3 Canonical Forms
3.1 Jordan Forms & Generalized Eigenvectors
Throughout this course we’ve concerned ourselves with variations of a general question: for a given
map T ∈ L(V), find a basis β such that the matrix [T]
β
is as close to diagonal as possible. In this
chapter we see what is possible when T is non-diagonalizable.
Example 3.1. The matrix A =
−8 4
−25 12
∈ M
2
(R ) has characteristic equation
p(t) = (−8 −t)(12 −t) + 4 ·25 = t
2
−4t + 4 = (t −2)
2
and thus a single eigenvalue λ = 2. It is non-diagonalizable since the eigenspace is one-dimensional
E
2
= N
−10 4
−25 10
= Span
2
5
However, if we consider a basis β = {v
1
, v
2
} where v
1
=
2
5
is an eigenvector, then [L
A
]
β
is upper-
triangular, which is better than nothing! How simple can we make this matrix? Let v
2
=
(
x
y
)
, then
Av
2
=
−8x + 4y
−25x + 12y
= 2
x
y
+
−10x + 4y
−25x + 10y
= 2v
2
+ (−5x + 2y)v
1
=⇒ [L
A
]
β
=
2 −5x + 2y
0 2
Since v
2
cannot be parallel to v
1
, the only thing we cannot have is a diagonal matrix. The next best
thing is for the upper right corner be 1; for instance we could choose
β = {v
1
, v
2
} =
2
5
,
1
3
=⇒ [L
A
]
β
=
2 1
0 2
Definition 3.2. A Jordan block is a square matrix of the form
J =
λ 1
λ
.
.
.
.
.
.
1
λ
where all non-indicated entries are zero. Any 1 ×1 matrix is also a Jordan block.
A Jordan canonical form is a block-diagonal matrix diag(J
1
, . . . , J
m
) where each J
k
is a Jordan block.
A Jordan canonical basis for T ∈ L(V) is a basis β of V such that [T]
β
is a Jordan canonical form.
If a map is diagonalizable, then any eigenbasis is Jordan canonical and the corresponding Jordan
canonical form is diagonal. What about more generally? Does every non-diagonalizable map have a
Jordan canonical basis? If so, how can we find such?
1
Example 3.3. It can easily be checked that β = {v
1
, v
2
, v
3
} =
n
1
0
1
,
1
2
0
,
1
1
1
o
is a Jordan canon-
ical basis for
A =
−1 2 3
−4 5 4
−2 1 4
(really L
A
∈ L(R
3
)). Indeed
Av
1
= 2v
1
, Av
2
= 3v
2
, Av
3
=
4
5
3
=
1 + 3
2 + 3
0 + 3
= v
2
+ 3v
3
=⇒ [L
A
]
β
=
2 0 0
0 3 1
0 0 3
Generalized Eigenvectors
Example 3.3 was easy to check, but how would we go about finding a suitable β if we were merely
given A? We brute-forced this in Example 3.1, but such is not a reasonable approach in general.
Eigenvectors get us some of the way:
• v
1
is an eigenvector in Example 3.1, but v
2
is not.
• v
1
and v
2
are eigenvectors in Example 3.3, but v
3
is not.
The practical question is how to fill out a Jordan canonical basis once we have a maximal independent
set of eigenvectors. We now define the necessary objects.
Definition 3.4. Suppose T ∈ L(V) has an eigenvalue λ. Its generalized eigenspace is
K
λ
:= {x ∈ V : (T −λI)
k
( x) = 0 for some k ∈ N} =
[
k∈N
N(T − λ I)
k
A generalized eigenvector is any non-zero v ∈ K
λ
.
As with eigenspaces, the generalized eigenspaces of A ∈ M
n
(F ) are those of the map L
A
∈ L(F
n
).
It is easy to check that our earlier Jordan canonical bases consist of generalized eigenvectors.
Example 3.1: We have one eigenvalue λ = 2. Since (A − 2I)
2
=
0 0
0 0
is the zero matrix, every
non-zero vector is a generalized eigenvector; plainly K
2
= R
2
.
Example 3.3: We see that
(A − 2I)v
1
= 0, (A − 3I)v
2
= 0, (A − 3I)
2
v
3
= (A −3I)v
2
= 0
whence β is a basis of generalized eigenvectors. Indeed
K
3
= Span{v
1
, v
2
}, K
2
= E
2
= Span{v
3
}
though verifying this with current technology is a little awkward. . .
2
In order to easily compute generalized eigenspaces, it is useful to invoke the main result of this
section. We postpone the proof for a while due to its meatiness.
Theorem 3.5. Suppose that the characteristic polynomial of T ∈ L(V) splits over F:
p(t) = (λ
1
−t)
m
1
···(λ
k
−t)
m
k
where the λ
j
are the distinct eigenvalues of T with algebraic multiplicities m
j
. Then:
1. For each eigenvalue; (a) K
λ
= N(T − λI)
m
and (b) dim K
λ
= m.
2. V = K
λ
1
⊕···⊕K
λ
k
: there exists a basis of generalized eigenvectors.
Compare this with the statement on diagonalizability from the start of the course.
With regard to part 2; we shall eventually be able to choose this to be a Jordan canonical basis. In
conclusion: a map has a Jordan canonical basis if and only if its characteristic polynomial splits.
