3 Invariant Subspaces

DEFINITIONS 3.1

Subspaces V1,...,Vt of V are called independent if

v1 + + vt = 0 v1 = 0, . . . , vt = 0 ··· ⇒ v1 V1, . . . , vt Vt. ∀ ∈ ∈

We say that V is the (internal) direct sum of the subspaces V1,...,Vt if

(a) V1,...,Vt are independent and

(b) V = V1 + + Vt. ··· I.e. every element v V is uniquely expressible as ∈

v = v1 + + vt ··· with vi Vi. ∈ Then V is isomorphic to the (external) direct sum V1 Vt under ⊕ · · · ⊕ the isomorphism v (v1, . . . , vt) and we write V = V1 Vt. 7→ ⊕ · · · ⊕ THEOREM 3.1 If V = V1 Vt (an internal direct sum) and β1, . . . , βt are bases for ⊕ · · · ⊕ V1,...,Vt respectively, then

β = β1 βt, ∪ · · · ∪ the sequence formed by juxtaposing the separate bases, is a basis for V . Also dim V = dim V1 + + dim Vt. ··· Proof: Left as an exercise. DEFINITION. Let T : V V be a LT and W a subspace of V . Then if 7→ w W T (w) W, ∈ ⇒ ∈ we say W is a T -invariant subspace of V . We can then consider the linear transformation TW : W W deﬁned by →

TW (w) = T (w) w W. ∀ ∈ 53 If β0 is a basis for W , 0 W V , and β is an extension to a basis of V , then { } ⊂ ⊂ β0 β [TW ] B1 " β0 # [T ]β = . 0 B2 A situation of great interest is when we have T -invariant subspaces W1,...,Wt and V = W1 Wt. For if β = β1 βt, where βi is ⊕ · · · ⊕ ∪ · · · ∪ a basis for Wi, we see that

β β1 βt [T ] = [TW ] [TW ] . β 1 β1 ⊕ · · · ⊕ t βt There are two important examples of T –invariant subspaces that arise in our study of Jordan and rational canonical forms - Ker pt(T ) and T –cyclic subspaces.

3.1 T –cyclic subspaces DEFINITION 3.1 The unique monic polynomial f in F [x] of least degree satisfying

f(T )(v) = 0 is called the minimum polynomial of the vector v V relative to the ∈ transformation T : V V and is denoted mT,v. 7→

Then f(T )(v) = 0 mT,v f, so mT,v mT . Also mT,v = 1 v = 0 and ⇒ | | ⇔ so if v = 0, deg mT,v 1. 6 ≥ 0 0 EXAMPLE. Let T = T , where A = . Also let A 1 0

1 0 v = and v = . 1 0 2 1

0 2 0 Then Av1 = = c0v1, so mT, v = x c0. Next A v1 = , so 1 6 1 6 − 0 0 m = x2. Also Av = , so m = x. T, v1 2 0 T, v2

DEFINITION 3.2 (T -cyclic subspace generated by v.) If v V , the set of all vectors of the form f(T )(v), f F [x], forms a subspace∈ of V called the T -cyclic subspace generated by v.∈ It is denoted by CT,v.

54 PROOF. Exercise. Also, CT,v is a T -invariant subspace of V . For

w CT,v w = f(T )(v) ∈ ⇒ T (w) = T (f(T )(v)) = (T f(T ))(v) = ((xf)(T ))(v) CT,v. ⇒ ∈ We see that v = 0 if and only if CT,v = 0 . { } THEOREM 3.2 Let v = 0, v V . Then CT,v has the basis β 6 ∈ 2 k 1 v, T (v),T (v),...,T − (v) where k = deg mT,v.(β is called the T -cyclic basis generated by v. ) Note that dim CT,v = deg mT,v. Finally, β TC = C(mT,v), T,v β the companion matrix of the minimum polynomial of v. PROOF.

1. The T -cyclic basis is a basis for CT,v: Spanning: Let w v, T (v),...,T k 1(v) , so ∈ h − i k 1 w = w0v + w1T (v) + + wk 1T − (v) − ··· k 1 = (w0IV + + wk 1T − )(v) ··· − = g(T )(v), k 1 where g = w0 + + wk 1x − , so w CT,v. Hence ··· − ∈ k 1 v, T (v),...,T − (v) CT,v. h i ⊆ Conversely, suppose that w CT,v so ∈ w = f(T )(v) and f = qmT,v + r k 1 where r = a0 + a1x + + ak 1x − and a0, . . . , ak 1 F . So ··· − − ∈ f(T )(v) = q(T )mT,v(T )(v) + r(T )(v) k 1 = q(T )mT,v(T )(v) + a0v + a1T (v) + + ak 1T − (v) − k 1 ··· = a0v + a1T (v) + + ak 1T − (v) − k ···1 v, T (v),...,T − (v) . ∈ h i

