simultaneous block-diagonalization of upper triangular commuting matrices

simultaneous block-diagonalization of upper triangular commuting matrices

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 1

.

Let

K

be a field, let

n

be a positive integer, and let

∼

be an

equivalence relation

on

[

n

]

such that if

i

∼

j

and

i

⩽

k

⩽

j

then

k

∼

i

. Let

A

1

,

…

,

A

r

∈

M

n

⁢

(

K

)

be pairwise commuting

upper triangular matrices

.
If these matrices and

∼

are related such that

i

∼

j

if and only if

d

i

⁢

(

A

k

)

=

d

j

⁢

(

A

k

)

⁢

for all

⁢

k

∈

[

r

]

⁢

,

then there exists a matrix

B

∈

GL

n

⁢

(

K

)

such that:

1.

If

𝐞

i

T

⁢

B

-

1

⁢

A

k

⁢

B

⁢

𝐞

j

≠

0

then

i

∼

j

and

i

⩽

j

.

2.

If

i

∼

j

then

𝐞

i

T

⁢

B

-

1

⁢

A

k

⁢

B

⁢

𝐞

j

=

𝐞

i

T

⁢

A

k

⁢

𝐞

j

.

Condition

1

says that if an element of

B

-

1

⁢

A

k

⁢

B

is nonzero then both its row and column indices must belong to the
same

equivalence class

of

∼

, i.e., the nonzero elements of

B

-

1

⁢

A

k

⁢

B

only

occur in

particular
blocks (

http://planetmath.org/PartitionedMatrix

)
along the diagonal, and these blocks
correspond to equivalence classes of

∼

.
Condition

2

says that within one of these blocks,

B

-

1

⁢

A

k

⁢

B

is equal to

A

k

.

The proof of the

theorem

requires the following lemma.

Lemma 2

.

Let a

sequence

A

1

,

…

,

A

r

∈

M

n

⁢

(

K

)

of upper
triangular matrices be given, and denote by

A

the
unital (

http://planetmath.org/unity

)

algebra

(

http://planetmath.org/Algebra

)
generated by these matrices. For every sequence

λ

1

,

…

,

λ

r

∈

K

of scalars there exists a
matrix

C

∈

A

such that

d

i

⁢

(

C

)

=

{

1

if

d

i

⁢

(

A

k

)

=

λ

k

for all

k

∈

[

r

]

,

0

𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

for all

i

∈

[

n

]

.

The proof of that lemma can be found in
this article (

http://planetmath.org/CharacteristicMatrixOfDiagonalElementCrossSection

).

Proof of theorem.

The proof is by

induction

on the number of equivalence classes of

∼

. If there is only one equivalence class then one can take

B

=

I

.

If there is more than one equivalence class, then let

S

be the
equivalence class that contains

n

. By Lemma

2

there exists a matrix

C

in the unital algebra generated by

A

1

,

…

,

A

r

(hence necessarily upper triangular) such that

d

i

⁢

(

C

)

=

1

for all

i

∈

S

and

d

i

⁢

(

C

)

=

0

for all

i

∈

[

n

]

∖

S

. Thus

C

has a

C

=

(

C

11

C

12

0

C

22

)

where

C

22

is a

|

S

|

×

|

S

|

matrix
that has all

1

s on the diagonal, and

C

11

is a

(

n

-

|

S

|

)

×

(

n

-

|

S

|

)

matrix that has all

0

s
on the diagonal.

Let

k

∈

[

r

]

be arbitrary and similarly decompose

C

n

=

(

D

11

D

12

0

D

22

)

⁢

,

A

k

=

(

A

11

A

12

0

A

22

)

⁢

.

One can identify

D

11

=

(

C

11

)

n

and

D

22

=

(

C

22

)

n

,
but due to the zero diagonal of

C

11

and the fact that the

of these matrices are smaller than

n

, the more striking

equality

D

11

=

0

also holds. As for

D

22

, one may conclude that it
is

invertible

.

Since the algebra that

C

n

belongs to was generated by pairwise
commuting

elements

, it is a

commutative

(

http://planetmath.org/Commutative

)
algebra, and in particular

C

n

⁢

A

k

=

A

k

⁢

C

n

. In

of the

individual

blocks, this

becomes

(

0

D

12

⁢

A

22

0

D

22

⁢

A

22

)

=

(

0

A

11

⁢

D

12

+

A

12

⁢

D

22

0

A

22

⁢

D

22

)

⁢

.

Now let

D

=

(

I

D

12

0

D

22

)

⁢

, so that

⁢

D

-

1

=

(

I

-

D

12

⁢

D

22

-

1

0

D

22

-

1

)

and consider the matrix

D

-

1

⁢

A

k

⁢

D

. Clearly

D

-

1

⁢

A

k

⁢

D

=

D

-

1

⁢

(

A

11

A

12

0

A

22

)

⁢

(

I

D

12

0

D

22

)

=

=

D

-

1

⁢

(

A

11

A

11

⁢

D

12

+

A

12

⁢

D

22

0

A

22

⁢

D

22

)

=

(

I

-

D

12

⁢

D

22

-

1

0

D

22

-

1

)

⁢

(

A

11

D

12

⁢

A

22

0

D

22

⁢

A

22

)

=

=

(

A

11

D

12

⁢

A

22

-

D

12

⁢

D

22

-

1

⁢

D

22

⁢

A

22

0

D

22

-

1

⁢

D

22

⁢

A

22

)

=

(

A

11

0

0

A

22

)

so that the positions with row not in

S

and column in

S

are all
zero, as requested for

B

-

1

⁢

A

k

⁢

B

. It should be observed that the
choice of

D

is

independent

of

k

, and that the same

D

thus
works for all the

A

k

.

In order to

complete

the proof, one applies the

induction hypothesis

to the

restriction

of

∼

to

[

n

]

∖

S

and the
corresponding

submatrices

of

D

-

1

⁢

A

k

⁢

D

, which

satisfy

the same
conditions but have one equivalence class less. This produces a
block-diagonalising matrix

B

′

for these submatrices, and thus the
sought

B

can be constructed as

D

⁢

(

B

′

0

0

I

)

.
∎

Title

simultaneous block-diagonalization of upper triangular commuting matrices

Canonical name

SimultaneousBlockdiagonalizationOfUpperTriangularCommutingMatrices

Date of creation

2013-03-22 15:29:42

Last modified on

2013-03-22 15:29:42

Owner

lars_h (9802)

Last modified by

lars_h (9802)

Numerical id

5

Author

lars_h (9802)

Entry type

Theorem

Classification

msc 15A21

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