7 Canonical Forms

7.1 Jordan Forms, part I As previously, we remain concerned with the question of diagonalizability, more properly: given 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 not diagonalizable. We start with an example.

−8 4 Example 7.1. A = −25 12 ∈ M2(R) has characteristic equation

p(t) = (−8 − t)(12 − t) + 4 · 25 = t2 − 4t + 4 = (t − 2)2 and thus only one eigenvalue λ = 2. We quickly observe that the eigenspace has dimension 1:

−10 4 2 E = N = Span 2 −25 10 5

We cannot diagonalize A since there are insufﬁcient independent eigenvectors. However, if we con- 2 sider a basis β = {v1, v2} where v1 = 5 is an eigenvector, we see that [LA]β is upper-triangular, x which is better than nothing. How simple can we make this matrix? Let v2 = ( y ), then

−8x + 4y x −10x + 4y Av = = 2 + = 2v + (−5x + 2y)v 2 −25x + 12y y −25x + 10y 2 1 2 −5x + 2y =⇒ [L ] = A β 0 2

Since v2 isn’t parallel to v1, the only thing we cannot have is a diagonal matrix. The next best thing is to have the upper right corner be 1; for instance,

2 1 2 1 β = {v , v } = , =⇒ [L ] = 1 2 5 3 A β 0 2

The matrix [LA]β in this example is of a special type:

Deﬁnition 7.2. A Jordan block is a square matrix of the form

λ 1   ..   λ .  J =  .   .. 1 λ where λ ∈ F and all non-indicated entries are zero. Every 1 × 1 matrix is also a Jordan block.

A Jordan canonical form is a block-diagonal matrix diag(J1,..., Jm) where each Jk 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. Our interest is in the situation when a map isn’t diagonalizable.

1 Example 7.3. It can easily be checked that if         −1 2 3  1 1 1  A = −4 5 4 and β = {v1, v2, v3} = 2 , 1 , 0 −2 1 4  0 1 1  then 3 1 0 Av1 = 3v1, Av2 = v1 + 3v2, Av3 = 2v3 =⇒ [LA]β = 0 3 0 0 0 2

Thus β is a Jordan canonical basis for A (that is, for LA).

Generalized Eigenvectors We have two goals: showing that a linear map has a Jordan canonical basis, and computing such. For instance, in Example 7.3, if we were merely given A, how could we ﬁnd β? We brute-forced this in Example 7.1, but such is not a reasonable approach in general. Eigenvectors get you some of the distance but not all the way:

• v1 is an eigenvector in Example 7.1, but v2 is not;

• v1, v3 are eigenvectors in Example 7.3, but v2 is not. The practical question is therefore how we ﬁll out a Jordan canonical basis once we have all the eigenvectors. We now deﬁne the necessary objects.

Deﬁnition 7.4. Let T ∈ L(V) have an eigenvalue λ: its generalized eigenspace is

k [ k Kλ := {x ∈ V : (T − λI) (x) = 0 for some k ∈ N} = N (T − λI) k∈N

A non-zero element of Kλ is called a generalized eigenvector. n If A ∈ Mn(F) then its generalized eigenspaces are those of the linear map LA ∈ L(F ).

It is easy to check that our earlier Jordan canonical bases consist of generalized eigenvectors:

2 0 0 • In Example 7.1, we have one eigenvalue λ = 2. Since (A − 2I) = 0 0 is the zero matrix, it 2 is immediate that we have a single generalized eigenspace: K2 = R . Every non-zero vector is therefore a generalized eigenvector.

• In Example 7.3, we easily check that

2 (A − 3I)v1 = 0, (A − 3I) v2 = 0, (A − 2I)v3 = 0

so that the basis β consists of generalized eigenvectors. Indeed it can be seen that

K3 = Span{v1, v2}, K2 = E2 = Span{v3}

though to verify this thoroughly using only the deﬁnition is a little awkward.

2 In order to compute generalized eigenspaces directly, it is useful to invoke the main result of this section; we’ll delay the proof until later.

Theorem 7.5. Suppose T is a linear operator on a ﬁnite-dimensional vector space V over F. Suppose that the characteristic polynomial of T splits over F:

m1 mk p(t) = (λ1 − t) ··· (λk − t) (∗) where the λj are the distinct eigenvalues of T with multiplicities mj. Then:

m 1. For each eigenvalue, Kλ = N (T − λI) ;

2. V = Kλ1 ⊕ · · · ⊕ Kλk so that dim Kλj = mj for each eigenvalue.

Compare, especially part 2, with the statement on diagonalizability from the start of the course. . .