Examples 3.6. 1. Observe how Example 3.3 works in this language:
A =
−1 2 3
−4 5 4
−2 1 4
=⇒ p(t) = (2 −t)
1
(3 −t)
2
K
2
= N(A −2I)
1
= Span
1
0
1
=⇒ dim K
2
= 1
K
3
= N(A −3I)
2
= N
2 −1 −1
0 0 0
2 −1 −1
= Span
1
2
0
,
1
1
1
=⇒ dim K
3
= 2
R
3
= K
2
⊕K
3
2. We find the generalized eigenspaces of the matrix A =
5 2 −1
0 0 0
9 6 −1
The characteristic polynomial is
p(t) = det(A −λI) = −t
5 − t −1
9 −1 −t
= −t(t
2
−5t + t −5 + 9) = −(0 −t)
1
(2 −t)
2
• λ = 0 has multiplicity 1; indeed K
0
= N(A − 0I)
1
= N(A) = Span
1
−1
3
is just the
eigenspace E
0
.
• λ = 2 has multiplicity 2,
K
2
= N(A −2I)
2
= N
3 2 −1
0 −2 0
9 6 −3
2
= N
0 −4 0
0 4 0
0 −12 0
= Span
1
0
0
,
0
0
1
In this case the corresponding eigenspace is one-dimensional, E
2
= Span
1
0
3
⪇ K
2
, and
the matrix is non-diagonalizable.
Observe also that R
3
= K
0
⊕K
2
in accordance with the Theorem.
3
Properties of Generalized Eigenspaces and the Proof of Theorem 3.5
A lot of work is required to justify our main result. Feel free to skip the proofs at first reading.
Lemma 3.7. Let λ be an eigenvalue of T ∈ L(V). Then:
1. E
λ
is a subspace of K
λ
, which is itself a subspace of V.
2. K
λ
is T-invariant.
3. Suppose K
λ
is finite-dimensional and µ = λ. Then:
(a) K
λ
is ( T −µI)-invariant and the restriction of T − µI to K
λ
is an isomorphism.
(b) If µ is another eigenvalue, then K
λ
∩ K
µ
= {0}. In particular K
λ
contains no eigenvectors
other than those in E
λ
.
Proof. 1. These are an easy exercise.
2. Let x ∈ K
λ
, then ∃k such that (T − λI)
k
( x) = 0. But then
(T −λI)
k
T(x)
= (T −λI)
k
T(x) − λx + λx
= (T −λI)
k+1
( x) + λ(T − λI)
k
( x) = 0
Otherwise said, T(x) ∈ K
λ
.
3. (a) Let x ∈ K
λ
. Part 2 tells us that
(T −µI)(x) = T(x) −µx ∈ K
λ
whence K
λ
is ( T −µI)-invariant.
Suppose, for a contradiction, that T − µI is not injective on K
λ
. Then
∃y ∈ K
λ
\{0} such that (T − µI)(y) = 0
Let k ∈ N be minimal such that (T − λI)
k
( y) = 0 and let z = (T − λI)
k−1
( y) . Plainly
z = 0, for otherwise k is not minimal. Moreover,
(T −λI)(z) = (T − λI)
k
( y) = 0 =⇒ z ∈ E
λ
Since T − µI and T −λI commute, we can also compute the effect of T − µI;
(T −µI)(z) = (T − µI)(T − λI)
k−1
( y) = (T − λI)
k−1
(T −µI)(x) = 0
which says that z is an eigenvector in E
µ
; if µ isn’t an eigenvalue, then we already have
our contradiction! Even if µ is an eigenvalue, E
µ
∩ E
λ
= {0} provides the desired contra-
diction.
We conclude that (T − µI)
K
λ
∈ L(K
λ
) is injective. Since dim K
λ
< ∞, the restriction is
automatically an isomorphism.
(b) This is another exercise.
4
Now to prove Theorem 3.5: remember that the characteristic polynomial of T is assumed to split.
Proof. (Part 1(a)) Fix an eigenvalue λ. By definition, we have N(T −λI)
m
≤ K
λ
.
For the converse, parts 2 and 3 of the Lemma tell us (why?) that
p
λ
( t) = (λ − t)
dim K
λ
from which dim K
λ
≤ m (∗)
By the Cayley–Hamilton Theorem, T
K
λ
satisfies its characteristic polynomial, whence
∀x ∈ K
λ
,
(
λI −T
)
dim K
λ
( x) = 0 =⇒ K
λ
≤ N(T − λI)
m
(Parts 1(b) and 2) We prove simultaneously by induction on the number of distinct eigenvalues of T.
(Base case) If T has only one eigenvalue, then p(t) = (λ − t)
m
. Another appeal to Cayley–
Hamilton says ( T −λI)
m
( x) = 0 for all x ∈ V. Thus V = K
λ
and dim K
λ
= m.
(Induction step) Fix k and suppose the results hold for maps with k distinct eigenvalues. Let T
have distinct eigenvalues λ
1
, . . . , λ
k
, µ, with multiplicities m
1
, . . . , m
k
, m respectively. Define
1
W = R(T −µI)
m
The subspace W has the following properties, the first two of which we leave as exercises:
• W is T-invariant.
• W ∩ K
µ
= {0} so that µ is not an eigenvalue of the restriction T
W
.
• Each K
λ
j
≤ W: since (T −µI)
K
λ
j
is an isomorphism (Lemma part 3), we can invert,
x ∈ K
λ
j
=⇒ x = (T −µI)
m
(T −µI)
−1
K
λ
j
m
( x) ∈ R( T −µI)
m
= W
We conclude that λ
j
is an eigenvalue of the restriction T
W
with generalized eigenspace K
λ
j
.
