Lecture 11: Proving the Jordan Decomposition Theorem
11 The Jordan Decomposition
11.1 Review
Recall this theorem from last time.
Theorem 11.1
Considering a transformation T : V → V, there must exist a basis #»v 1 , · · · , #»
v n such that the matrix of T
(in this basis) is
Ja1 (λ1 ) 0 ··· 0
..
.
0 Ja2 (λ2 ) 0
A= .
,
.. ..
.
Jai (λi ) 0
0 0 0 Jan (λn )
where Jai (λi ) are the Jordan blocks.
A special case is when all the ai = 1. Then,
λ1 ··· 0
.. .. ..
A= . . .
0 ··· λr
is a diagonal matrix.35
11.2 The Jordan Decomposition, Continued
The characteristic polynomial of the matrix A will be
pA (t) = (t − λ1 )a1 · · · (t − λr )ar ,
where it is possible to have repeated λi . As a result, it is not possible to determine the Jordan decomposition
simply from the characteristic polynomial, since there are diferent ways to take a repeated root and split it up
into Jordan blocks. (If all the roots of the characteristic polynomial are distinct, the Jordan form is uniquely
determined.)
However, the characteristic polynomial does provide some information. For a fxed eigenvalue λ,
X
ai = exponent of (t − λ) in pA (t).
Jai (λ)
Example 11.2 (n = 4)
For example, when n = 4, consider a matrix where pA (t) = t4 . There are multiple possible Jordan forms;
in particular, it can be split up as 4, 3 + 1, 2 + 2, 2 + 1 + 1, or 1 + 1 + 1 + 1 :
0 1 0 0 0 1 0 0 1 0 1 0
0 0 1 0 0 0 1 0 0 0 0 0
0 0 0 1 , 0 0 0
, , , .
0 1 0 0
0 0 0 0 1 0 0 0 0
For a given Jordan block, there is one eigenvector. Fixing λ again, this tells us that
dim(ker(λI − A))
is equal to the number of blocks with λ along the diagonal.
Up to reordering of the basis vectors, the Jordan decomposition is unique.
35 In the textbook, Artin puts the 1s below the diagonal in a Jordan block. Conventionally, the 1s are above the diagonal, but it
doesn’t make a diference, because reversing the order of the vectors #»
e 1 , · · · #»
e a to #»
e a , · · · #»
e 1 moves the 1s from above the diagonal
to below the diagonal. The diference is notational.
49
, Lecture 11: Proving the Jordan Decomposition Theorem
Example 11.3
0 1 0 0
0 0 1 0
Take J4 (0) =
0 0
. Under J4 (0), each basis vector maps to the next basis vector, and there is
0 1
0 0 0 0
one chain of length 4:
⃗e4 7→ ⃗e3 7→ ⃗e2 7→ ⃗e1 7→ ⃗0.
As a result, applying J4 (0) multiple times will eventually send all vectors to zero; that is, in this case,
J4 (0)4 = 0.
0 1 0 0
0 0 0 0
On the other hand, consider J2,2 (0) =
= J2 (0) 0
.
0 0 0 1 0 J2 (0)
0 0 0 0
Applying the operator to the basis vectors yields two chains of length 2:
⃗e2 7→ ⃗e1 7→ ⃗0
⃗e4 7→ ⃗e3 7→ ⃗0
In this case as well, the operator will map every vector to zero upon repeated application.
In general, for λ ̸= 0, (λI − T ) #»
e i is not necessarily zero (it is zero only if #e»i is an eigenvector), but for some
large enough n,
(λI − T )n #e»i = 0.36
In Example 11.3, there was a chain of length 4 for the frst matrix, while in the second matrix, we had two
chains of length 2.
Note 11.4
The Jordan decomposition theorem is powerful because any square matrix has a Jordan decomposition.
On the other hand, most matrices are diagonalizable, and any matrix will be ε away from a diagonalizable
matrix, and the Jordan decomposition is unnecessary. Only in the zero percent of the timea when the
characteristic polynomial has repeated roots is it necessary.
a This concept is feshed out in measure theory.
11.3 Proof of Jordan Decomposition Theorem
The proof of the Jordan decomposition theorem is quite involved and relatively tricky, so the important part for
the rest of class is understanding the style of proof, rather than the exact details. This proof will break down
the theorem inductively into smaller and smaller pieces.
Let’s start with a couple of defnitions that will help us with the proof.
Defnition 11.5
Given a vector space V and a linear transformation T : V → V, a subspace W ⊆ V is called T -invariant if
⃗ ) ∈ W for all w
T (w ⃗ ∈ W.
For example, if the vector space V is the space of polynomials of degree at most 3, and the subspace W is the
space of polynomials of degree at most 2, W will be T −invariant under the linear operator T that is taking the
derivative.
36 A vector that is killed not necessarily immediately but eventually by λI − T is known as a generalized eigenvector; there is
a question about them on the problem set.