Observe how Example 7.3 works in this language:

p(t) = det(A − tI) = (3 − t)2(2 − t)1 2 −1 −1 1 1 2   N (A − 3I) = N 0 0 0  = Span 2 , 1 = K3 =⇒ dim K3 = 2 2 −1 −1  0 1  1 1 N (A − 2I) = Span 0 = E2 = K2 =⇒ dim K2 = 1 1 3 R = K3 ⊕ K2

5 2 −1 Example 7.6. We ﬁnd the generalized eigenspaces of the matrix A = 0 0 0 9 6 −1 The characteristic polynomial is

5 − t −1 p(t) = det(A − λI) = −t = −t(t2 − 5t + t − 5 + 9) = −(0 − t)1(2 − t)2 9 −1 − t whence there are two eigenvalues.

1 1 • λ = 0 has multiplicity1 and so K0 = N (A − 0I) = N (A) = Span −1 . This is in fact E0. 3 • λ = 2 has multiplicity2, and so

2 3 2 −1 0 −4 0 1 0 2   K2 = N (A − 2I) = N 0 −2 0  = N 0 4 0 = Span 0 , 0 9 6 −3 0 −12 0  0 1 

1 Observe that the corresponding eigenspace is only one-dimensional; E2 = Span 0 , and so the 3 matrix is not diagonalizable.

3 Observe also that R = K0 ⊕ K2 in accordance with the Theorem.

3 Properties of Generalized Eigenspaces and the Proof of Theorem 7.5 Quite a lot of work is required in order to justify our main result. You might want to skip the proofs at ﬁrst reading.

Lemma 7.7. Let λ be an eigenvalue of an operator T on a vector space V. Then:

1. Eλ is a subspace of Kλ, which is indeed a subspace of V;

2. Kλ is T-invariant;

3. If µ 6= λ and Kλ is ﬁnite-dimensional, then T − µI restricted to Kλ is an isomorphism of Kλ. In particular, Kλ does not contain the eigenspace Eµ.

Proof. Part 1 is an easy exercise.

k 2. Let x ∈ Kλ, then ∃k ∈ N such that (T − λI) (x) = 0. But then

(T − λI)kT(x) = (T − λI)kT(x) − λx + λx = (T − λI)k+1(x) + λ(T − λI)k(x)

= 0 =⇒ T(x) ∈ Kλ

3. Since Kλ is T-invariant, it is also (T − µI)-invariant. It is therefore enough to show that T − µI is injective on Kλ. Suppose not, then ∃x ∈ Kλ non-zero such that (T − µI)(x) = 0. Let k ∈ N be k k−1 minimal such that (T − λI) (x) = 0, and let y = (T − λI) (x). Clearly y ∈ Eλ. But then

(T − µI)(y) = (T − λI + (λ − µ)I(T − λI)k−1(x) = (T − λI)k−1(T − λI + (λ − µ)I(x) = (T − λI)k−1(T − µI)(x) = 0

Thus y ∈ Eµ ∩ Eλ =⇒ y = 0, contradicting the minimality of k.

Now we come to the main proof: it’s a bit of a monster, but things become much easier once we’re through with it.

Proof of Theorem 7.5. Fix an eigenvalue λ. Since Kλ is T-invariant, the characteristic polynomial pλ(t) of the restriction TKλ divides that of T. Since Kλ contains no eigenvectors except those in Eλ, we see that

dim Kλ pλ(t) = (λ − t) =⇒ dim Kλ ≤ m

By the Cayley–Hamilton Theorem, TKλ satisﬁes its characteristic polynomial, whence

dim Kλ m (λI − T) (x) = 0, ∀x ∈ Kλ =⇒ Kλ ⊆ N (T − λI)

m That N (T − λI) ⊆ Kλ is by deﬁnition of Kλ. We’ve therefore established part 1.

For part 2, we prove by induction on the number of distinct eigenvalues of T. Let dim V = n and suppose the characteristic polynomial of T splits as in (∗). (Base case) If T has one eigenvalue, then p(t) = (λ − t)n. By Cayley–Hamilton, (T − λI)n(x) = 0 for all x ∈ V and so V = Kλ.

