common eigenvector of a diagonal element cross-section

common eigenvector of a diagonal element cross-section

Denote by

M

n

⁢

(

𝒦

)

the set of all

n

×

n

matrices over

𝒦

. Let

d

i

:

M

n

⁢

(

𝒦

)

⟶

𝒦

be
the

function

which extracts the

i

th

diagonal

element of a matrix,
and let

ε

i

:

𝒦

n

⟶

𝒦

be
the function which extracts the

i

th
of a vector. Finally denote by

[

n

]

the set

{

1

,

…

,

n

}

.

Theorem 1

.

Let

K

be a field.
For any

sequence

A

1

,

…

,

A

r

∈

M

n

⁢

(

K

)

of

upper
triangular

pairwise

commuting matrices

and every row index

i

∈

[

n

]

, there exists

u

∈

K

n

∖

{

0

}

such that

A

k

⁢

𝐮

=

d

i

⁢

(

A

k

)

⁢

𝐮

for all

k

∈

[

r

]

.

(1)

Proof.

Let

λ

k

=

d

i

⁢

(

A

k

)

for all

k

∈

[

r

]

, so that the
problem is to find a common

eigenvector

𝐮

of

A

1

,

…

,

A

r

whose corresponding

eigenvalue

for

A

k

is

λ

k

. It is

sufficient

to find such a common eigenvector in the case that

i

=

n

is the least

i

∈

[

n

]

for which

d

i

⁢

(

A

k

)

=

λ

k

for all

k

∈

[

r

]

, because if some smaller

i

also
has this property then one can solve the corresponding problem for
the

i

×

i

submatrices

consisting of rows and columns

1

through

i

of

A

1

,

…

,

A

r

, and then pad the common eigenvector
of these submatrices with zeros to get a common eigenvector of the
original

A

1

,

…

,

A

r

.

By the

existence

of a
characteristic matrix of a diagonal

element

cross-section (

http://planetmath.org/CharacteristicMatrixOfDiagonalElementCrossSection

)
there exists a matrix

C

in the unital

algebra

generated by

A

1

,

…

,

A

r

such that

d

i

⁢

(

C

)

=

1

if

d

i

⁢

(

A

k

)

=

λ

k

for all

k

∈

[

r

]

, and

d

i

⁢

(

C

)

=

0

otherwise; in other words that matrix

C

satisfies

d

n

⁢

(

C

)

=

1

and

d

i

⁢

(

C

)

=

0

for all

i

<

n

. Since it is also upper triangular it
follows that the matrix

I

-

C

has
rank (

http://planetmath.org/RankLinearMapping

)

n

-

1

, so the kernel of this
matrix is one-dimensional. Let

𝐮

=

(

u

1

,

…

,

u

n

)

∈

ker

⁡

(

I

-

C

)

be such that

u

n

=

1

; it is easy to see that
this is always possible (indeed, the only vector in this

nullspace

with

n

th

0

is the

zero vector

). This

𝐮

is the
wanted eigenvector.

To see that it is an eigenvector of

A

k

, one may first observe
that

C

commutes with this

A

k

, since the unital algebra of
matrices to which

C

belongs is

commutative

(

http://planetmath.org/Commutative

). This implies that

A

k

⁢

𝐮

∈

ker

⁡

(

I

-

C

)

since

𝟎

=

A

k

⁢

𝟎

=

A

k

⁢

(

I

-

C

)

⁢

𝐮

=

(

I

-

C

)

⁢

A

k

⁢

𝐮

. As

ker

⁡

(

I

-

C

)

is
one-dimensional it follows that

A

k

⁢

𝐮

=

λ

⁢

𝐮

for some

λ

∈

𝒦

. Since

A

k

is upper triangular
and

u

n

=

1

this

λ

must furthermore satisfy

λ

=

λ

⁢

u

n

=

ε

n

⁢

(

λ

⁢

𝐮

)

=

ε

n

⁢

(

A

k

⁢

𝐮

)

=

d

n

⁢

(

A

k

)

⁢

u

n

=

d

n

⁢

(

A

k

)

, which is
indeed what the eigenvalue was claimed to be.
∎

Title

common eigenvector of a diagonal element cross-section

Canonical name

CommonEigenvectorOfADiagonalElementCrosssection

Date of creation

2013-03-22 15:30:41

Last modified on

2013-03-22 15:30:41

Owner

lars_h (9802)

Last modified by

lars_h (9802)

Numerical id

4

Author

lars_h (9802)

Entry type

Theorem

Classification

msc 15A18

Related topic

CommutingMatrices

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