simultaneous triangularisation of commuting matrices over any field

simultaneous triangularisation of commuting matrices over any field

Let

𝐞

i

denote the (column) vector whose

i

th position is

1

and where all other positions are

0

. Denote by

[

n

]

the set

{

1

,

…

,

n

}

. Denote by

M

n

⁢

(

𝒦

)

the set of all

n

×

n

matrices over

𝒦

, and by

GL

n

⁢

(

𝒦

)

the set of all

invertible elements

of

M

n

⁢

(

𝒦

)

. Let

d

i

be the

function

which extracts the

i

th

diagonal

element of a matrix, i.e.,

d

i

⁢

(

A

)

=

𝐞

i

T

⁢

A

⁢

𝐞

i

.

Theorem

.

Let

K

be a field, let

A

1

,

…

,

A

r

∈

M

n

⁢

(

K

)

be pairwise

commuting matrices

, and let

L

be a

field extension

of

K

in which the

characteristic polynomials

of all

A

k

split (

http://planetmath.org/SplittingField

). Then there exists some

P

∈

GL

n

⁢

(

L

)

such that

1.

P

-

1

⁢

A

k

⁢

P

is

upper triangular

for all

k

=

1

,

…

,

r

,

and

2.

if

i

,

j

,

l

∈

[

n

]

are such that

i

⩽

l

⩽

j

and

d

i

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

d

j

⁢

(

P

-

1

⁢

A

k

⁢

P

)

for all

k

=

1

,

…

,

r

, then

d

l

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

d

j

⁢

(

P

-

1

⁢

A

k

⁢

P

)

for all

k

=

1

,

…

,

r

as well.

The proof relies on two lemmas.

Lemma 1

.

Let

K

be a field, let

A

1

,

…

,

A

r

∈

M

n

⁢

(

K

)

be pairwise commuting matrices, and let

L

be a field extension
of

K

in which the characteristic polynomials of all

A

k

split. Then there exists some nonzero

u

∈

L

n

which
is an

eigenvector

of

A

k

for all

k

=

1

,

…

,

r

.

Lemma 2

.

For any

sequence

R

1

,

…

,

R

r

∈

M

n

⁢

(

L

)

of upper
triangular pairwise commuting matrices and every row index

i

∈

[

n

]

, there exists

v

∈

L

n

∖

{

0

}

such that

R

k

⁢

𝐯

=

d

i

⁢

(

R

k

)

⁢

𝐯

for all

k

∈

[

r

]

.

Proof.

This is by

induction

on

n

. The

induction hypothesis

is that given
pairwise commuting matrices

A

1

,

…

,

A

r

∈

M

n

⁢

(

ℒ

)

,
whose characteristic polynomials all split in

ℒ

, and a
sequence of arbitrary scalars

μ

1

,

…

,

μ

r

∈

ℒ

,
there exists some

P

∈

GL

n

⁢

(

ℒ

)

such that:

1.

P

-

1

⁢

A

k

⁢

P

is upper triangular for all

k

=

1

,

…

,

r

.

2.

If some

i

,

j

∈

[

n

]

are such that

i

<

j

and

d

j

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

d

i

⁢

(

P

-

1

⁢

A

k

⁢

P

)

for all

k

∈

[

r

]

,
then

d

i

+

1

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

d

i

⁢

(

P

-

1

⁢

A

k

⁢

P

)

.

3.

If some

j

∈

[

n

]

is such that

d

j

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

μ

k

for all

k

∈

[

r

]

, then

d

1

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

μ

k

for all

k

∈

[

r

]

.

For

n

=

1

this

hypothesis

is trivially fulfilled (all

1

×

1

matrices are upper triangular). Assume that it holds for

n

=

m

and
consider the case

n

=

m

+

1

.

It is easy to see that condition

1

implies
that

P

⁢

𝐞

1

must be an eigenvector that is common to all the
matrices. If there exists a nonzero vector

𝐮

1

∈

ℒ

n

such that

A

k

⁢

𝐮

1

=

μ

k

⁢

𝐮

1

for all

k

=

1

,

…

,

r

then this is such a common eigenvector, and in that case let

λ

k

=

μ

k

for all

k

=

1

,

…

,

r

. Otherwise there by
Lemma

1

exists a vector

𝐮

1

∈

ℒ

n

∖

{

𝟎

}

such that

A

k

⁢

𝐮

1

=

λ

k

⁢

𝐮

1

for some

{

λ

k

}

k

=

1

r

⊆

ℒ

. Either way, one gets a
suitable candidate

𝐮

1

for

P

⁢

𝐞

1

and

eigenvalues

λ

1

,

…

,

λ

r

that incidentally will

satisfy

d

1

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

λ

k

for all

k

∈

[

r

]

.

Let

𝐮

2

,

…

,

𝐮

n

∈

ℒ

n

be arbitrary
vectors such that

{

𝐮

i

}

i

=

1

n

is a basis of

ℒ

n

. Let

U

be the

n

×

n

matrix whose

i

th column
is

𝐮

i

for

1

⩽

i

⩽

n

.