4 (Induction step) Fix k ∈ N and suppose the result is true for any map with k − 1 distinct eigenvalues. Suppose T ∈ L(V) has k distinct eigenvalues. Let λk have multiplicity mk and deﬁne

mk W = R(T − λkI)

The subspace W has the following properties:

• W is T-invariant,

mk w ∈ W =⇒ ∃x such that w = (T − λkI) (x)

mk+1 =⇒ T(w) = (T − λkI) (x) + λkw ∈ W

whence the characteristic polynomial of the restriction TW divides that of T.

• By Lemma 7.7, we have isomorphisms (T − λ I)K ∈ L(K ) for each j 6= k. Thus each K ≤ W k λj λj λj

so that each λj is an eigenvalue of the restriction TW with generalized eigenspace Kλj .

• W ∩ Kλk = {0} so that λk is not an eigenvalue of TW.

Since TW has k − 1 distinct eigenvalues, the induction hypothesis applies and we see that

dim K dim K = ⊕ · · · ⊕ ( ) = ( − ) λ1 ··· ( − ) λk−1 W Kλ1 Kλk−1 , pTW t λ1 t λk−1 t

where each dim Kλj = mj. Since W ∩ Kλk = {0} it is enough ﬁnally for us to use the rank–nullity theorem to count dimensions:

mk mk n = rank(T − λkI) + null(T − λkI) = dim W + dim Kλk

whence dim Kλk = n − ∑j6=k dim Kλj = n − ∑j6=k mj = mk. We conclude that

V = Kλ1 ⊕ · · · ⊕ Kλk

This completes the induction step and thus the proof of part 2. Whew!

Cycles of Generalized Eigenvectors By Theorem 7.5, for a linear map whose characteristic polynomial splits, there exists a basis of generalized eigenvectors. This isn’t quite enough to guarantee a Jordan canonical form. To see how things work, suppose, for instance, that β = {v1, v2, v3} is a basis of a generalized eigenspace K5 for which the restriction of a linear map T has Jordan block matrix 5 1 0

[TK5 ]β = 0 5 1 0 0 5

This means that T(v1) = 5v1,T(v2) = v1 + 5v2,T(v3) = v2 + 5v3, or equivalently

(T − 5I)(v1) = 0, (T − 5I)(v2) = v1, (T − 5I)(v3) = v2 The crucial observation is that the basis β is generated by repeatedly applying T − 5I to the generalized eigenvector v3.

5 Deﬁnition 7.8. A cycle of generalized eigenvectors for a linear operator T is a set

n p−1 o βx := (T − λI) (x),..., (T − λI)(x), x

p where x ∈ Kλ is non-zero and p is minimal such that (T − λI) (x) = 0.

The following result tells us everything we need regarding the existence of a Jordan canonical basis. The idea is straightforward but the arguments are somewhat tedious, so we omit them.

Theorem 7.9. Given a cycle βx of generalized eigenvectors as in the Deﬁnition:

1. βx is linearly independent and thus a basis of Span βx.

2. Span βx is T-invariant. The matrix of T restricted to Span βx is the p × p Jordan block with λ’s down the diagonal.

3. If Kλ is ﬁnite dimensional, then there exists a basis βλ = βx1 ∪ · · · ∪ βxj of Kλ consisting of ﬁnitely many linearly independent cycles of generalized eigenvectors.

4. If V is ﬁnite dimensional and the characteristic polynomial of T splits, then there exists a Jordan canonical basis, being the union of the bases βλ.

The basic idea for finding a Jordan canonical basis is therefore to find the generalized eigenspaces and to play with cycles until you find a basis of each Kλ. It should be clear that many choices of canonical basis exist for a given map, especially if any generalized eigenspace has dimension ≥ 2!

