

B

⁢

A

is injective.

Proof :

1.

Suppose that

I

-

A

⁢

B

is invertible. We shall prove that

B

⁢

(

I

-

A

⁢

B

)

-

1

⁢

A

+

I

is the

inverse

of

I

-

B

⁢

A

.
In fact

(

I

-

B

⁢

A

)

⁢

(

B

⁢

(

I

-

A

⁢

B

)

-

1

⁢

A

+

I

)

=

B

⁢

(

I

-

A

⁢

B

)

-

1

⁢

A

+

I

-

B

⁢

A

⁢

B

⁢

(

I

-

A

⁢

B

)

-

1

⁢

A

-

B

⁢

A

=

B

⁢

(

(

I

-

A

⁢

B

)

-

1

-

A

⁢

B

⁢

(

I

-

A

⁢

B

)

-

1

)

⁢

A

+

I

-

B

⁢

A

=

B

⁢

(

(

I

-

A

⁢

B

)

⁢

(

I

-

A

⁢

B

)

-

1

)

⁢

A

+

I

-

B

⁢

A

=

B

⁢

A

+

I

-

B

⁢

A

=

I

A

similar

computation shows that

(

B

⁢

(

I

-

A

⁢

B

)

-

1

⁢

A

+

I

)

⁢

(

I

-

B

⁢

A

)

=

I

, i.e.

I

-

B

⁢

A

is invertible.

Exchanging the roles of

A

and

B

we can prove the ”if” part. So

I

-

A

⁢

B

is invertible if and only if

I

-

B

⁢

A

is invertible.

2.

Let us first
recall that a

linear map

between vector spaces is
invertible if and only if its kernel

ker

is the

zero vector

(see this page (

http://planetmath.org/KernelOfALinearTransformation

)).

Suppose

I

-

A

⁢

B

is not injective, i.e. there exists

u

≠

0

such that

(

I

-

A

⁢

B

)

⁢

u

=

0

. Then

(

I

-

B

⁢

A

)

⁢

B

⁢

u

=

B

⁢

(

I

-

A

⁢

B

)

⁢

u

=

0

i.e.

B

⁢

u

∈

ker

⁡

(

I

-

B

⁢

A

)

. Notice that

B

⁢

u

≠

0

because

u

=

A

⁢

B

⁢

u

(by

definition

of

u

), so

I

-

B

⁢

A

is also not injective.

Similarly, if

I

-

B

⁢

A

is not injective then

I

-

A

⁢

B

is not injective.

□

Remark -

It is known that for finite dimensional vector spaces a linear endomorphism is invertible if and only if it is injective. This does not remain true for infinite dimensional spaces, hence 1 and 2 are two different statements.

0.1

Comments

The result stated in 1 can be proven in a more general

context

— If

A

and

B

are elements of a

ring with unity

, then

I

-

A

⁢

B

is invertible if and only if

I

-

B

⁢

A

is invertible. See the entry on

techniques in mathematical proofs

, in which this result is proven using several different techniques.

This entry is based on

http://planetmath.org/?op=getmsg&id=5088

this discussion on PM.

Title

I-AB is invertible if and only if I-BA is invertible

Canonical

name

IABIsInvertibleIfAndOnlyIfIBAIsInvertible

Date of creation

2013-03-22 14:44:43

Last modified on

2013-03-22 14:44:43

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

16

Author

asteroid (17536)

Entry type

Theorem

Classification

msc 16B99

Classification

msc 15A04

Classification

msc 47A10

Related topic

TechniquesInMathematicalProofs

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