55 Independence: Assume k 1 a0v + a1T (v) + + ak 1T − (v) = 0, ··· − where a0, . . . , ak 1 F ; that is, f(T )(v) = 0 where − ∈ k 1 f = a0 + a1x + + ak 1x − . ··· −

Hence mT,v f and since |

deg f = k 1 < k = deg mT, v, −

we have f = 0 and thus ai = 0 i. ∀ β 2. [TCT,v ]β = C(mT,v):

Let L = TCT,v , the restriction of T to CT,v. β We want to ﬁnd [L]β. So

2 k 1 L(v) = T (v) = 0v + 1T (v) + 0T (v) + + 0T − (v) 2 2 ··· k 1 L(T (v)) = T (v) = 0v + 0T (v) + 1T (v) + + 0T − (v) ··· . . k 2 k 1 2 k 1 L(T − (v)) = T − (v) = 0v + 0T (v) + 0T (v) + + 1T − (v) ··· k 1 k Finally, to calculate L(T − (v)) = T (v), we let

k 1 k mT,v = a0 + a1x + + ak 1x − + x . ··· −

Then mT,v(T )(v) = 0 and hence

k 1 k k 1 L(T − (v)) = T (v) = a0v a1T (v) ak 1T − (v). − − − · · · − − Hence 0 0 ... 0 a0  −  1 0 a1 [L]β = − = C(m ), β  . .. .  T,v  . . .     0 0 1 ak 1  ··· − − as required.

56 THEOREM 3.3 k Suppose that mT, v = (x c) . Then the vectors − k 1 v, (T cIV )(v),..., (T cIV ) − (v) − − form a basis β for W = CT, v which we call the elementary Jordan basis. Also β [TW ]β = Jk(c). k More generally suppose mT,v = p , where p is a monic irreducible polynomial in F [x], with n = deg p. Then the vectors n 1 v, T (v),...,T − (v) n 1 p(T )(v), T p(T )(v),...,T − p(T )(v) ...... k 1 k 1 n 1 k 1 p − (T )(v), T p − (T )(v),...,T − p − (T )(v), form a basis for W = CT,v, which reduces to the elementary Jordan basis when p = x c. Also − β k [TW ]β = H(p ), where H(pk) is a hypercompanion matrix, which reduces to the elementary Jordan matrix Jk(c) when p = x c: − C(p) 0 0  NC(p) ··· 0  ··· H(pk) =  0 N 0  ,  ···   . . . .   . . . .     0 NC(p)  ··· where there are k blocks on the diagonal and N is a square matrix of same size as C(p) which is everywhere zero, except in the top right–hand corner, where there is a 1. The overall eﬀect is an unbroken subdiagonal of 10s.

3.1.1 A nice proof of the Cayley-Hamilton theorem (From Insel, Friedberg and Spence.) Let f = chT , for some T : V V . We must show that f(T ) = 0V —i.e. that f(T )(v) = 0 v V . 7→ ∀ ∈ This is immediate if v = 0, so assume v = 0 and let W = CT,v. Let β 6 0 be a basis of W and β be an extension of β0 to a basis of V . Then

β0 β [TW ] B1 " β0 # [T ]β = 0 B2

57 and chT = chT chB . So chT chT and since we know that W · 2 W |

chTW = mT,v, we have mT,v chT . | Hence chT = gmT,v and

chT (T )(v) = (g(T )mT,v(T ))(v) = g(T )(mT, v(T )(v)) = g(T )(0) = 0.

3.2 An Algorithm for Finding mT

We use the factorization of chT into monic irreducibles. THEOREM 3.4 Suppose T : V V , 7→ b1 bt mT = p1 . . . pt where b1, . . . , bt 1, and p1, . . . , pt are distinct monic irreducibles. Then for i =≥ 1, . . . , t we have (a) bi 1 bi V Im pi(T ) Im p − (T ) Im p (T ) = ⊃ ⊃ · · · ⊃ i ⊃ i ··· (b) bi 1 bi 0 Ker pi(T ) Ker p − (T ) Ker p (T ) = . { } ⊂ ⊂ · · · ⊂ i ⊂ i ··· Note: In terms of nullities, conclusion (b) says that

bi 1 bi 0 < ν(pi(T )) < < ν(p − (T )) < ν(p (T )) = ··· i i ··· so this gives us a method of calculating bi. a1 at Presently we’ll show that if chT = p1 . . . pt , then

bi nullity (pi (T )) = ai deg pi.