7 1 −4 Example 7.10. Consider the matrix A = 0 3 0 ∈ M3(R). This has characteristic equation 8 1 −5

7 − t −4 p(t) = (3 − t) = (3 − t)(t2 − 2t − 3) = −(t + 1)(t − 3)2 8 −5 − t Since p(t) splits, there must exist a Jordan canonical basis. We treat the eigenvalues separately:

• dim K−1 = 1 =⇒ K−1 = E−1 is the eigenspace. The usual calculation shows that 1 K−1 = Span 0 2

• Since dim K3 = 2, we compute 2 2 4 1 −4 −16 0 16 n 1 0 o K2 = N (A − 3I) = N 0 0 0 = N 0 0 0 = Span 0 , 1 8 1 −8 −32 0 32 1 0 The ﬁrst is an eigenvector, while the second satisﬁes 0 1 (A − 3I) 1 = 0 0 1 0 whence the cycle generated by 1 forms a basis of K3. 0 n 1 1 0 o We conclude that β = 0 , 0 , 1 is a Jordan canonical basis for A, and that 2 1 0 −1 −1 1 1 0 −1 0 0 1 1 0 A = QJQ = 0 0 1 0 3 1 0 0 1 2 1 0 0 0 3 2 1 0

6 Even such a simple example turned out to be a lot of work. Here are two more.

1 0 2 Examples 7.11. 1. The matrix A = 0 1 6 ∈ M3(R) has characteristic equation 6 −2 1

1 − t 6 0 1 − t p(t) = (1 − t) + 2 = (1 − t) (1 − t)2 + 12 − 12 = (1 − t)3 −2 1 − t 6 −2 3 With only one eigenvalue and a splitting characteristic equation, we see that K1 = R . Simply choose any vector in R3 and see what A − λI does to it. For instance, with x = i,

βx = {12i + 36j, 6k, i}

a 12 results in a Jordan canonical basis of R3 where only 36 is an eigenvector. We conclude 0

−1 −1 12 0 1 1 1 0 12 0 1 A = QJQ = 36 0 0 0 1 1 36 0 0 0 6 0 0 0 1 0 6 0

d 2 3 2. Let T = dx on P3(R). With respect to the standard basis e = {1, x, x , x } we have

0 1 0 0 0 0 2 0 A = [T]e = 0 0 0 3 0 0 0 0

This has only one eigenvalue λ = 0 and thus a single generalized eigenspace K0 = P3(R). Now choose f (x) = x3 to generate a Jordan canonical basis:

0 1 0 0 6 0 0 0 −1 0 1 0 0 6 0 0 0 2 3 0 0 1 0 0 6 0 0 0 0 2 0 0 6 0 0 β = {6, 6x, 3x , x } =⇒ [T]β = 0 0 0 1 = 0 0 3 0 0 0 0 3 0 0 3 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1

aRemember to list the basis in the reverse order to how it is generated!

Exercises. 7.1.1. For each matrix A, find a basis for each generalized eigenspace Kλ consisting of a union of disjoint cycles of generalized eigenvectors, and thus find a Jordan canonical form J and invertible Q such that A = QJQ−1. 1 1 1 2 (a) A = (b) A = −1 3 3 2 2 1 0 0 11 −4 −5  0 2 1 0 (c) A = 21 −8 −11 (d) A =   0 0 3 0 3 −1 0 0 1 −1 3 7.1.2. Find a Jordan canonical basis for each linear map T: 0 (a)T ∈ L(P2(R)) defined by T( f (x)) = 2 f (x) − f (x) (b)T ( f ) = f 0 defined on Span{1, t, t2, et, tet} T (c)T (A) = 2A + A defined on M2(R) 7.1.3. Prove part 1 of Lemma 7.7.

7.1.4. In the induction step of the proof of Theorem 7.5 part 2, explain why W ∩ Kλk = {0}. 7.1.5. Prove parts 1 and 2 of Theorem 7.9.

7 7.2 Jordan Forms, part II 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 7.12. Consider the following three Jordan canonical forms A, B, C ∈ M3(R) corresponding to the characteristic polynomial p(t) = (5 − t)3:

5 0 0 5 1 0 5 1 0 A = 0 5 0 B = 0 5 0 C = 0 5 1 0 0 5 0 0 5 0 0 5

The difference between A, B, C lies in the possible lengths of cycles of generalized eigenvectors, and therefore in how many cycles we require to construct a Jordan basis.

3 3 • A has eigenspace E5 = K5 = R : since (A − 5I)v = 0 for all v ∈ R , we have maximum cycle-length one. We therefore need three cycles to construct a Jordan basis, for instance,

βe1 = {e1}, βe2 = {e2}, βe3 = {e3} =⇒ β = βe1 ∪ βe2 ∪ βe3 = {e1, e2, e3}

• B has eigenspace E5 = Span{e1, e3}, the maximum length of a cycle of generalized eigenvectors is two: for instance, if b 6= 0 (v 6∈ E5), then

a b v = b =⇒ (B − 5I)v = 0 =⇒ (B − 5I)2v = 0 c 0

generates a cycle of length two. We therefore need two cycles to construct a Jordan basis: e.g.