Since T
W
has k distinct eigenvalues, the induction hypotheses apply:
W = K
λ
1
⊕···⊕K
λ
k
and p
W
( t) = (λ
1
−t)
dim K
λ
1
···(λ
k
−t)
dim K
λ
k
Since W ∩K
µ
= {0} it is enough finally to use the rank–nullity theorem and count dimensions:
dim V = rank(T −µI)
m
+ null(T − µI)
m
= dim W + dim K
µ
=
k
∑
j=1
dim K
λ
j
+ dim K
µ
(∗)
≤ m
1
+ ···+ m
k
+ m = deg(p(t)) = dim V
The inequality is thus an equality; each dim K
λ
j
= m
j
and dim K
µ
= m. We conclude that
V = K
λ
1
⊕···⊕K
λ
k
⊕K
µ
which completes the induction step and thus the proof. Whew!
1
This is yet another argument where we consider a suitable subspace to which we can apply an induction hypothesis;
recall the spectral theorem, Schur’s lemma, bilinear form diagonalization, etc. Theorem 3.12 will provide one more!
5
Cycles of Generalized Eigenvectors
By Theorem 3.5, for every linear map whose characteristic polynomial splits there exists generalized
eigenbasis. This isn’t the same as a Jordan canonical basis, but we’re very close!
Example 3.8. The matrix A =
5 1 0
0 5 1
0 0 5
∈ M
3
(R ) is a single Jordan block, whence there is a single
generalized eigenspace K
5
= R
3
and the standard basis ϵ = {e
1
, e
2
, e
3
} is Jordan canonical.
The crucial observation for what follows is that one of these vectors e
3
generates the others via re-
peated applications of A −5I:
e
2
= (A −5I)e
3
, e
1
= (A −5I)e
2
= (A −5I)
2
e
3
Definition 3.9. A cycle of generalized eigenvectors for a linear operator T is a set
β
x
:=
n
(T −λI)
k−1
( x) , . . . , (T − λI)(x), x
o
where the generator x ∈ K
λ
is non-zero and k is minimal such that (T − λI)
k
( x) = 0.
Note that the first element ( T −λI)
k−1
( x) is an eigenvector.
Our goal is to show that K
λ
has a basis consisting of cycles of generalized eigenvectors; putting these
together results in a Jordan canonical basis.
Lemma 3.10. Let β
x
be a cycle of generalized eigenvectors of T with length k. Then:
1. β
x
is linearly independent and thus a basis of Span β
x
.
2. Span β
x
is T-invariant. With respect to β
x
, the matrix of the restriction of T is the k × k Jordan
block [T
Span β
x
]
β
x
=
λ 1
λ
.
.
.
.
.
.
1
λ
!
.
In what follows, it will be useful to consider the linear map U = T −λI. Note the following:
• The nullspace of U is the eigenspace: N(U) = E
λ
≤ K
λ
.
• T commutes with U: that is TU = UT.
• β
x
= {U
k−1
( x) , . . . , U(x), x}; that is, Span β
x
=
⟨
x
⟩
is the U-cyclic subspace generated by x.
Proof. 1. Feed the linear combination
∑
k−1
j=0
a
j
U
j
( x) = 0 to U
k−1
to obtain
a
0
U
k−1
( x) = 0 =⇒ a
0
= 0
Now feed the same combination to U
k−2
, etc., to see that all coefficients a
j
= 0.
2. Since T and U commute, we see that
T
U
j
( x)
= U
j
T(x)
= U
j
(U + λI)(x)
= U
j+1
( x) + λU
j
( x) ∈ Span β
x
This justifies both T-invariance and the Jordan block claim!
6
The basic approach to finding a Jordan canonical basis is to find the generalized eigenspaces and play
with cycles until you find a basis for each K
λ
. Many choices of canonical basis exist for a given map!
We’ll consider a more systematic method in the next section.
Examples 3.11. 1. The characteristic polynomial of A =
1 0 2
0 1 6
6 −2 1
∈ M
3
(R ) splits:
p(t) = (1 − t)
1 − t 6
−2 1 −t
+ 2
0 1 −t
6 −2
= (1 −t )
(1 −t)
2
+ 12 −12
= (1 −t )
3
With only one eigenvalue we see that K
1
= R
3
. Simply choose any vector in R
3
and see what
U = A − I does to it! For instance, with x = e
1
,
β
x
=
n
U
2
1
0
0
, U
1
0
0
,
1
0
0
o
=
n
12
36
0
,
0
0
6
,
1
0
0
o
provides a Jordan canonical basis of R
3
. We conclude
A = QJQ
−1
=
12 0 1
36 0 0
0 6 0
1 1 0
0 1 1
0 0 1
12 0 1
36 0 0
0 6 0
−1
In practice, almost any choice of x ∈ R
3
will generate a cycle of length three!
2. The matrix B =
7 1 −4
0 3 0
8 1 −5
∈ M
3
(R ) has characteristic equation
p(t) = (3 − t)(t
2
−2t − 3) = −(t + 1)
1
( t −3)
2
dim K
−1
= 1 =⇒ K
−1
= E
−1
= Span
1
0
2
, spanned by a cycle of length one.