1

1

By
imposing extra conditions on the choice of the basis

{

𝐮

i

}

i

=

1

n

(such as for example requesting that
it is orthonormal) at this point, one can often prove a stronger
claim where the choice of

P

is restricted to some smaller
group of matrices (for example the group of

orthogonal
matrices

), but this requires assuming additional things about
the fields

𝒦

and

ℒ

.

Then

U

is

invertible

and for each

k

the first column of

B

k

=

U

-

1

⁢

A

k

⁢

U

is

U

-

1

⁢

A

k

⁢

U

⁢

𝐞

1

=

U

-

1

⁢

A

k

⁢

𝐮

1

=

λ

k

⁢

U

-

1

⁢

𝐮

1

=

λ

k

⁢

𝐞

1

⁢

.

Furthermore

B

j

⁢

B

k

=

U

-

1

⁢

A

j

⁢

U

⁢

U

-

1

⁢

A

k

⁢

U

=

U

-

1

⁢

A

j

⁢

A

k

⁢

U

=

=

U

-

1

⁢

A

k

⁢

A

j

⁢

U

=

U

-

1

⁢

A

k

⁢

U

⁢

U

-

1

⁢

A

j

⁢

U

=

B

k

⁢

B

j

for all

j

and

k

.

Now let

A

k

′

be the matrix formed from rows and columns

2

though

n

of

B

k

. Since

det

⁡

(

A

k

-

x

⁢

I

)

=

det

⁡

(

B

k

-

x

⁢

I

)

=

(

λ

k

-

x

)

⁢

det

⁡

(

A

k

′

-

x

⁢

I

)

by
expansion (

http://planetmath.org/LaplaceExpansion

) along the first column,
it follows that the characteristic polynomial of

A

k

′

splits in

ℒ

. Furthermore all the

A

k

′

have side

m

=

n

-

1

and
commute pairwise with each other, whence by the induction hypothesis
there exists some

P

′

∈

GL

n

-

1

⁢

(

ℒ

)

such that every

P

′

⁣

-

1

⁢

A

k

′

⁢

P

′

is upper triangular. Let

P

=

U

⁢

(

1

0

0

P

′

)

. Then the

submatrix

consisting of rows and columns

2

through

n

of

P

-

1

⁢

A

k

⁢

P

is equal to

P

′

⁣

-

1

⁢

A

k

′

⁢

P

′

and hence
contains no nonzero

subdiagonal

elements

. Furthermore the first
column of

P

-

1

⁢

A

k

⁢

P

is equal to the first column of

B

k

and
thus the

P

-

1

⁢

A

k

⁢

P

are all upper triangular, as claimed.

It also follows from the induction hypothesis that

P

can be chosen
such that

d

2

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

d

1

⁢

(

P

′

⁣

-

1

⁢

A

k

′

⁢

P

′

)

=

λ

k

=

d

1

⁢

(

P

-

1

⁢

A

k

⁢

P

)

for all

k

∈

[

r

]

if there is any

j

⩾

2

for which

d

j

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

d

j

-

1

⁢

(

P

′

⁣

-

1

⁢

A

k

′

⁢

P

′

)

=

λ

k

=

d

1

⁢

(

P

-

1

⁢

A

k

⁢

P

)

for all

k

∈

[

r

]

and
more generally if

2

⩽

i

<

j

are such that

d

j

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

d

i

⁢

(

P

-

1

⁢

A

k

⁢

P

)

for all

k

∈

[

r

]

then similarly

d

i

+

1

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

d

i

⁢

(

P

-

1

⁢

A

k

⁢

P

)

for all

k

∈

[

r

]

. This has verified
condition

2

of the induction hypothesis.
For the remaining condition

3

, one may first observe
that if there is some

i

∈

[

n

]

such that

d

i

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

μ

k

for all

k

∈

[

r

]

then by Lemma

2

there exists a nonzero

𝐯

∈

ℒ

n

such that

P

-

1

⁢

A

k

⁢

P

⁢

𝐯

=

μ

k

⁢

𝐯

for all

k

∈

[

r

]

. This
means

P

⁢

𝐯

will fulfill the condition for choice of

𝐮

1

, and hence

d

1

⁢

(

P

-

1

⁢

A

k

⁢

P

)

=

λ

k

=

μ

k

as
claimed.

The

theorem

now follows from the principle of induction.
∎

Title

simultaneous triangularisation of commuting matrices over any field

Canonical

name

SimultaneousTriangularisationOfCommutingMatricesOverAnyField

Date of creation

2013-03-22 15:29:38

Last modified on

2013-03-22 15:29:38

Owner

lars_h (9802)

Last modified by

lars_h (9802)

Numerical id

4

Author

lars_h (9802)

Entry type

Theorem

Classification

msc 15A21

Related topic

CommutingMatrices

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