9.3. CYCLIC SETS 239

where x occupies the positions between∑k−1

i=1 |βi| + 1 and∑k

i=1 |βi|. Then M will be thematrix of A with respect to β if and only if a similar diagram to the above commutes.Thus it is required that Aq = qM . However, from the description of q just made, and theinvariance of each Vk,

Aq



0...

x...

0

= Akqkx = qkM

kx = q



M1 0. . .

Mk

. . .

0 Mq





0...

x...

0

It follows that the above block diagonal matrix is the matrix of A with respect to the givenordered basis. ■

An examination of the proof of the above theorem yields the following corollary.

Corollary 9.2.6 If any βk in the above consists of eigenvectors, then Mk is a diagonalmatrix having the corresponding eigenvalues down the diagonal.

It follows that it would be interesting to consider special bases for the vector spaces inthe direct sum. This leads to the Jordan form or more generally other canonical forms suchas the rational canonical form.

9.3 Cyclic Sets

It was shown above that for A ∈ L (V, V ) for V a finite dimensional vector space over thefield of scalars F, there exists a direct sum decomposition

V = V1 ⊕ · · · ⊕ Vq

whereVk = ker (ϕk (A)

mk)

and ϕk (λ) is an irreducible polynomial. Here the minimal polynomial of A was

q∏k=1

ϕk (λ)mk

Next I will consider the problem of finding a basis for Vk such that the matrix of Arestricted to Vk assumes various forms.

Definition 9.3.1 Letting x ̸= 0 denote by βx the vectors{x,Ax,A2x, · · · , Am−1x

}where

m is the smallest such that Amx ∈ span(x, · · · , Am−1x

). This is called an A cyclic set.

The vectors which result are also called a Krylov sequence. For such a sequence of vectors,|βx| ≡ m.

The first thing to notice is that such a Krylov sequence is always linearly independent.

Lemma 9.3.2 Let βx ={x,Ax,A2x, · · · , Am−1x

}, x ̸= 0 where m is the smallest such

that Amx ∈ span(x, · · · , Am−1x

). Then

{x,Ax,A2x, · · · , Am−1x

}is linearly independent.

9.3. CYCLIC SETS 239where x occupies the positions between yet |6;| +1 and ye |6;|. Then M will be thematrix of A with respect to @ if and only if a similar diagram to the above commutes.Thus it is required that Aq = qM. However, from the description of q just made, and theinvariance of each Vz,0 M! 0 0Aq| x | = Araex = o.M*x = ¢ Mk x0 0) M14 0It follows that the above block diagonal matrix is the matrix of A with respect to the givenordered basis. llAn examination of the proof of the above theorem yields the following corollary.Corollary 9.2.6 If any 6, in the above consists of eigenvectors, then M* is a diagonalmatriz having the corresponding eigenvalues down the diagonal.It follows that it would be interesting to consider special bases for the vector spaces inthe direct sum. This leads to the Jordan form or more generally other canonical forms suchas the rational canonical form.9.3. Cyclic SetsIt was shown above that for A € £(V,V) for V a finite dimensional vector space over thefield of scalars F, there exists a direct sum decompositionV=V,0-:-8V,whereVe = ker (@; (A)"*)and ¢, (A) is an irreducible polynomial. Here the minimal polynomial of A wasTT ea)”k=1Next I will consider the problem of finding a basis for V, such that the matrix of Arestricted to V; assumes various forms.Definition 9.3.1 Letting x £ 0 denote by 8, the vectors {x, Ax, A?x, ee Am lah wherem is the smallest such that A™x © span (x, ee ,A™!z) . This is called an A cyclic set.The vectors which result are also called a Krylov sequence. For such a sequence of vectors,|6,,| =m.The first thing to notice is that such a Krylov sequence is always linearly independent.Lemma 9.3.2 Let 6, = {x, Ax, Ax, tee A™lyh ,« # 0 where m is the smallest suchthat A™x € span (x, vee ,A™!z) . Then {a, Ax, A?x,+-: ,Amtyh is linearly independent.