Since dim K
3
= 2, we have
K
3
= N(B −3I)
2
= N
4 1 −4
0 0 0
8 1 −8
2
= N
−16 0 16
0 0 0
−32 0 32
= Span
n
1
0
1
,
0
1
0
o
This is spanned by a cycle of length two:
1
0
1
is an eigenvector and
1
0
1
= (B −3I)
0
1
0
We conclude that β =
n
1
0
2
,
1
0
1
,
0
1
0
o
is a Jordan canonical basis for B, and that
B = QJQ
−1
=
1 1 0
0 0 1
2 1 0
−1 0 0
0 3 1
0 0 3
1 1 0
0 0 1
2 1 0
−1
3. Let T =
d
dx
on P
3
(R ). With respect to the standard basis ϵ = {1, x, x
2
, x
3
},
A = [T]
ϵ
=
0 1 0 0
0 0 2 0
0 0 0 3
0 0 0 0
With only one eigenvalue λ = 0, we have a single generalized eigenspace K
0
= P
3
(R ). It is
easy to check that f (x) = x
3
generates a cycle of length three and thus a Jordan canonical basis:
β = {6, 6x , 3x
2
, x
3
} =⇒ [T]
β
=
0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
=
6 0 0 0
0 6 0 0
0 0 3 0
0 0 0 1
−1
0 1 0 0
0 0 2 0
0 0 0 3
0 0 0 0
6 0 0 0
0 6 0 0
0 0 3 0
0 0 0 1
7
Our final results state that this process works generally.
Theorem 3.12. Let T ∈ L(V) have an eigenvalue λ. If dim K
λ
< ∞, then there exists a basis
β
λ
= β
x
1
∪···∪ β
x
n
of K
λ
consisting of finitely many linearly independent cycles.
Intuition suggests that we create cycles β
x
j
by starting with a basis of the eigenspace E
λ
and extending
backwards: for each x, if x = (T −λI)(y), then x ∈ β
y
; now repeat until you have a maximum length
cycle. This is essentially what we do, though a sneaky induction is required to make sure we keep
track of everything and guarantee that the result really is a basis of K
λ
.
Proof. We prove by induction on m = dim K
λ
.
(Base case) If m = 1, then K
λ
= E
λ
= Span x for some eigenvector x. Plainly {x} = β
x
.
(Induction step) Fix m ≥ 2. Write n = dim E
λ
≤ m and U = (T − λI)
K
λ
.
(i) For the induction hypothesis, suppose every generalized eigenspace with dimension < m (for
any linear map!) has a basis consisting of independent cycles of generalized eigenvectors.
(ii) Define W = R(U) ∩ E
λ
: that is
w ∈ W ⇐⇒
(
U(w) = 0 and
w = U(v) for some v ∈ K
λ
Let k = dim W, choose a complementary subspace X such that E
λ
= W ⊕ X and select a basis
{x
k+1
, . . . , x
n
} of X. If k = 0, the induction step is finished (why?). Otherwise we continue. . .
(iii) The calculation in the proof of Lemma 3.10 (take j = 1) shows that R( U) is T-invariant; it is
therefore the single generalized eigenspace
˜
K
λ
of T
R(U)
.
(iv) By the rank–nullity theorem,
dim R(U) = rank U = dim K
λ
−null U = m −dim E
λ
< m
By the induction hypothesis, R(U) has a basis of independent cycles. Since the last non-zero
element in each cycle is an eigenvector, this basis consists of k distinct cycles β
ˆ
x
1
∪ ··· ∪ β
ˆ
x
k
whose terminal vectors form a basis of W.
(v) Since each
ˆ
x
j
∈ R(U), there exist vectors x
1
, . . . , x
k
such that
ˆ
x
j
= U(x
j
). Including the length-
one cycles generated by the basis of X, the cycles β
x
1
, . . . , β
x
n
now contain
dim R(U) + k + (n − k) = rank U + null U = m
vectors. We leave as an exercise the verification that these vectors are linearly independent.
Corollary 3.13. Suppose that the characteristic polynomial of T ∈ L(V) splits (necessarily dim V <
∞). Then there exists a Jordan canonical basis, namely the union of bases β
λ
from Theorem 3.12.
Proof. By Theorem 3.5, V is the direct sum of generalized eigenspaces. By the previous result, each
K
λ
has a basis β
λ
consisting of finitely many cycles. By Lemma 3.10, the matrix of T
K
λ
has Jordan
canonical form with respect to β
λ
. It follows that β =
S
β
λ
is a Jordan canonical basis for T.
8
Exercises 3.1 1. For each matrix, find the generalized eigenspaces K
λ
, find bases consisting of
unions of disjoint cycles of generalized eigenvectors, and thus find a Jordan canonical form J
and invertible Q so that the matrix may be expressed as QJQ
−1
.
(a) A =
1 1
−1 3
(b) B =
1 2
3 2
(c) C =
11 −4 −5
21 −8 −11
3 −1 0
(d) D =
2 1 0 0
0 2 1 0
0 0 3 0
0 1 −1 3
2. If β = {v
1
, . . . , v
n
} is a Jordan canonical basis, what can you say about v
1
? Briefly explain why
the linear map L
A
∈ L(R
2
) where A =
0 −1
1 0
has no Jordan canonical form.
3. Find a Jordan canonical basis for each linear map T:
(a) T ∈ L(P
2
(R )) defined by T( f (x)) = 2 f (x) − f
′
(x)
(b) T( f ) = f
′
defined on Span{1, t, t
2
, e
t
, te
t
}
(c) T(A) = 2A + A
T
defined on M
2
(R )
4. In Example 3.11.1, suppose x =
a
b
c
. Show that almost any choice of a, b, c produces a Jordan
canonical basis β
x
.
5. We complete the proof of Lemma 3.7.
(a) Prove part 1: that E
λ
≤ K
λ
≤ V.
(b) Verify that T − µI and T − λI commute.
(c) Prove part 3(b): generalized eigenspaces for distinct eigenvalues have trivial intersection.
6. Consider the induction step in the proof of Theorem 3.5.
(a) Prove that W is T-invariant.
(b) Explain why W ∩ K
µ
= {0}.