βe2 = {(B − 5I)e2, e2} = {e1, e2}, βe3 = {e3} =⇒ β = βe2 ∪ βe3 = {e1, e2, e3}

• C has eigenspace E5 = Span{e1}, the maximum length of a cycle of generalized eigenvectors is three: if c 6= 0, then

a b c v = b =⇒ (C − 5I)v = c , (C − 5I)2v = 0 , (C − 5I)3v = 0 c 0 0

generates a three-cycle which is Jordan basis: e.g.

2 βe3 = {(C − 5I) e3, (C − 5I)e3, e3} = {e1, e2, e3} =⇒ β = βe3 = {e1, e2, e3}

3 Now suppose that dim VR = 3 and that T ∈ L(V) has characteristic polynomial p(t) = (5 − t) . By the previous section, one of the above matrices A, B, C is a Jordan canonical form for T. We will be able to decide which by observing the possible patterns of cycle-lengths. To do this generally requires some terminology. . .

8 Deﬁnition 7.13. Let Kλ be a generalized eigenspace of a linear operator T ∈ L(V) on a ﬁnite- dimensional vector space. Assume:

1. βλ is a Jordan canonical basis of TKλ with Jordan canonical form J = [TKλ ]βλ .

2. βλ = γ1 ∪ · · · ∪ γk is a union of disjoint cycles of generalized eigenvectors, with each γj corresponding to a Jordan block Jj.

3. pj := γj deﬁnes a non-increasing sequence (the Jordan blocks do not increase in size).

pj−1 4. γj = {(T − λI) vj,..., vj} has ‘end vector’ vj.

Represent the cycles γj as consecutive sequences of vertical dots: we call this the dot diagram of TKλ , where each column represents the elements of γj arranged vertically with vj at the bottom.

While computing the dot diagram for a given linear map is the main goal, converting dot diagrams to a Jordan form is essentially trivial.

Example 7.14. Suppose T ∈ L(V) has the following eigenvalues and dot diagrams:

λ1 = −4 λ2 = 7 λ3 = 12 •••• •• ••• •• •• •

Then generalized eigenspaces of T satisfy:

2 • K−4 = N (T + 4I) and dim K−4 = 6;

3 • K7 = N (T − 7I) and dim K7 = 5;

• K12 = N (T − 12I) = E12 and dim K12 = 3; Finally, T has a Jordan canonical basis β with respect to which its Jordan canonical form is   −4 1  0 −4     −   4 1   −   0 4     −4     −4     7 1 0  [T] =   β  0 7 1     0 0 7       7 1     0 7     12     12  12

9 Theorem 7.15. Suppose βλ is a Jordan canonical basis of TKλ as described in the deﬁnition, and th suppose the j row of the dot diagram has rj 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. r1 = null(T − λI) = dim V − rank(T − λI) j−1 j 3. When j > 1, rj = rank(T − λI) − rank(T − λI)

Example 7.16. We describe the dot diagrams of the three matrices from our earlier example, along with the corresponding vectors in the Jordan canonical basis β and the numbers rj:

A : ••• v1 v2 v3 e1 e2 e3

Since A − 5I is the zero matrix, rank(A − 5I) = 0 and r1 = 3. The dot diagram has one row.

B : •• (B − 5I)v1 v2 e1 e3 • v1 e2

0 1 0 Row 1: B − 5I = 0 0 0 =⇒ rank(B − 5I) = 1 and r1 = 3 − 1 = 2. The ﬁrst row {e1, e3} is 0 0 0 a basis of E5 = N (B − 5I). 2 2 Row 2: (B − 5I) is the zero matrix, whence rank(B − 5I) = 0 and r2 = 1 − 0 = 1.

