Chapter 10
Jordan Canonical Form
10.1 Finding the Jordan Canonical Form with a Dot Diagram
*10.2 Jordan Canonical Form and Generalized Eigenvectors
10.3 Jordan Canonical Form and CAS
10.4 Exercises
If a matrix is diagonalizable, every thing is much easier. But most of matrices are not diagonalizable. The Jordan canonical form or Jordan normal form is an upper triangular matrix of a particular form called a Jordan matrix (a simple block diagonal matrix) representing an operator with respect to some basis. The diagonal entries of the normal form are the eigenvalues of the operator, with the number of times each one occurs given by its algebraic multiplicity.
Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. Since each matrix has a corresponding Jordan canonical form which is similar to it, all computations can be done with this simple upper triangular matrix. The Jordan normal form is named after Camille Jordan, a French mathematician renowned for his work in various branches of mathematics. In this chapter, we will study how to find a Jordan matrix which is similar to any given matrix and how to find generalized eigenvectors.
Book http://matrix.skku.ac.kr/2015-Album/Big-Book-LinearAlgebra-Eng-2015.pdf
Book http://matrix.skku.ac.kr/2015-Album/BigBook-LinearAlgebra-SGLee-New-2015.pdf
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Theorem10.1.2 |
|
|
The number of dots in the first the dimension of solution space of |
|
rows of the dot diagram for 
(i.e. the nullity of 
).
nullity
=nullity
.
|
Theorem10.1.3 |
|
|
For Then, the following are true.
(1) (2) If |
|
, let 
denote the number of dots in the 
th row of the dot diagram of 
. 
.
, 
.
By Theorem 10.1.2,
(provided 
)
Also, 
and
(The number of dots in each row,
, means the number of blocks of size at least 
). ■
From Theorem 10.1.3, let’s see how the dot diagram for each 
is completely determined by the matrix 
.
Find the Jordan Canonical Form of 
.
, 
, 
.
has characteristic polynomial 
, so 
has two distinct eigenvalues 
, 
. 
has algebraic multiplicity 1, and 
has algebraic multiplicity 3. Thus, the dot diagram for 
has 1 dot
has one 
Jordan block. That is, 
.
has 3 dots, and
,
.
is the following.
: 
: 
has one 
Jordan block and one 
Jordan block. That is,
is
.
has characteristic polynomial 
,
, 
and 
, each with algebraic multiplicity 2. For 
,
.
is the following.
: • (Number of Jordan blocks: 
)
: •
.
, 
. 
is 0(∵The number of dots is 
). Therefore, the dot diagram for 
is the following:
: • • (Number of Jordan blocks = 2)
is
. 
Reference video: 
Practice site: 
,
.
in the above equation.
.
. The Jordan normal form is obtained
, i.e. 
.
have column vectors 
, 
, then
.
.
, we have 
,
is an eigenvector of 
corresponding to the eigenvalue 
.
, multiplying both sides by 
gives 
.
, so 
. Thus, 
.
are called 
.
, its corresponding Jordan block gives rise to a 
corresponding to an eigenvalue 
is a sequence of non zero vectors, 
such that
, 
, 
.
) of the chain is a generalized eigenvector such that 
, where 
is the size of the Jordan block.
is an eigenvector corresponding to 
. In general, 
is the pre-image of 
under 
.
.
can be put in Jordan normal form is equivalent to the claim that
.
Reference video: 
Practice site: 
10 matrix, you need to find the roots of a characteristic polynomial of degree 10.
10 matrix requires us to calculate many exponents and coefficients.
be a 5×5 matrix with the only one eigenvalue 
with algebraic multiplicity of 5.
when the number of linearly independent eigenvectors corresponding 
is 2.
is the following.
=2 : ⦁⦁
=1 : ⦁
=1 : ⦁
=1 : ⦁
is the following.
=2 : ⦁⦁
=2 : ⦁⦁
=2 : ⦁
, calculate the following:
■
has its characteristic polynomial 
= 
.
has two distinct eigenvalues 
, 
.