(c) The assumption p
W
( t) = ( λ
1
− t)
dim K
λ
1
···(λ
k
− t)
dim K
λ
k
near the end of the proof is the
induction hypothesis for part 1(b). Why can’t we also assume that dim K
λ
j
= m
j
and thus
tidy the inequality argument near the end of the proof?
7. We finish some of the details of Theorem 3.12.
(a) In step (ii), suppose dim W = k = 0. Explain why {x
1
, . . . , x
n
} is in fact a basis of K
λ
, so
that the rest of the proof is unnecessary.
(b) In step (v), prove that the m vectors in the cycles β
x
1
, . . . , β
x
n
are linearly independent.
(Hint: model your argument on part 1 of Lemma 3.10)
9
3.2 Cycle Patterns and the Dot Diagram
In this section we obtain a useful result that helps us compute Jordan forms more efficiently and
systematically. To give us some clues how to proceed, here is a lengthy example.
Example 3.14. Precisely three Jordan canonical forms A, B, C ∈ M
3
(R ) correspond to the charac-
teristic polynomial p(t) = (5 −t)
3
:
A =
5 0 0
0 5 0
0 0 5
B =
5 1 0
0 5 0
0 0 5
C =
5 1 0
0 5 1
0 0 5
In all three cases the standard basis β = {e
1
, e
2
, e
3
} is Jordan canonical, so how do we distinguish
things? By considering the number and lengths of the cycles of generalized eigenvectors.
• A has eigenspace E
5
= K
5
= R
3
. Since (A − 5I)v = 0 for all v ∈ R
3
, we have maximum
cycle-length one. We therefore need three distinct cycles to construct a Jordan basis, e.g.
β
e
1
= {e
1
}, β
e
2
= {e
2
}, β
e
3
= {e
3
} =⇒ β = β
e
1
∪ β
e
2
∪ β
e
3
= {e
1
, e
2
, e
3
}
• B has eigenspace E
5
= Span{e
1
, e
3
}. By computing
v =
a
b
c
=⇒ (B −5I)v =
b
0
0
=⇒ (B −5I)
2
v = 0
we see that β
v
is a cycle with maximum length two, provided b = 0 (v ∈ E
5
). We therefore
need two distinct cycles, of lengths two and one, to construct a Jordan basis, e.g.
β
e
2
=
(B −5I)e
2
, e
2
= {e
1
, e
2
}, β
e
3
= {e
3
} =⇒ β = β
e
2
∪ β
e
3
= {e
1
, e
2
, e
3
}
• C has eigenspace E
5
= Span e
1
. This time
v =
a
b
c
=⇒ (C −5I)v =
b
c
0
, ( C −5I)
2
v =
c
0
0
, ( C −5I)
3
v = 0
generates a cycle with maximum length two provided c = 0. Indeed this cycle is a Jordan basis,
so one cycle is all we need:
β = β
e
3
=
(C −5I)
2
e
3
, (C −5I)e
3
, e
3
= {e
1
, e
2
, e
3
}
Why is the example relevant? Suppose that dim
R
V = 3 and that T ∈ L(V) has characteristic polyno-
mial p(t) = (5 −t)
3
. Theorem 3.12 tells us that T has a Jordan canonical form, and that is is moreover
one of the above matrices A, B, C. Our goal is to develop a method whereby the pattern of cycle-
lengths can be determined, thus allowing us to be able to discern which Jordan form is correct. As a
side-effect, this will also demonstrate that the pattern of cycle lengths for a given T is independent of
the Jordan basis so that, up to some reasonable restriction, the Jordan form of T is unique. To aid us
in this endeavor, we require some terminology. . .
10
Definition 3.15. Let V be finite dimensional and K
λ
a generalized eigenspace of T ∈ L(V). Follow-
ing the Theorem 3.12, assume that β
λ
= β
x
1
∪ ··· ∪ β
x
n
is a Jordan canonical basis of T
K
λ
, where the
cycles are arranged in non-increasing length. That is:
1. β
x
j
= {(T −λI)
k
j
−1
( x
j
), . . . , x
j
} has length k
j
, and
2. k
1
≥ k
2
≥ ··· ≥ k
n
The dot diagram of T
K
λ
is a representation of the elements of β
λ
, one dot for each vector: the j
th
column
represents the elements of β
x
j
arranged vertically with x
j
at the bottom.
Given a linear map, our eventual goal is to identify the dot diagram as an intermediate step in the
computation of a Jordan basis. First, however, we observe how the conversion of dot diagrams to a
Jordan form is essentially trivial.
Example 3.16. Suppose dim V = 14 and that T ∈ L(V) has the following eigenvalues and dot
diagrams:
λ
1
= −4 λ
2
= 7 λ
3
= 12
• • • •
• •
• •
• •
•
• • •
Then generalized eigenspaces of T satisfy:
• K
−4
= N(T + 4I)
2
and dim K
−4
= 6;
• K
7
= N(T − 7I)
3
and dim K
7
= 5;
• K
12
= N(T − 12I) = E
12
and dim K
12
= 3;
T has a Jordan canonical basis β with respect to which its Jordan canonical form is
[T]
β
=
−4 1
0 −4
−4 1
0 −4
−4
−4
7 1 0
0 7 1
0 0 7
7 1
0 7
12
12
12
Note how the sizes of the Jordan blocks are non-increasing within each eigenvalue. For instance, for
λ
1
= −4, the sequence of cycle lengths (k
j
) is 2 ≥ 2 ≥ 1 ≥ 1.
11
Theorem 3.17. Suppose β
λ
is a Jordan canonical basis of T
K
λ
as described in Definition 3.15, and
suppose the i
th
row of the dot diagram has r
i
entries. Then:
1. For each r ∈ N, the vectors associated to the dots in the first r rows form a basis of N(T −λI)
r
.