2 C : • (C − 5I) v1 e1 • (C − 5I)v1 e2 • v1 e3

0 1 0 Row 1: C − 5I = 0 1 0 =⇒ rank(C − 5I) = 2 and r1 = 3 − 2 = 1: the ﬁrst row {e1} is a 0 0 0 basis of E5 = N (C − 5I). 2 0 0 1 Row 2: (C − 5I) = 0 0 0 =⇒ rank(C − 5I) = 2 and r2 = 2 − 1 = 1: the ﬁrst two rows 0 0 0 2 {e1, e2} form a basis of N (C − 5I) . 3 3 Row 3: (C − 5I) is the zero matrix, whence rank(C − 5I) = 0 and r3 = 1 − 0 = 1.

pj−1 Proof. 1. Every vector represented in the ﬁrst row of the dot diagram has the form (T − λI) (vj) and so lies in the nullspace of T − λI. All other vectors in βλ simply get moved up to the next row by T − λI. The result follows immediately.

2. By part 1, r1 = null(T − λI) = dim V − rank(T − λI) 3. More generally,

j j−1 rj = (r1 + ··· + rj) − (r1 + ··· + rj−1) = null(T − λI) − null(T − λI) = rank(T − λI)j−1 − rank(T − λI)j

10 Corollary 7.17. 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.

Armed with our technology, we now have a slightly more systematic method for ﬁnding Jordan canonical bases.

6 2 −4 −6 0 3 0 0 Example 7.18. The matrix A = 0 0 3 0 is easily seen to have characteristic equation 2 1 −2 −1

6 − t −6 p(t) = (3 − t)2 = (2 − t)(3 − t)3 2 −1 − t

We have two generalized eigenspaces:

4 2 −4 −6 3 0 1 0 0 0 • K2 = E2 = N (A − 2I) = N 0 0 1 0 = Span 0 . The dot diagram is simply • which 2 1 −2 −3 2 corresponds to this single eigenvector.

3 • K3 = N (A − 3I) . To ﬁnd the dot diagram, compute powers of A − 3I:

3 2 −4 −6 0 0 0 0 Row 1: A − 3I = 0 0 0 0 has rank 2 and the ﬁrst row has r1 = 4 − 2 = 2 entries. 2 1 −2 −4 −3 0 0 6 2 0 0 0 0 Row 2: (A − 3I) = 0 0 0 0 has rank 1 and the second row has r2 = 2 − 1 = 1 entries. −2 0 0 4

Since we now have three dots (equalling dim K3), the dot diagram for K3 is •• • 2 0 0 2 The ﬁrst row is a basis of E3 = N (A − 3I) = Span 0 , 1 . We may choose anything in 1 0 2 0 0 2 0 2 1 the second row to complete N (A − 3I) = Span 0 , 1 , 0 . Indeed, since 1 0 0

0 2 1 0 (A − 3I) 0 = 0 0 1 we now have suitable cycles and a Jordan canonical basis/form:

3 2 0 0 3 2 0 0 2 0 0 0 3 2 0 0 −1 0 0 1 2 0 0 1 2 0 3 1 0 0 0 1 2 β = 0 , 0 , 0 , 1 A = 0 0 0 1 0 0 3 0 0 0 0 1 2 1 0 0 2 1 0 0 0 0 0 3 2 1 0 0 Other choices are available. For instance, in the ﬁnal step if we’d chosen the two-cycle generated by e3, we’d instead obtain a different Jordan basis, but the same canonical form

3 −4 0 0 3 −4 0 0 2 0 0 0 3 −4 0 0 −1 ˜ 0 0 0 2 0 0 0 2 0 3 1 0 0 0 0 2 β = 0 , 0 , 1 , 1 A = 0 0 1 1 0 0 3 0 0 0 1 1 2 −2 0 0 2 −2 0 0 0 0 0 3 2 −2 0 0

11 Finally we do one example without matrices.

2 2 ∂ f ∂ f Example 7.19. Let e = {1, x, y, x , y , xy} and deﬁne T( f (x, y)) = 2 ∂x − ∂y to be a linear operator on V = Span e. Then

 0 2 −1 0 0 0   0 0 0 8 2 −4  0 0 0 4 0 −1 0 0 0 0 0 0 [ ] = 0 0 0 0 −2 2 =⇒ ( ) = 6 [ 2] = 0 0 0 0 0 0 T e  0 0 0 0 0 0  p t t , T e  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

There is exactly one eigenvalue λ = 0 and therefore one generalized eigenspace K0 = V. We could keep working with matrices, but it is easy instead to translate the nullspaces of the matrices back to subspaces of V, from which the necessary data can be read off:

2 2 N (T − 0I) = Span{1, x + 2y, x + 4y + 4xy} null T = 3, rank T = 3, r1 = 3 2 2 2 2 2 N (T ) = Span{1, x, y, x + 2xy, 2y + xy} null T = 5, rank T = 1, r2 = 3 − 1 = 2

We now have five dots; since dim K0 = 6, the last row has one, and the dot diagram is ••• •• • 2 2 Since the first two rows span N (T ), we may choose any v1 6∈ N (T ) for the final dot: for instance v1 = xy is suitable, from which the first column of the dot diagram becomes

T2(xy) •• −4 •• T(xy) • 2y − x • xy xy

Now choose the second dot on the second row to be anything in N (T2) such that the ﬁrst two rows 2 2 2 span N (T ): this time v2 = x − 4y is suitable, and the diagram becomes:

T2(xy) T(x2 − 4y2) • −4 4x + 8y • T(xy) x2 − 4y2 2y − x x2 − 4y2 xy xy

2 2 The ﬁnal dot is now chosen so that the ﬁrst row spans N (T): this time v3 = x + 4y + 4xy works. The result is a Jordan canonical basis and form for T

 0 1 0   0 0 1     0 0 0  β = {−4, 2y − x, xy, 4x + 8y, x2 − 4y2, x2 + 4y2 + 4xy} [T] =   β  0 1     0 0  0

As previously, many other choices of the cycle-generators v1, v2, v3 are available: these result in different Jordan canonical bases, but the same canonical form [T]β.

12 Exercises. 7.2.1. Let T be a linear operator whose characteristic polynomial splits. Suppose the

eigenvalues and the dot diagrams for the generalized eigenspaces Kλi are as follows:

λ1 = 2 λ2 = 4 λ3 = −3 ••• •• •• •• • • •

Find the Jordan form J of T. 7.2.2. Suppose T has Jordan canonical form   2 1 0  0 2 1     0 0 2      J =  2 1     0 2     3  3

(a) Find the characteristic polynomial of T. (b) Find the dot diagram for each eigenvalue. pi (c) For each eigenvalue find the smallest pi such that Kλi = N (T − λiI) . 7.2.3. For each matrix A find a Jordan canonical form and an invertible Q such that A = QJQ−1. ! − − − 0 −3 1 2 3 3 2 0 1 1 −2 1 −1 2 (a) A = −7 6 −3 (b) A = −4 4 −2 (c) A = − − 1 −1 2 − 2 1 1 2 2 1 1 −2 −3 1 4 7.2.4. For each linear operator T, find a Jordan canonical form J and basis β: t t 2 t 2t 0 (a) V = SpanR{e , te , t e , e } and T( f ) = f 00 (b)T ( f (x)) = x f (x) on P3(R) 2 2 (c)T ( f (x, y)) = a fx + b fy on V = SpanR{1, x, y, x , y , xy}. How does your answer depend on a, b? 7.2.5. We apply the method of generalized eigenvectors to the solution of a linear system of differen-

tiatial equations Let A ∈ Mn(R) have an eigenvalue λ and suppose βv0 = {vp−1,..., v1, v0} p is a cycle of generalized eigenvectors for this eigenvalue (note (A − λI) v0 = 0). Show that

p−1 λt x(t) := e ∑ bj(t)vj j=0

0 0 0 satisﬁes x (t) = Ax ⇐⇒ b0(t) = 0 and bj(t) = bj−1(t) for all j = 1, . . . , p − 1. Hence solve the system of differential equations

3 1 0 0 0 0 3 1 0 x =   x 0 0 3 0 0 0 0 2

13 7.4 The Rational Canonical Form (non-examinable) We ﬁnish the course with a very quick discussion of what can be done when the characteristic polynomial of a linear map does not split. In such a situation, we may assume that

n m1 mk p(t) = (−1) (φ1(t)) ··· (φk(t)) (∗) where each φj(t) is an irreducible monic polynomial over the ﬁeld.

Example 7.20. The following matrix has characteristic equation p(t) = (t2 + 1)2(3 − t)

0 −1 0 0 0 ! 1 0 0 0 0 A = 0 0 0 −1 0 ∈ M5(R) 0 0 1 0 0 0 0 0 0 3

This doesnt split over R since t2 + 1 = 0 has no real roots. It is, however, diagonalizable over C.

