Kenneth Kuttler

7.3. CYCLIC DECOMPOSITION 155

where P is an invertible n×n matrix of entries of D, then {e′1, ...,e′n} is linearly independentin the same way. Indeed, if 0= ∑k σ ke

′k then

0= ∑k

σ ke′k = ∑

kσ k ∑

je jPk j = ∑

(∑k

σ kPk j

)e j

and so ∑k σ kPk j = 0 for each j. Since P is invertible, this gives each σ k = 0. This observa-tion is useful in the following proof.

7.3 Cyclic DecompositionTheorem 7.3.1 Let V be a finite dimensional vector space and let L ∈ L (V,V ). Thenthere are vectors m1, ...,mp such that V = Dm1⊕·· ·⊕Dmp where D = F [x] and for p ∈D,v ∈V, pv≡ p(L)(v).

Proof: Let V = Db1 + · · ·+Dbn for bk ∈V . This is possible because V is finite dimen-sional. In fact, we could pick a basis and let this be a direct sum of Fbk. However, the pointhere is that there are more vectors in Dbk than in Fbk. Thus p will likely be smaller than nif the dimension of V is n. Now define η : Dn→V by

∑i=1

σ iei

)≡ η (σ)≡

∑i=1

σ ibi, σ ibi ≡ σ i (L)bi

Here ei has 1 in the ith position and 0 elsewhere and σ i will denote a polynomial. Then itfollows that η is a morphism. For v, v̂ ∈ Dn and σ ∈ D,

η (v+ v̂) = η (v)+η (v̂) , η (σv) = ση (v) (7.1)

Let K ≡ ker(η) .Now use Lemma 7.2.5 to define the one to one and onto morphism η̂ : Dn/K → V as

η̂ ([v])≡ η (v).It was shown above in Proposition 7.2.10 that Dn is Noetherian and so the submodule

K = ker(η) is span(z1, · · · ,zm) for some m. Out of all such spans, let m be as small aspossible. Let the matrix A be defined by zk = ∑ j A jke j written as(

z1 · · · zm

)=(

e1 · · · en

)An×m

By Theorem 7.1.1, there are invertible P,Q such that PAQ = B for Bn×m a matrix 0 offthe main diagonal, and so A = P−1BQ−1 where B is n×m. Therefore,(

z′1 · · · z′m

)≡(

z1 · · · zm

)Q =

(e1 · · · en

)P−1B (7.2)

Thus span(z′1, ...,z′m) is the same as the span of the zk which is K. Then if m≤ n the above

gives≡(

e′1 · · · e′n

)B =

(δ 1e

′1 · · · δ me

′m

)(7.3)

where δ k = B̂kk,k ≤ m. If m > n then some of the z′k equals 0 and so m would not be assmall as possible.

7.3. CYCLIC DECOMPOSITION 155where P is an invertible n x n matrix of entries of D, then {e/,...,e/, } is linearly independentin the same way. Indeed, if 0 = ), xe, then0- Sack Ea bens =¥( ots) ejk k J J kand so ), 0; P<; = 0 for each j. Since P is invertible, this gives each 0, = 0. This observa-tion is useful in the following proof.7.3 Cyclic DecompositionTheorem 7.3.1 Let V be a finite dimensional vector space and let LE £(V,V). Thenthere are vectors m,...,Mp such that V = Dm, ®---® Dmp where D = F [x] and for p €D,v €V, pv= p(L)(v).Proof: Let V = Db, +---+ Db, for by € V. This is possible because V is finite dimen-sional. In fact, we could pick a basis and let this be a direct sum of Fb;. However, the pointhere is that there are more vectors in Db, than in Fb,. Thus p will likely be smaller than nif the dimension of V is n. Now define n : D” — V byn (z ox) =n (0) =) o(b;, ojb; = 6; (L) bji= i=lHere e; has | in the i” position and 0 elsewhere and 0; will denote a polynomial. Then itfollows that 1 is a morphism. For v,& € D” and o € D,n(v+6)=n(v) +n (6), N(ov) =o7n (v) (7.1)Let K =ker(n).Now use Lemma 7.2.5 to define the one to one and onto morphism fj : D"/K — V as7 ([e]) =n (w).It was shown above in Proposition 7.2.10 that D” is Noetherian and so the submoduleK =ker(m) is span(z1,--:,2m) for some m. Out of all such spans, let m be as small aspossible. Let the matrix A be defined by z; = )};Aj,e; written as(21 en )=(er en )AnmBy Theorem 7.1.1, there are invertible P,Q such that PAQ = B for B,x a matrix O offthe main diagonal, and so A = P-!BQ™! where B is n x m. Therefore,( 2 Le 2, )=( 21 i zm )O=(e1 i en )P IB (7.2)Thus span (z/,..., Z/,,) is the same as the span of the z, which is K. Then if m <n the abovegives=(e, -- € )B=( die, Suet, ) (7.3)where 6, = Byx,k <m. If m > n then some of the Zh equals 0 and so m would not be assmall as possible.