2. r
1
= null(T −λI) = dim V −rank(T −λI)
3. When i > 1, r
i
= null(T −λI)
i
−null(T − λI)
i −1
= rank(T −λI)
i −1
−rank(T − λI)
i
Example (3.14 cont). We describe the dot diagrams of the three matrices A, B, C, along with the
corresponding vectors in the Jordan canonical basis β and the values r
i
.
A :
• • • x
1
x
2
x
3
e
1
e
2
e
3
Since A − 5I is the zero matrix, r
1
= 3 −rank(A − 5I) = 3. The dot diagram has one row,
corresponding to three independent cycles of length one: β = β
e
1
∪ β
e
2
∪ β
e
3
.
B : • •
•
(B −5I)x
1
x
2
x
1
e
1
e
3
e
2
Row 1: B −5I =
0 1 0
0 0 0
0 0 0
=⇒ rank(B −5I) = 1 and r
1
= 3 −1 = 2. The first row {e
1
, e
3
} is
a basis of E
5
= N(B −5I).
Row 2: (B −5I)
2
is the zero matrix, whence r
2
= rank(B −5I) −rank(B −5I)
2
= 1 −0 = 1.
The dot diagram corresponds to β = β
e
2
∪ β
e
3
= {e
1
, e
2
}∪{e
3
}.
C : •
•
•
(C −5I)
2
x
1
(C −5I)x
1
x
1
e
1
e
2
e
3
Row 1: C − 5I =
0 1 0
0 0 1
0 0 0
=⇒ r
1
= 3 −rank(C − 5I) = 1. The first row {e
1
} is a basis of
E
5
= N(C − 5I).
Row 2: (C −5I)
2
=
0 0 1
0 0 0
0 0 0
=⇒ r
2
= rank( C −5I) −rank(C −5I)
2
= 2 − 1 = 1. The first
two rows {e
1
, e
2
} form a basis of N(C −5I)
2
.
Row 3: (C −5I)
3
is the zero matrix, whence r
3
= rank(C −5I)
2
−rank(C −5I)
3
= 1 −0 = 1.
Proof. As previously, let U = T − λI.
1. Since each dot represents a basis vector U
p
( v
j
), any v ∈ K
λ
may be written uniquely as a linear
combination of the dots. Applying U simply moves all the dots up a row and all dots in the top
row to 0. It follows that v ∈ N(U
r
) ⇐⇒ it lies in the span of the first r rows. Since the dots
are linearly independent, they form a basis.
2. By part 1, r
1
= dim N(U) = null(T −λI) = dim V −rank( T −λI).
3. More generally,
r
i
= (r
1
+ ···+ r
i
) −(r
1
+ ···+ r
i −1
) = dim N(U
i
) −dim N(U
i −1
)
= null(U
i
) −null(U
i −1
) = rank(T −λI)
i −1
−rank(T − λI)
i
12
Since the ranks of maps ( T −λI)
i
are independent of basis, so also is the dot diagram. . .
Corollary 3.18. For any eigenvalue λ, the dot diagram is uniquely determined by T and λ. If we
list Jordan blocks for each eigenspace in non-increasing order, then the Jordan form of a linear map
is unique up to the order of the eigenvalues.
We now have a slightly more systematic method for finding Jordan canonical bases.
Example 3.19. The matrix A =
6 2 −4 −6
0 3 0 0
0 0 3 0
2 1 −2 −1
has characteristic equation
p(t) = (3 − t)
2
6 − t −6
2 −1 −t
= (2 −t )(3 −t)
3
We have two generalized eigenspaces:
• K
2
= E
2
= N(A −2I) = N
4 2 −4 −6
0 1 0 0
0 0 1 0
2 1 −2 −3
= Span
3
0
0
2
. The trivial dot diagram • corresponds
to this single eigenvector.
• K
3
= N(A −3I)
3
. To find the dot diagram, compute powers of A −3I:
Row 1: A −3I =
3 2 −4 −6
0 0 0 0
0 0 0 0
2 1 −2 −4
has rank 2 and the first row has r
1
= 4 −2 = 2 entries.
Row 2: (A −3I)
2
=
−3 0 0 6
0 0 0 0
0 0 0 0
−2 0 0 4
has rank 1 and the second row has r
2
= 2 −1 = 1 entry.
Since we now have three dots (equalling dim K
3
), the algorithm terminates and the dot diagram
for K
3
is • •
•
For the single dot in the second row, we choose something in N(A −3I)
2
which isn’t an eigen-
vector; perhaps the simplest choice is x
1
= e
2
, which yields the two-cycle
β
x
1
=
{
(A − 3I)x
1
, x
1
}
=
2
0
0
1
,
0
1
0
0
To complete the first row, choose any eigenvector to complete the span: for instance x
2
=
0
2
1
0
.
We now have suitable cycles and a Jordan canonical basis/form:
β =
3
0
0
2
,
2
0
0
1
,
0
1
0
0
,
0
2
1
0
, A = QJQ
−1
=
3 2 0 0
0 0 1 2
0 0 0 1
2 1 0 0
2 0 0 0
0 3 1 0
0 0 3 0
0 0 0 3
3 2 0 0
0 0 1 2
0 0 0 1
2 1 0 0
−1
Other choices are available! For instance, if we’d chosen the two-cycle generated by x
1
= e
3
, we’d
obtain a different Jordan basis but the same canonical form J:
˜
β =
3
0
0
2
,
−4
0
0
−2
,
0
0
1
0
,
0
2
1
0
, A =
3 −4 0 0
0 0 0 2
0 0 1 1
2 −2 0 0
2 0 0 0
0 3 1 0
0 0 3 0
0 0 0 3
3 −4 0 0
0 0 0 2
0 0 1 1
2 −2 0 0
−1
13
We do one final example for a non-matrix map.