Hence bi is also characterised as the smallest integer h such that h nullity (pi (T )) = ai deg pi. Also note that (a) and (b) are equivalent, and it is the latter that we prove. A notational simpliﬁcation—the left F [x]-module notation. If f F [x] and v V , we deﬁne ∈ ∈ fv = f(T )(v). It is easy to verify that

58 1. (f + g)v = fv + gv f, g F [x], v V ; ∀ ∈ ∈ 2. f(v + w) = fv + fw f F [x], v, w V ; ∀ ∈ ∈ 3. (fg)v = f(gv) f, g F [x], v V ; ∀ ∈ ∈ 4. 1v = v v V. ∀ ∈ These axioms, together with the four axioms for addition on V , turn V into what is called a “left F [x]-module”. (So there are deeper considerations lurking in the background—ideas of greater generality which make the algorithm we unravel for the rational canonical form also apply to other things such as the theorem that any ﬁnite abelian group is a direct product of cyclic prime power subgroups.)

(i) We ﬁrst prove that 0 Ker pi(T ). We write p = pi, b = bi for brevity; no confusion{ should} ⊂ arise since i is ﬁxed.

PROOF. mT = pf, f F [x] and f(T ) = 0V . Hence v V such that fv = 0. Then ∈ 6 ∃ ∈ 6 p(fv) = (pf)v = mT v = 0, so fv Ker p(T ). ∈ (ii) We next prove that

Ker pb(T ) = Ker pb+1(T ).

The containment Ker pb(T ) Ker pb+1(T ) ⊆ is obvious, so we need only show that

Ker pb(T ) Ker pb+1(T ). ⊇

b+1 b+1 b b Let w Ker p (T ), i.e. p w = 0. Now if mT = p q, then gcd(p , q) = 1. So u,∈ v F [x] such that 1 = upb + vq. Hence ∃ ∈ b 2b p = up + vmT .

59 Hence

b 2b p (T ) = (up )(T ) + v(T )mT (T ) = (up2b)(T )

and thus b b 1 b+1 b 1 p w = (p − )(p w) = p − 0 = 0 and w Ker pb, as required. ∈ (iii)

Ker ph(T ) = Ker ph+1(T ) Ker ph+1(T ) = Ker ph+2(T ), ⇒ i.e. Ker ph(T ) Ker ph+1(T ) Ker ph+1(T ) Ker ph+2(T ). ⊇ ⇒ ⊇ PROOF. Suppose that Ker ph(T ) Ker ph+1(T ). Then ⊇ v Ker ph+2(T ) ph+2v = 0 ∈ ⇒ ph+1(pv) = 0 pv Ker ph+1(T ) ⇒ ⇒ ∈ pv Ker ph(T ) ph(pv) = 0 ⇒ ∈ ⇒ ph+1v = 0 v Ker ph+1(T ). ⇒ ⇒ ∈ So it follows by induction from (ii) that

Ker pbi (T ) = Ker pbi+1 (T ) = . i i ··· (iv) b 1 b Ker p − (T ) Ker p (T ) ⊂ and this forces a chain of proper inclusions:

b 1 b 0 Ker p(T ) Ker p − (T ) Ker p (T ) = { } ⊂ ⊂ · · · ⊂ ⊂ ··· which is the desired result. For b 1 b 1 p q(T ) = 0V , so v such that p qv = 0. Then − 6 ∃ − 6 b 1 qv Ker p − (T ), 6∈ but qv Ker pb(T ) as ∈ b p qv = mT v = 0.

60 3.3 Primary Decomposition Theorem THEOREM 3.5 (Primary Decomposition) b1 bt If T : V V is a LT with mT = p1 . . . pt , where p1, . . . , pt are monic irreducibles,7→ then

V = Ker pb1 (T ) Ker pbt (T ), 1 ⊕ · · · ⊕ t a direct sum of T -invariant subspaces. Moreover for 1 i t, ≤ ≤ bi ν(pi(T ) ) = ai deg pi,

a1 at where chT = p p . 1 ··· t REMARK. The same proof gives a slightly more general result: If p = pb1 pbt , then 1 ··· t Ker p(T ) = Ker pb1 (T ) Ker pbt (T ). 1 ⊕ · · · t We subsequently give an application of the decomposition theorem in this form to the solution of the n–th order linear differential equations with constant coefficients. (See Hoffman and Kunze, pages 184–185.) bi PROOF. Let mT = p qi i = 1, . . . , t. Then i ∀

