Introduction
Preliminaries
Invariant Subspaces
Eigenspaces
The Jordan Normal Form and its Applications
Jeremy West
IMPACT
Brigham Young University
West
Jordan Normal Form
JNF
Introduction
Preliminaries
Invariant Subspaces
Eigenspaces
A square matrix A is a linear operator on {R, C}n .
A is diagonalizable if and only if it has n linearly independent
eigenvectors.
What happens if A does not have n linearly independent
eigenvectors?
When does this happen?
What general form can we obtain in this case?
West
Jordan Normal Form
JNF
Introduction
Preliminaries
Invariant Subspaces
Eigenspaces
The Jordan Normal Form
The Jordan Normal Form is one decomposition of a matrix,
A = P −1 JP
where J is the normal form. It has the advantage of corresponding
to the eigenspaces and of being as“close” to diagonal as possible.
More specifically, if a matrix is diagonal then its Jordan Normal
Form is the diagonalization.
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Jordan Normal Form
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Introduction
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Invariant Subspaces
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Complementary Subspaces
Definition
Two subspaces U, W of a vector space V are complementary if
U ∩ W = {0} and for all v ∈ V , there exist u ∈ U, w ∈ W such
that v = u + w . In fact, u, w are the unique vectors that satisfy
this property. We denote this V = U ⊕ W .
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Complementary Subspaces
Remark
This idea extends to finite collections: V = W1 ⊕ W2 ⊕ · · · ⊕ Wm .
Remark
If U, W are subspaces of V with dim U + dim W = dim V and
U ∩ W = {0} then it can be shown that U ⊕ W = V .
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Jordan Normal Form
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Introduction
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The Index of a Matrix
Recall
N(A) ≤ N(A2 ) ≤ N(A3 ) ≤ . . .
and
R(A) ≥ R(A2 ) ≥ R(A3 ) ≥ . . .
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Introduction
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The Index of a Matrix
Definition
The index of a matrix is the smallest nonnegative integer
k = Ind(A) such that
N(Ak ) = N(Ak+1 ) = . . .
R(Ak ) = R(Ak+1 ) = . . .
where A0 = I . Note that Ind(A) = 0 if A is invertible.
West
Jordan Normal Form
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Introduction
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The Index of a Matrix
Theorem
Let A be a square matrix and let k = Ind(A). Then
V = N(Ak ) ⊕ R(Ak ).
West
Jordan Normal Form
Eigenspaces
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Introduction
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The Index of a Matrix
Proof.
Suppose x ∈ N(Ak ) ∩ R(Ak ). Then Ak x = 0 and there exists y
such that x = Ak y . Therefore, Ak Ak y = A2k y = 0 so that
y ∈ N(A2k ). But N(A2k ) = N(Ak ) so that x = Ak y = 0. The
rank-nullity theorem implies that
dim N(Ak ) + dim R(Ak ) = n = dim(V )
so V = N(Ak ) ⊕ R(Ak ).
West
Jordan Normal Form
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Introduction
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Invariant Subspaces
Eigenspaces
Invariant Subspaces
Definition
A subspace W ≤ V is said to be invariant (with respect to a
matrix A) if AW ≤ W .
West
Jordan Normal Form
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Introduction
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Invariant Subspaces
Example
Notice that for any matrix A, the range R(A) is invariant since for
x ∈ R(A), Ax ∈ R(A) by definition. It follows that R(Ak ) is
invariant for any k. Also, N(A) is invariant since Ax = 0 ∈ N(A).
So is N(Ak ). Another example is an eigenspae N(A − λI ) because
any vector satisfies Ax = λx ∈ N(A − λI ).
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Jordan Normal Form
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Introduction
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Decomposing a matrix
If V = U ⊕ W and U and W are A-invariant subspaces then there
exists an invertible matrix P such that
0
−1 AU
A=P
P.
0 AW
In fact, P = [p1 , . . . , pr , pr +1 , . . . pn ] where {p1 , . . . , pr } is a basis
for U and {pr +1 , . . . , pn } is a basis for W . Furthermore, AU = A|U
is the restriction of A to the subspace U.
West
Jordan Normal Form
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Introduction
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Matrix Diagonalization
When we diagonalize A, we are simply using complementary
invariant spaces. These are the eigenspaces:
V = N(A − λ1 I ) ⊕ N(A − λ2 I ) ⊕ · · · ⊕ N(A − λr I ).
The matrix that diagonalizes A is P containing bases for the
eigenspaces (the columns are eigenvectors) and the blocks Aλi are
diagonal because on the space N(A − λI ), the action of A is simply
that of λI .
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Introduction
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Matrix Diagonalization
How do we know that
V =
n
M
N(A − λi I )?
i=1
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Jordan Normal Form
Eigenspaces
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Introduction
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Matrix Diagonalization
Example
Consider the matrix
2 1
A=
0 2
Since A is upper diagonal, its only eigenvalue is 2. What are the
eigenvectors?
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Matrix Diagonlization
Clearly
V 6= N(A − 2I )
because the dimensions do not match. This matrix cannot be
diagonalized because it doesn’t have a full set of linearly
independent eigenvectors.
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Generalized Eigenspaces
Notice that we had repeated eigenvalues. Remember that if we
have n distinct eigenvalues we know there are n linearly
independent eigenvectors. This problem only occurs when we have
repeated eigenvalues.
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Introduction
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Generalized Eigenspaces
What if we could make N(A − 2I ) “bigger” so that it covered all
of V ? Let’s try N(A − 2I )2 for example. It is easy to show that
V = N(A − 2I )2 . Notice that
2 1
2 0
0
1
−
=
0 2
0 2
1
0
So we found a vector not in N(A − 2I ) such that
(A − 2I )x ∈ N(A − 2I ). This is called an generalized eigenvector of
second order.
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Introduction
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Generalized Eigenspaces
This can be repeated. In fact, we can show that if λ1 , . . . , λr are
the distinct eigenvalues of A and ki = Ind(A − λi I ) then
V = N(A − λ1 I )k1 ⊕ · · · ⊕ N(A − λr I )kr .
N(A − λi I )ki is called the generalized eigenspace of A
corresponding to λi .
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Diagonalization Revisited
If we can’t diagonalize a matrix, how do we choose a basis that
gets us “close”? Remember that if x is a generalized eigenvector of
order k then (A − λI )x is a generalized eigenvector of order k − 1.
Repeating we may obtain a sequence x1 , x2 , . . . , xk such that
0 = (A − λI )x1
x1 = (A − λI )x2
..
.
xk−1 = (A − λI )xk
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Introduction
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Diagonalization Revisited
What is the action of A on the space spanned by {x1 , . . . , xk }?
Well, we know what A − λI looks like relative to this basis:


0 1 0 ... 0
0 0 1 . . . 0




..
A − λI = 

.


0 0 0 . . . 1
0 0 0 ... 0
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Diagonlization Revisited
So then A must be

λ
0


A=

0
0
1 0 ...
λ 1 ...
..
.
0 0 ...
0 0 ...
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
0
0




1
λ
Jordan Normal Form
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Diagonalization Revisited
Of course, {x1 , . . . , xk } may not span all of N(A − λI )k . So, to get
a basis for N(A − λI )k we follows this same idea. Take a basis
{x1 , . . . xd1 }
for N(A − λI ). Extend this to a basis
{x1 , . . . , xd1 , xd1 +1 , . . . , xd2 }
for N(A − λI )2 so that
(A − λI ){xd1 +1 , . . . , xd2 } = {x1 , . . . , xd2 −d1 }.
Then the portion of P corresponding to N(A − λI ) is be
[x1 , xd1 +1 , xd2 +1 , . . . , x2 , xd1 +2 , . . . , xd1 ].
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Introduction
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Jordan Normal Form
If we choose our basis this way, we can decompose A into the
following form:


J(λ1 )
0
...
0
 0
J(λ2 ) . . .
0 


A=
.
..


.
0
0
. . . J(λr )
The block J(λi ) is called a Jordan segment for λi .
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Jordan Normal Form
A Jordan segment is a matrix of the form

J1 (λi )
0
...
 0
J
(λ
)
.
..
2 i

J(λi ) = 
..

.
0
0
...
0
0
Jri (λi )
Each Jl (λi ) is called a Jordan block for λi .
West
Jordan Normal Form



.

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Jordan Normal Form
A Jordan block for λi is a matrix

λi 1
 0 λi


Jl (λi ) = 

0 0
0 0
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0 ...
1 ...
..
.
0 ...
0 ...

0
0




1
λi
Jordan Normal Form
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