Example 3.20. Let ϵ = {1, x, y, x
2
, y
2
, xy} and define T
f (x, y)
= 2
∂ f
∂x
−
∂ f
∂y
as a linear operator on
V = Span
R
ϵ. The matrix and characteristic polynomial of T is easy to compute:
[T]
ϵ
=
0 2 −1 0 0 0
0 0 0 4 0 −1
0 0 0 0 −2 2
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
=⇒ p(t) = t
6
, [T
2
]
ϵ
=
0 0 0 8 2 −4
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
, [T
3
]
ϵ
= O
There is only one eigenvalue λ = 0 and therefore one generalized eigenspace K
0
= V. We could keep
working with matrices, but it is easy to translate the nullspaces of the matrices back to subspaces of
V, from which the necessary data can be read off:
N(T) = Span{1, x + 2y, x
2
+ 4y
2
+ 4xy} null T = 3, rank T = 3, r
1
= 3
N(T
2
) = Span{1, x, y, x
2
+ 2xy, 2y
2
+ xy} null T
2
= 5, rank T
2
= 1, r
2
= 3 −1 = 2
We now have five dots; since dim K
0
= 6, the last row has one, and the dot diagram is • • •
• •
•
Since the first two rows span N(T
2
), we may choose any f
1
∈ N(T
2
) for the final dot: f
1
= xy is
suitable, from which the first column of the dot diagram becomes
T
2
(xy) • •
T(xy) •
xy
−4 • •
2y − x •
xy
Now choose the second dot on the second row to be anything in N(T
2
) such that the first two rows
span N(T
2
): this time f
2
= x
2
−4y
2
is suitable, and the diagram becomes:
T
2
(xy) T(x
2
−4y
2
) •
T(xy) x
2
−4y
2
xy
−4 4x + 8y •
2y − x x
2
−4y
2
xy
The final dot is now chosen so that the first row spans N(T): this time f
3
= x
2
+ 4y
2
+ 4xy works.
The result is a Jordan canonical basis and form for T
β =
−4, 2y − x, xy, 4x + 8y, x
2
−4y
2
, x
2
+ 4y
2
+ 4xy
, J = [T]
β
=
0 1 0
0 0 1
0 0 0
0 1
0 0
0
As previously, many other choices of cycle-generators f
1
, f
2
, f
3
are available; while these result in
different Jordan canonical bases, Corollary 3.18 assures us that we’ll always obtain the same canonical
form J.
14
Exercises 3.2 1. Let T be a linear operator whose characteristic polynomial splits. Suppose the
eigenvalues and the dot diagrams for the generalized eigenspaces K
λ
i
are as follows:
λ
1
= 2 λ
2
= 4 λ
3
= −3
• • •
• •
•
• •
•
•
• •
Find the Jordan form J of T.
2. Suppose T has Jordan canonical form
J =
2 1 0
0 2 1
0 0 2
2 1
0 2
3
3
(a) Find the characteristic polynomial of T.
(b) Find the dot diagram for each eigenvalue.
(c) For each eigenvalue find the smallest k
j
such that K
λ
j
= N(T − λ
j
I)
k
j
.
3. For each matrix A find a Jordan canonical form and an invertible Q such that A = QJQ
−1
.
(a) A =
−3 3 −2
−7 6 −3
1 −1 2
(b) A =
0 1 −1
−4 4 −2
−2 1 1
(c) A =
0 −3 1 2
−2 1 −1 2
−2 1 −1 2
−2 −3 1 4
4. For each linear operator T, find a Jordan canonical form J and basis β:
(a) T( f ) = f
′
on Span
R
{e
t
, te
t
, t
2
e
t
, e
2t
}
(b) T
f (x)
= x f
′′
(x) on P
3
(R )
(c) T( f ) = a f
x
+ b f
y
on Span
R
{1, x, y, x
2
, y
2
, xy}. How does your answer depend on a, b?
5. (Generalized Eigenvector Method for ODEs) Let A ∈ M
n
(R ) have an eigenvalue λ and sup-
pose β
v
0
= {v
k−1
, . . . , v
1
, v
0
} is a cycle of generalized eigenvectors for this eigenvalue. Show
that
x(t) := e
λt
k−1
∑
j=0
b
j
( t)v
j
satisfies x
′
( t) = Ax ⇐⇒
(
b
′
0
( t) = 0, and
b
′
j
( t) = b
j−1
( t) when j ≥ 1
Use this method to solve the system of differential equations
x
′
=
3 1 0 0
0 3 1 0
0 0 3 0
0 0 0 2
x
15
3.3 The Rational Canonical Form (non-examinable)
We finish the course with a very quick discussion of what can be done when the characteristic poly-
nomial of a linear map does not split. In such a situation, we may assume that
p(t) = (−1)
n
ϕ
1
( t)
m
1
···
ϕ
k
( t)
m
k
(∗)
where each ϕ
j
( t) is an irreducible monic polynomial over the field.
Example 3.21. The following matrix has characteristic equation p(t) = (t
2
+ 1)
2
(3 −t)
A =
0 −1 0 0 0
1 0 0 0 0
0 0 0 −1 0
0 0 1 0 0
0 0 0 0 3
!
∈ M
5
(R )
This doesn’t split over R since t
2
+ 1 = 0 has no real roots. It is, however, diagonalizable over C.
