Notes on the Jordan canonical form Introduction - wiki - University of ...

MAT 240, University of Toronto, notes by Marco Gualtieri 

Introduction 

<strong>Notes</strong> on the Jordan canonical form 

Let V be a finite-dimensional vector space over a field F, and let T : V −→ V be a linear 

operator such that 

(T − a1) k1 · · · (T − am) km = 0, (1) 

for a1, . . . , am distinct numbers in F. The purpose of this note is to explain how to find a 

is block diagonal, where each block looks like 

Jordan basis, i.e. a basis β for V such that [T] β 

β 

⎡ 

a 1 

⎢ 

⎣ 

a 1 

. .. 

⎤ 

⎥ 

. .. 

⎥ 

a 1⎦ 

a 

The matrix above is called a Jordan block with eigenvalue a. Note that a basis (v1, . . . , vl) 

giving rise to a Jordan block is “cyclic” in a certain sense: from the shape of the matrix, we 

see that if we apply T − a to vi, we obtain vi−1. We call vl a “cyclic” vector, since the rest of 

the basis can be generated from vl, via vi = (T − a) l−ivl. The first thing we must do to simplify our problem is to apply the decomposition theorem, 

which tells us that the null space of m i=1 (T − ai) ki, which is all of V by (1), is a direct 

sum of the generalized1 eigenspaces null(T − ai) ki. In other words, we have a direct sum 

decomposition 

V = V(a1) ⊕ · · · ⊕ V(am), (3) 

where V(ai) = null(T − ai) ki. So, we can focus on finding a Jordan basis for each of the 

V(ai) separately. Then by putting the bases together we obtain a Jordan basis for V. For 

this reason, we may assume from now on that T is a linear operator on a vector space V(a) 

and that (T − a) k = 0, and we are trying to find a basis in which T has diagonal blocks of the 

form (2). 

1 Recall that v is a generalized eigenvector with eigenvalue a when there exists some i with (T − a) i v = 0. 

1 

(2)

What’s the idea? 

To better illustrate the kind of problem we are trying to solve, here are two linear operators, 

each with an 8-dimensional generalized eigenspace V(a): 

⎡ 

a 1 

⎤ 

⎡ 

a 1 

⎤ 

⎢ 

T1 = ⎢ 

⎣ 

a 1 

a 

a 1 

a 

a 1 

a 

⎥ 

⎦ 

⎢ 

T2 = ⎢ 

⎣ 

a 1 

a 1 

a 

a 1 

a 

a 

⎥ 

⎦ 

a 

a 

So, T1 has Jordan blocks of size (3, 2, 2, 1), while T2 has blocks of size (4, 2, 1, 1). They happen 

to have the same number of blocks, but the size of the blocks is different. In general, we 

could have blocks with sizes making up any partition of 8, for example 8 = 3 + 2 + 1 + 1 + 1. 

Given T, we would like to figure out how many Jordan blocks there are, what their sizes 

are, and how to find the actual basis giving this Jordan form. The idea is simple: 

The key is the sequence of null spaces null(T − a) i , i = 1, . . . , k. 

It is convenient to define ti = dim null(T − a) i . 

To see how this works, let’s assume that T has a Jordan form, but we don’t know what it is. 

• The dimension of the eigenspace null(T − a) tells you exactly how many Jordan blocks 

there are, since each Jordan block has a 1-dimensional eigenspace. In other words, t1 is 

the number of Jordan blocks. 

• If T has only Jordan blocks of size 1, then t2 = dim null(T − a) 2 is the same as t1. But in 

general, t2 will be bigger than t1, because each Jordan block of size > 1 will contribute 

+1. Therefore t2 − t1 tells us the number of Jordan blocks of size > 1, and we can 

conclude that the number of Jordan blocks of size 1 is t1 − (t2 − t1). 

• In general, tk+1 − tk is the number of Jordan blocks of size > k, and so the number of 

Jordan blocks of size exactly k is (tk − tk−1) − (tk+1 − tk) = 2tk − tk−1 − tk+1. 

We can phrase the above discussion in a very clever way 2 as follows: 

Proposition 1. Let t = (t0, t1, . . .), where ti = dim null(T − a) i . Then T has si Jordan blocks of 

size i, where s = (s0, s1, . . .) is given by 

s = −R(L − 1) 2 t, 

where R, L are the right and left shift operators on sequences. 

For example, for T1 as above, we have t = (0, 4, 7, 8, 8, . . .). Hence we have s = (0, 1, 2, 1, 0, 0, . . .) 

indicating one block of size 1, two of size 2, and one of size 3, as is indeed the case. For T2 we 

have t = (0, 4, 6, 7, 8, 8, . . .), so that s = (0, 2, 1, 0, 1, 0, 0, . . .), yielding two Jordan blocks of size 

1, one of size 2, and one of size 4. 