50
11 The Jordan Decomposition
11.1 Review
Recall this theorem from last time.
Theorem 11.1
Considering a transformation T : V → V, there must exist a basis #»v 1 , · · · , #»
v n such that the matrix of T
(in this basis) is
Ja1 (λ1 ) 0 ··· 0
..
.
0 Ja2 (λ2 ) 0
A= .
,
.. ..
.
Jai (λi ) 0
0 0 0 Jan (λn )
where Jai (λi ) are the Jordan blocks.
A special case is when all the ai = 1. Then,
λ1 ··· 0
.. .. ..
A= . . .
0 ··· λr
is a diagonal matrix.35
11.2 The Jordan Decomposition, Continued
The characteristic polynomial of the matrix A will be
pA (t) = (t − λ1 )a1 · · · (t − λr )ar ,
where it is possible to have repeated λi . As a result, it is not possible to determine the Jordan decomposition
simply from the characteristic polynomial, since there are diferent ways to take a repeated root and split it up
into Jordan blocks. (If all the roots of the characteristic polynomial are distinct, the Jordan form is uniquely
determined.)
However, the characteristic polynomial does provide some information. For a fxed eigenvalue λ,
X
ai = exponent of (t − λ) in pA (t).
Jai (λ)
Example 11.2 (n = 4)
For example, when n = 4, consider a matrix where pA (t) = t4 . There are multiple possible Jordan forms;
in particular, it can be split up as 4, 3 + 1, 2 + 2, 2 + 1 + 1, or 1 + 1 + 1 + 1 :
0 1 0 0 0 1 0 0 1 0 1 0
0 0 1 0 0 0 1 0 0 0 0 0
0 0 0 1 , 0 0 0
, , , .
0 1 0 0
0 0 0 0 1 0 0 0 0
For a given Jordan block, there is one eigenvector. Fixing λ again, this tells us that
dim(ker(λI − A))
is equal to the number of blocks with λ along the diagonal.
Up to reordering of the basis vectors, the Jordan decomposition is unique.
35 In the textbook, Artin puts the 1s below the diagonal in a Jordan block. Conventionally, the 1s are above the diagonal, but it
doesn’t make a diference, because reversing the order of the vectors #»
e 1 , · · · #»
e a to #»
e a , · · · #»
e 1 moves the 1s from above the diagonal
to below the diagonal. The diference is notational.
49
, Lecture 11: Proving the Jordan Decomposition Theorem
Example 11.3
0 1 0 0
0 0 1 0
Take J4 (0) =
0 0
. Under J4 (0), each basis vector maps to the next basis vector, and there is
0 1
0 0 0 0
one chain of length 4:
⃗e4 7→ ⃗e3 7→ ⃗e2 7→ ⃗e1 7→ ⃗0.
As a result, applying J4 (0) multiple times will eventually send all vectors to zero; that is, in this case,
J4 (0)4 = 0.
0 1 0 0
0 0 0 0
On the other hand, consider J2,2 (0) =
= J2 (0) 0
.
0 0 0 1 0 J2 (0)
0 0 0 0
Applying the operator to the basis vectors yields two chains of length 2:
⃗e2 7→ ⃗e1 7→ ⃗0
⃗e4 7→ ⃗e3 7→ ⃗0
In this case as well, the operator will map every vector to zero upon repeated application.
In general, for λ ̸= 0, (λI − T ) #»
e i is not necessarily zero (it is zero only if #e»i is an eigenvector), but for some
large enough n,
(λI − T )n #e»i = 0.36
In Example 11.3, there was a chain of length 4 for the frst matrix, while in the second matrix, we had two
chains of length 2.
Note 11.4
The Jordan decomposition theorem is powerful because any square matrix has a Jordan decomposition.
On the other hand, most matrices are diagonalizable, and any matrix will be ε away from a diagonalizable
matrix, and the Jordan decomposition is unnecessary. Only in the zero percent of the timea when the
characteristic polynomial has repeated roots is it necessary.
a This concept is feshed out in measure theory.
11.3 Proof of Jordan Decomposition Theorem
The proof of the Jordan decomposition theorem is quite involved and relatively tricky, so the important part for
the rest of class is understanding the style of proof, rather than the exact details. This proof will break down
the theorem inductively into smaller and smaller pieces.
Let’s start with a couple of defnitions that will help us with the proof.
Defnition 11.5
Given a vector space V and a linear transformation T : V → V, a subspace W ⊆ V is called T -invariant if
⃗ ) ∈ W for all w
T (w ⃗ ∈ W.
For example, if the vector space V is the space of polynomials of degree at most 3, and the subspace W is the
space of polynomials of degree at most 2, W will be T −invariant under the linear operator T that is taking the
derivative.
36 A vector that is killed not necessarily immediately but eventually by λI − T is known as a generalized eigenvector; there is
a question about them on the problem set.
50