A couple of basic facts from algebra:
• Every polynomial splits over C: every A ∈ M
n
(C ) therefore has a Jordan form.
• Every polynomial over R factorizes into linear or irreducible quadratic factors.
The question is how to deal with non-linear irreducible factors in the characteristic polynomial.
Definition 3.22. The monic polynomial t
k
+ a
k−1
t
k−1
+ ···+ a
0
has companion matrix
0 0 0 ··· 0 −a
0
1 0 0 0 −a
1
0 1 0 0 −a
2
.
.
.
.
.
.
.
.
.
0 0 0 0 −a
k−2
0 0 0 ··· 1 −a
k−1
(when k = 1, this is the 1 ×1 matrix (−a
0
))
If T ∈ L(V) has characteristic polynomial (∗), then a rational canonical basis is a basis for which
[T]
β
=
C
1
O ··· O
O C
2
O
.
.
.
.
.
.
.
.
.
O O ··· C
r
where each C
j
is a companion matrix of some (ϕ
j
( t))
s
j
where s
j
≤ m
j
. We call [T]
β
a rational canonical
form of T.
We state the main result without proof:
Theorem 3.23. A rational canonical basis exists for any linear operator T on a finite-dimensional
vector space V. The canonical form is unique up to ordering of companion matrices.
Example (3.21 cont). The matrix A is already in rational canonical form: the standard basis is rational
canonical with three companion blocks,
C
1
= C
2
=
0 −1
1 0
, C
3
= (3)
16
Example 3.24. Let A =
4 −3
2 2
∈ M
2
(R ). Its characteristic polynomial
p(t) = t
2
−6t + 14 = (t −3)
2
+ 5
doesn’t split over R and so it has no eigenvalues. Instead simply pick a vector, x =
1
0
(say), define
y = Ax =
4
2
, let β = {x, y} and observe that
[L
A
]
β
=
0 −14
1 6
is a rational canonical form. Indeed this works for any x = 0: if β := {x, Ax}, then Cayley–Hamilton
forces
A
2
x = (6A −14I)x = −14x + 6Ax =⇒ [L
A
]
β
=
0 −14
1 6
whence β is a rational canonical basis and the form [L
A
]
β
is independent of x!
A systematic approach to finding rational canonical forms is similar to that for Jordan forms: for each
irreducible divisor of p(t), the subspace K
ϕ
= N
ϕ(T)
m
plays a role analogous to a generalized
eigenspace; indeed K
λ
= K
ϕ
for the linear irreducible factor ϕ(t) = λ −t!
We finish with two examples; hopefully the approach is intuitive, even without theoretical justifica-
tion.
Examples 3.25. If the characteristic polynomial of T ∈ L(R
4
) is
p(t) = (ϕ(t))
2
= (t
2
−2t + 3)
2
= t
4
−4t
3
+ 10t
2
−12t + 9
then there are two possible rational canonical forms; here is an example of each.
1. If A =
0 −15 0 −9
2 2 −3 0
0 −9 0 −6
−3 0 5 2
!
, then ϕ(A) = O is the zero matrix, whence N(ϕ(A)) = R
4
. Since ϕ(t)
isn’t the full characteristic polynomial, we expect there to be two independent cycles of length
two in the canonical basis. Start with something simple as a guess:
x
1
=
1
0
0
0
=⇒ x
2
= Ax
1
=
0
2
0
−3
=⇒ Ax
2
=
−3
4
0
−6
= −3x
1
+ 2x
2
Now make another choice that isn’t in the span of {x
1
, x
2
}:
x
3
=
0
0
1
0
=⇒ x
4
= Ax
3
=
0
−3
0
5
=⇒ Ax
4
=
0
−6
−3
10
= −3x
3
+ 2x
4
We therefore have a rational canonical basis β = {x
1
, x
2
, x
3
, x
4
} and
A =
1 0 0 0
0 2 0 −3
0 0 1 0
0 −3 0 5
0 −3 0 0
1 2 0 0
0 0 0 −3
0 0 1 2
1 0 0 0
0 2 0 −3
0 0 1 0
0 −3 0 5
−1
Over C, this example is diagonalizable. Indeed each of the 2 ×2 companion matrices is diago-
nalizable over C.
17
2. Let B =
0 0 2 1
1 1 −1 −1
0 1 −2 −16
0 0 1 5
. This time
ϕ(B) = B
2
−2B + 3I =
3 2 −7 −29
−1 1 4 13
1 −3 −6 −17
0 1 1 2
!
=⇒ N(ϕ(B)) = Span
3
−1
1
0
,
11
−2
0
1
Anything not in this span will suffice as a generator for a single cycle of length four: e.g.,
x
1
=
1
0
0
0
, x
2
= Bx
1
=
0
1
0
0
, x
3
= Bx
2
=
0
1
1
0
, x
4
= Bx
3
=
2
0
−1
1
Bx
4
=
−1
2
−14
4
= −9
1
0
0
0
+ 12
0
1
0
0
−10
0
1
1
0
+ 4
2
0
−1
1
We therefore have a rational canonical basis β = {x
1
, x
2
, x
3
, x
4
} and
B =
1 0 0 2
0 1 1 0
0 0 1 −1
0 0 0 1
0 0 0 −9
1 0 0 12
0 1 0 −10
0 0 1 4
1 0 0 2
0 1 1 0
0 0 1 −1
0 0 0 1
−1
In contrast to the first example, B isn’t diagonalizable over C. It has Jordan form J =
λ 1 0 0
0 λ 0 0
0 0 λ 1
0 0 0 λ
!
where λ = 1 + i
√
2.
18