2 I learned this from Prof. John Labute, I’d like to know if appears elsewhere. 

2

How to find a Jordan basis 

In this section we will give an algorithm for finding a Jordan basis, i.e. a basis β for V(a) 

such that [T] β 

β is in Jordan form. Beware: there are many possible choices of Jordan basis, 

even though the Jordan form will always consist of the same number of Jordan blocks of 

each size. For example, the identity operator is in Jordan form in any basis. 

To make our life easier, we would like to introduce the following terminology: 

Definition 1. Let U ⊂ V be a linear subspace. Then we say that the list of vectors (v1, . . . , vl) in V is 

“linearly independent mod U” when 

i αivi ∈ U implies αi = 0 for all i. We say that (v1, . . . , vl) 

is a “basis of V mod U” when it is linearly independent mod U and V = U + span(v1, . . . , vl). 

It is easy to find a basis of V mod U: simply choose a basis (u1, . . . , uk) for U and extend 

it to a basis (u1, . . . , uk, v1, . . . , vl) for V. The vectors (v1, . . . , vl) are then a basis of V mod U. 

We now give the algorithm for finding a Jordan basis for V(a). The idea is to find a 

cyclic vector for each Jordan block, which then generates the Jordan basis for that block. This 

algorithm finds the cyclic vectors for the largest Jordan blocks first, and then proceeds to find 

the cyclic vectors for the smaller blocks. 

Step 1: Choose a basis (v 1 k 

, . . . , vsk 

our cyclic vectors for the Jordan blocks of size k. There are sk = tk − tk−1 of them. 

k ) of V(a) = null(T − a)k mod null(T − a) k−1 . These will be 

Step 2: If k = 1, stop. Else, apply T − a to the cyclic vectors from the previous step, obtaining 

(T − a)v i k in null(T − a)k−1 . Then the key point is that 

((T − a)v 1 k , . . . , (T − a)vsk k ) is linearly independent mod null(T − a)k−2 . 

Extend to a basis of null(T − a) k−1 mod null(T − a) k−2 , by choosing (v1 k−1 , . . . , vsk−1 

k−1 ), 

which are then cyclic vectors for the Jordan blocks of size k−1. Since we have extended 

a list of size sk to reach a length tk−1 −tk−2, there are sk−1 = (tk −tk−1)−(tk−1 −tk−2) 

of these new cyclic vectors. 

Step 3: Repeat Step 2, with k replaced by k − 1. 

Once the algorithm terminates, we may arrange the chosen basis as follows: 

β = (v 1 1 , . . . , vs1 

1 , (T−a)v12 , v12 , . . . , (T−a)vs2 2 , vs2 

2 , . . . , (T−a)i−1v 1 i , . . . v1i , . . . , (T−a)i−1v si 

i 

this would put [T] β 

β into Jordan form with blocks of increasing size down the diagonal. 

3 

, . . . , vsi 

i , . . .)

An example 

Find a Jordan basis for the following operator: 

⎡ 

1 

⎢ 

⎢0 

⎢ 

T = ⎢0 

⎢ 

⎣0 

0 

1 

0 

0 

1 

0 

1 

0 

0 

1 

0 

1 

⎤ 

1 

1 

⎥ 

0 ⎥ 

1⎦ 

0 0 0 0 1 

Clearly the only eigenvalue is 1. Let ɛ = (e1, . . . , e5) be the standard basis. First we calculate 

null(T − 1) i : 

null(T − 1) = span(e1, e2) 

null(T − 1) 2 = span(e1, e2, e3, e4) 

null(T − 1) 3 = span(e1, e2, e3, e4, e5). 

Since t = (0, 2, 4, 5, 5, . . .), we have s = (0, 0, 1, 1, 0, 0, . . .), so we will have a Jordan form with 

a block of size 2 and one of size 3. The algorithm proceeds as follows: 

Step 1: Pick v 1 3 = e5, it will be a cyclic vector for a Jordan block of size 3. 

Step 2: We must extend (T − 1)e5 = e1 + e2 + e4 to a basis of null(T − 1) 2 mod null(T − 1). We 

can do this by adding v 1 2 = e3. This will be a cyclic vector for a Jordan block of size 2, 

and we are done. 

The Jordan basis is 

β = ((T − 1)v 1 2 , v1 2 , (T − 1)2 v 1 3 , (T − 1)v1 3 , v1 3 ) = (e1, e3, e2, e1 + e2 + e4, e5). 

Therefore, we have that T = PJP−1 , where P = [I] ɛ β and J = [T]β β is the Jordan form. Explicitly 

writing P (its columns are the Jordan basis in terms of the standard basis), we have 

⎡ 

1 

⎢ 

⎢0 

⎢ 

P = ⎢0 

⎢ 

⎣0 

0 

0 

1 

0 

0 

1 

0 

0 

1 

1 

0 

1 

⎤ 

0 

0 

⎥ 

0 ⎥ 

0⎦ 

0 0 0 0 1 

4

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