A couple of basic facts from algebra:

• Every polynomial splits over C: thus every A ∈ Mn(C) has a Jordan form.

• Every polynomial over R factorizes into linear or irreducible quadratic factors.

The question is how to deal with irreducible polynomial factors in the characteristic polynomial.

k k−1 Deﬁnition 7.21. The monic polynomial t + ak−1t + ··· + a0 has companion matrix

 0 0 0 ··· 0 −a0  1 0 0 0 −a1  0 1 0 0 −a2   . .  (when k = 1, this is the 1 × 1 matrix (−a0))  . .  0 0 0 0 −ak−2 0 0 0 ··· 1 −ak−1

Suppose T ∈ L(V) has characteristic polynomial (∗). A rational canonical basis β for T is a basis of V such that

 C1 O ··· O  OC2 O [T]β =  . .  . . OO ··· Cr

sj where each Cj is a companion matrix of some (φj(t)) . We call [T]β a rational canonical form of T.

We state the main result without proof:

Theorem 7.22. 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 7.23. The matrix A in Example 7.20 is already in rational canonical form: the standard basis is rational canonical with three companion blocks,

0 −1 C1 = C2 = 1 0 , C3 = (3)

14 A systematic approach to finding rational canonical forms is similar to that for Jordan forms as seen in the previous section. The Cayley–Hamilton Theorem can be used to establish the existence of m subspaces Kφ = N (φ(T)) for each irreducible divisor of p(t). These play a role analogous to generalized eigenspaces: indeed Kφ is a generalized eigenspace if deg φ = 1. We finish with two examples: hopefully the approach is intuitive, even without full theoretical justification.

Examples 7.24. If the characteristic equation of A ∈ M4(R) is p(t) = (φ(t))2 = (t2 − 2t + 3)2 = t4 − 4t3 + 10t2 − 12t + 9 then there are two possible rational canonical forms: here is an example of each.

0 −15 0 −9 ! 2 2 −3 0 1. Let A = 0 −9 0 −6 . We can check that φ(A) = O is the zero matrix. Since φ(t) isn’t the −3 0 5 2 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:

1 0 −3 0 2 4 v1 = 0 =⇒ v2 = Av1 = 0 =⇒ Av2 = 0 = −3v1 + 2v2 0 −3 −6

Now make another choice that isn’t in the span of {v1, v2}: 0 0 0 0 −3 −6 v3 = 1 =⇒ v4 = Av3 = 0 =⇒ Av4 = −3 = −3v3 + 2v4 0 5 10

We therefore have a rational canonical basis β = {v1, v2, v3, v4} and

1 0 0 0 0 −3 0 0 1 0 0 0 −1 0 2 0 −3 1 2 0 0 0 2 0 −3 A = 0 0 1 0 0 0 0 −3 0 0 1 0 0 −3 0 5 0 0 1 2 0 −3 0 5 Over C, this example is in fact diagonalizable. Indeed each of the 2 × 2 companion matrices is diagonalizable over C. 0 0 2 1 1 1 −1 −1 2. Let B = 0 1 −2 −16 . This time 0 0 1 5 3 2 −7 −29 ! 3 11 ( ) = 2 − + = −1 1 4 13 =⇒ N ( ( )) = −1 −2 φ B B 2B 3I 1 −3 −6 −17 φ B Span 1 , 0 0 1 1 2 0 1 Anything not in this span will sufﬁce as a generator for a single cycle of length four: e.g., 1 0 0 2 0 1 1 0 v1 = 0 , v2 = Bv1 = 0 , v3 = Bv2 = 1 , v4 = Bv3 = −1 0 0 0 1 −1 1 0 0 2 2 0 1 1 0 Bv4 = −14 = −9 0 + 12 0 − 10 1 + 4 −1 4 0 0 0 1

We therefore have a rational canonical basis β = {v1, v2, v3, v4} and

1 0 0 2 0 0 0 −9 1 0 0 2 −1 0 1 1 0 1 0 0 12 0 1 1 0 B = 0 0 1 −1 0 1 0 −10 0 0 1 −1 0 0 0 1 0 0 1 4 0 0 0 1 λ 1 0 0 ! C 0 λ 0 0 In contrast to the√ ﬁrst example, B isn’t diagonalizable over . It has Jordan form J = 0 0 λ 1 where λ = 1 + i 2. 0 0 0 λ