=25 has algebraic multiplicity 1, 
=0 has algebraic multiplicity 4.
has 1 dot. 
=1 : ⦁ (number of Jordan block: 1)
has one Jordan block of 1
1. That is, 
=[25].
has 4 dot, and
= 5-rank(
-
) = 5-1 = 4
= rank(
) - rank[
] = 1-1 =0
= rank[
] - rank[
] = 1-1 =0
= rank[
] - rank[
] = 1-1 =0
is the following. 
=4 : ⦁⦁⦁⦁ (number of Jordan block: 4)
has 4 Jordan blocks of 1
1 . That is, 
=
.
is 
=
=
.
has characteristic polynomial 
= 
.
, 
, 
. Here 
=3 has algebraic multiplicity 1, 
=2 has algebraic multiplicity 1 and 
=1 has algebraic multiplicity 3.
has 1 dot.
=1 ; ⦁
has one 1
1 Jordan block. That is, 
=[3].
has 1 dot.
=1 : ⦁
has one 1
1 Jordan block. That is, 
=[2].
has 3 dot, and
= 5-rank(
-
) = 5-4 = 1
= rank(
) - rank[
] = 4-3 =1
= rank[
] - rank[
] = 3-2 =1
is the following.
=1 : ⦁ (number of Jordan block: 1)
=1 : ⦁
=1 : ⦁
has one 3
3 Jordan block. That is, 
=
.
is
=
=
has characteristic polynomial 
= 
.
, 
, 
.
=5 has algebraic multiplicity 1, 
=1 has algebraic multiplicity 1 and 
=-3 has algebraic multiplicity 1.
has 1 dot.
=1 ; ⦁ (number of Jordan block: 1)
has 
1 Jordan block. That is, 
=[5].
has 1 dot.
=1 : ⦁ (number of Jordan block: 1)
has 
1 Jordan block. That is, 
=[1].
has 1 dot, and.
=1 : ⦁ (number of Jordan block: 1)
has 
1 Jordan block. That is, 
=
.
is
=
=
has characteristic polynomial 
= 
.
, 
=2 and 
=0, each with 
For 
=2, 
= 4 - rank(
) = 4-2 = 2
= rank(
) - rank[
] = 2-2 = 0
is the following.
=1 : ⦁ ( Number of Jordan blocks : 1)
=1 : ⦁ 
So, 
=
=0
= 4 - rank
= 4 - 3 = 1
= rank
- rank[
] = 3 - 2 = 1
=1 : ⦁ ( Number of Jordan blocks : 1)
=1 : ⦁ 
=
. 
is
=
=
.
has characteristic polynomial 
= 
. So A has three distinct eigenvalues 
, 
, 
.
=4 has 
=2 has algebraic multiplicity 1 and 
=1 has algebraic multiplicity 3.
has 1 dot.
has 1
1 Jordan block. That is, 
=[4].
has 1 dot.
has 1
1 Jordan block. That is, 
=[2].
has 3 dot, and
= 5-rank(
-
) = 5-4 = 1
= rank(
) - rank[
] = 4-3 =1
= rank[
] - rank[
] = 3-2 =1
is the following. (number of Jordan block: 1)
=1 : ⦁
=1 : ⦁
=1 : ⦁
has ons 3
3 Jordan block. That is, 
=
. Hence, the Jordan Canonical form of 
is
=
=
has characteristic polynomial 
= 
. So A has two distinct eigenvalues of 
, 
=4 and 
=2, each with alegebraic multiplicity 2.
= 4,
= 4 - rank(
) = 4-3 = 1
= rank(
) - rank[
] = 3-2 = 1
is the following. ( Number of Jordan blocks : 1)
=1 : ⦁ 
=2 : ⦁
=
.
=2,
= 4 - rank(
) = 4-2 = 2
= rank(
) - rank[
] = 2-2 = 0
Therefore, the dot diagram for 
is the following. 
=2 : ⦁⦁ (Number of Jordan blocks = 2 )
=
.
is 
=
.