(qiqj)(T ) = 0V if i = j as mT qiqj if i = j. 6 | 6

Now gcd(q1, . . . , qt) = 1, so f1, . . . , ft F [x] such that ∃ ∈

1 = f1q1 + + ftqt ··· and with Ti = (fiqi)(T ) we have

IV = T1 + + Tt. (6) ··· Also

TiTj = (fiqi)(T )(fjqj)(T )

= (fifj)(T )(qiqj)(T )

= 0V if i = j, 6 Then t V = Im T . M i i=1

61 2 For T = Ti(T1 + + Tt) = TiIV = Ti. Next, V = Im T1 + + Im Tt. For i ··· ··· v V v = IV (v) = T1(v) + + Tt(v) Im T1 + + Im Tt. ∈ ⇒ ··· ∈ ··· Next assume v1 + + vt = 0, vi Im Ti, 1 i t. Then vi = Ti(ui) and ··· ∈ ≤ ≤

T1(u1) + + Tt(ut) = 0 ··· Ti(T1(u1) + + Tt(ut)) = T (0) = 0 ··· TiTi(ui) = 0

vi = Ti(ui) = 0.

We now show that bi Im Ti = Ker pi (T ).

“ ” Let v Im Ti. Then ⊆ ∈ v = fiqiw bi bi p v = p fiqiw ⇒ i i bi = fi(pi qi)w = 0.

“ ” Suppose pbi v = 0. ⊇ i bi Now if j = i, we have p fjqj, so 6 i | Tj(v) = fjqjv = 0.

v = IV (v) = T1(v) + + Tt(v) = Ti(v) ··· Im Ti, ∈ as required.

bi Finally, let Vi = Ker pi (T ) and Li = TVi . Then because V1,...,Vt are T –invariant subspaces of V , we have

chT = chL chL . 1 ··· t bi bi Now pi (T )(v) = 0 if v Vi, so pi (Li) = 0Vi . Hence mLi has the form ei ∈ di mLi = pi . Hence chLi has the form chLi = pi . Hence

a1 at d1 dt chT = p p = p p 1 ··· t 1 ··· t 62 and consequently di = ai. Finally, ai dim Vi = deg chLi = deg pi = ai deg pi.

(Incidentally, we mention that mT = lcm (mL1 , . . . , mLt ). Hence

b1 bt e1 et mT = p p = p p 1 ··· t 1 ··· t bi and consequently ei = bi. Hence mLi = pi .) THEOREM 3.6 (Commuting diagonable linear transformations) If T1,...,Tn : V V are commuting diagonable linear transformations, → β β then there exists a basis β for V such that each of [T1]β,... [Tm]β are each diagonal. (Matrix version) If A1,...,Am are commuting diagonable matrices of the same size, then there exists a non-singular matrix P such that the matrices

1 1 P − A1P,...P − AmP are simultaneously diagonal.

PROOF. (From Samelson page 158). We prove the result when m = 2, the general case follows by an easy iteration. Suppose T1 and T2 are commuting diagonable linear transformations on V . Because mT1 splits as a product of distinct linear factors, the primary decomposition theorem gives a direct sum decomposition as a sum of the T1–eigenspaces:

V = U1 Ut. ⊕ · · · ⊕ It turns out that not only are the subspaces Ui T1–invariant, they are T2– invariant. For if Ui = Ker (T1 cIV ), then − v Ui T1(v) = cv ∈ ⇒ T2(T1(v)) = cT2(v) ⇒ T1(T2(v)) = cT2(v) ⇒ T2(v) Ui. ⇒ ∈ Now because T2 is diagonable, V has a basis consisting of T2–eigenvectors and it is an easy exercise to show that in a direct sum of T2–invariant subspaces, each non-zero ”component” of a T2–eigenvector is itself a T2– eigenvector; moreover each non–zero component is a T1– eigenvector. Hence V is spanned by a family of vectors which are simultaneously T1–eigenvectors β and T2–eigenvectors. If β is a subfamily which forms a basis for V , then [T1]β β and [T2]β are diagonal.

63 THEOREM 3.7 (Fitting’s lemma) Suppose T : V V is a linear transformation over T and → Ker Ker T 2 Ker T n = Ker T n+1 = ⊂ ⊂ · · · ··· Then V = Im T n Ker T n. ⊕ COROLLARY 3.1 If T : V V is an indecomposable linear transformation (that is the only T –invariant→ subspaces of V are 0 and V ), then T is either nilpotent n { } (that is T = 0V for some n 1) or T is an isomorphism. ≥