for V a finite dimensional vector space over the field of scalars F,
there exists a direct sum decomposition
V = V1 ⊕ ⋅⋅⋅⊕ Vq (7.1)
(7.1)
where
Vk = ker(ϕk(A)mk)
and ϕ_{k}
(λ)
is an irreducible monic polynomial. Here the minimum polynomial of A was
∏q
ϕk(λ)mk
k=1
Next I will consider the problem of finding a basis for V_{k} such that the matrix of A restricted to V_{k}
assumes various forms.
Definition 7.1.1Letting x≠0 denote by β_{x}the vectors
{ }
x,Ax,A2x,⋅⋅⋅,Am−1x
where m is thesmallest such that A^{m}x ∈span
( )
x,⋅⋅⋅,Am −1x
. This is calledan A cyclic set. The vectors whichresult are also called a Krylov sequence.For such a sequence of vectors,
|βx|
≡ m, the number ofvectors in β_{x}.
Note that for such an A cyclic set, there exists a unique monic polynomial η
(λ)
of degree
|βx|
with
η
(A )
x = 0 such that if ϕ
(A )
x = 0 for any other polynomial, then η
(λ)
must divide ϕ
(λ)
. Such sequences
have some very useful properties.
Lemma 7.1.2Let β_{x} =
{x,Ax, A2x,⋅⋅⋅,Ad −1x}
,x≠0 where d is the smallestsuch thatA^{d}x ∈span
(x,⋅⋅⋅,Ad−1x)
. Then
{x,Ax,A2x, ⋅⋅⋅,Ad−1x}
is linearly independent. If x ∈
ker
(ϕ (A ))
,x≠0, where ϕ
(λ)
is irreducible and has degree d, then
|βx|
= d. If x ∈ ker
(ϕ (A))
whereϕ
(λ )
is just some polynomial, there is a unique monic polynomial of minimum degree η
(λ)
such thatη
(A )
x = 0. Alsospan
(βx)
is always A invariant. Thus A^{n}z ∈span
(βx)
whenever z ∈span
(βx)
.
Proof: Suppose that there are scalars a_{k}, not all zero such that
d∑−1 k
akA x = 0
k=0
Then letting a_{r} be the last nonzero scalar in the sum, you can divide by a_{r} and solve for A^{r}x as a linear
combination of the A^{j}x for j < r ≤ d − 1 contrary to the definition of d.
For the second claim, if
|βx|
< d, then A^{m}x is a linear combination of A^{k}x for k < m for some m < d.
That is, there is a polynomial η
(λ)
of degree less than d such that η
(A)
x = 0 and we can let its degree be
as small as possible. But then
ϕ(λ) = η(λ)q (λ) +r (λ )
where the degree of r
(λ)
is less than the degree of η
(λ)
or else is 0. It can’t be zero because ϕ
(λ)
is
irreducible. Hence r
(A)
x = 0 which is a contradiction. Thus
|βx|
= d because ϕ
(A )
x = 0.
For the third claim, let η
(λ)
be a monic polynomial of smallest degree such that η
(A)
x = 0. Then the
same argument just given shows that if
, the terms of A^{n}z are scalar
multiples of A^{m}x for various m so A^{n}z ∈span
(βx)
if z ∈span
(βx)
. ■
These sequences occur whenever x ∈ ker
(ϕ (A ))
where ϕ
(λ)
is a polynomial so by definition,
ϕ
(A )
x = 0. Out of all polynomials η
(λ)
such that η
(A)
x = 0, take the one with smallest degree. Say
η
(λ)
= a_{m}λ^{m} + a_{r−1}λ^{m−1} +
⋅⋅⋅
+ a_{1}λ + a_{0}. Then A^{m}x is in the span of
{x,Ax, A2x,⋅⋅⋅,Am −1x}
. Could
this happen for A^{k}x for some k < m? If so, there would be a polynomial ψ
(λ)
having degree smaller than
m such that ψ
(A )
x = 0 which doesn’t happen. Thus you get such a sequence whenever x ∈ ker
(ϕ(A ))
for
some polynomial ϕ
(λ)
.
Now here is a nice lemma which will be used in what follows.
Lemma 7.1.3Suppose W is a subspace of V where V is a vector space and L ∈ ℒ
(V,V)
andsuppose LW = LV. Then V = W + ker
(L)
.
Proof:Let v ∈ V . Then there exists w_{v}∈ W such that Lv = Lw_{v} by assumption. Then
L
(v− wv)
= 0. Letting z_{v} = v − w_{v}, it follows that z_{v}∈ ker
(L )
and v = w_{v} + z_{v} showing that
V = W + ker
(L)
. ■
For more on the next lemma and the following theorem, see Hofman and Kunze [18]. I am following the
presentation in Friedberg Insel and Spence [13]. See also Herstein [16] or Problem 11 on Page 381 for
a different approach to canonical forms. To help organize the ideas in the lemma, here is a
diagram.
PICT
Lemma 7.1.4Let V be a vector space, A ∈ℒ
(V,V)
, and W an A invariant
(AW ⊆ W )
subspaceof V for m a positive integer where ϕ
(λ)
is an irreducible monic polynomial of degree d. Let U bean A invariant subspace of ker
(ϕ (A ))
and assume ker
(ϕ (A ))
is finite dimensional.
If
{v1,⋅⋅⋅,vs}
is a basis for W then if x ∈ U ∖ W,
{v1,⋅⋅⋅,vs,βx}
is linearly independent. (In other words, we know that
{v1,⋅⋅⋅,vs,x}
is linearly independent by earliertheorems, but this says that you can include, not just x but the entire A cyclic set beginning withx.)
There exist vectors x_{1},
⋅⋅⋅
,x_{p}each in U such that
{ }
v1,⋅⋅⋅,vs,βx1,⋅⋅⋅,βxp
is a basis for
U + W.
|βxi|
= d and p is uniquely determined by U. Also, if x ∈ ker
m
(ϕ (A ) )
,
|βx|
= kd where k ≤ m. Here
|βx|
is the length of β_{x}, the degree of the monic polynomial η
so x ∈ W but this is assumed not to take place. Hence
z = 0 and so the linear independence of the
{v1,⋅⋅⋅,vs}
implies each a_{i} = 0. Then the linear
independence of
{x,Ax,⋅⋅⋅,Ad −1x}
, which follows from Lemma 7.1.2, shows each d_{j} = 0.
Thus
{v1,⋅⋅⋅,vs,x,Ax,⋅⋅⋅,Ad−1x}
is linearly independent as claimed.
Let x ∈ U ∖W ⊆ ker
(ϕ(A))
. Then it was just shown that
{v ,⋅⋅⋅,v,β }
1 s x
is linearly independent. Let
W_{1} be given by
y ∈ span(v1,⋅⋅⋅,vs,βx) ≡ W1
Then W_{1} is A invariant. If W_{1} equals U + W, then you are done. If not, let W_{1} play the role of W and
pick x_{1}∈ U ∖ W_{1} and repeat the argument. Continue till
span(v1,⋅⋅⋅,vs,βx1,⋅⋅⋅,βxn) = U +W
The process stops because ker
m
(ϕ(A) )
is finite dimensional.
Finally, letting x ∈ ker
m
(ϕ (A ) )
, there is a monic polynomial η
(λ)
such that η
(A)
x = 0 and η
(λ)
is of
smallest possible degree, which degree equals