Let V be a finite dimensional vector space with field of scalars F. Here I will make no assumption
on F. Also suppose A ∈ℒ
(V,V)
.
Recall Lemma 8.4.3 which gives the existence of the minimal polynomial for a linear
transformation A. This is the monic polynomial p which has smallest possible degree such that
p(A) = 0. It is stated again for convenience.
Lemma 9.2.1Let A ∈ℒ
(V,V )
where V is a finite dimensional vector space of dimension n withfield of scalars F. Then there exists a unique monic polynomial of the form
m m −1
p(λ) = λ + cm− 1λ + ⋅⋅⋅+ c1λ + c0
such that p
(A )
= 0 and m is as small as possible for this to occur.
Now it is time to consider the notion of a direct sum of subspaces. Recall you can always assert
the existence of a factorization of the minimal polynomial into a product of irreducible
polynomials. This fact will now be used to show how to obtain such a direct sum of
subspaces.
Definition 9.2.2For A ∈ℒ
(V,V )
where dim
(V )
= n, suppose the minimal polynomialis
∏q
p(λ) = (ϕk(λ))rk
k=1
where the polynomials ϕ_{k}have coefficients in F and are irreducible. Now define the generalizedeigenspaces
Vk ≡ ker((ϕk(A))rk)
Note that if one of these polynomials (ϕ_{k}(λ))^{rk}is a monic linear polynomial, then the generalizedeigenspace would be an eigenspace.
Proof:It is clear V_{k} is a subspace which is A invariant because A commutes with ϕ_{k}
(A)
^{mk}.
It is clear the operators ϕ_{k}
(A)
^{rk} commute. Thus if v ∈ V_{
k},
ϕ (A)rk ϕ (A)rl v = ϕ (A)rl ϕ (A)rk v = ϕ (A )rl 0 = 0
k l l k l
and so ϕ_{l}
(A)
^{rl} : V_{
k}→ V_{k}.
I claim ϕ_{l}
(A )
is one to one on V_{k} whenever k≠l. The two polynomials ϕ_{l}
(λ)
and ϕ_{k}
(λ)
^{rk} are
relatively prime so there exist polynomials m
(λ)
,n
(λ)
such that
m (λ)ϕ (λ)+ n (λ)ϕ (λ)rk = 1
l k
It follows that the sum of all coefficients of λ raised to a positive power are zero and the constant
term on the left is 1. Therefore, using the convention A^{0} = I it follows
m (A)ϕl(A)+ n (A)ϕk(A )rk = I
If v ∈ V_{k}, then from the above,
m(A )ϕl(A )v+ n (A)ϕk(A )rk v = v
Since v is in V_{k}, it follows by definition,
m (A)ϕl(A )v = v
and so ϕ_{l}
(A)
v≠0 unless v = 0. Thus ϕ_{l}
(A)
and hence ϕ_{l}
(A )
^{rl} is one to one on V_{
k} for every k≠l.
By Lemma 9.1.7 and the fact that ker
(∏q ϕk (λ)rk)
k=1
= V, 9.2 is obtained. The claim about the
bases follows from Lemma 9.1.6. ■
You could consider the restriction of A to V_{k}. It turns out that this restriction has minimal
polynomial equal to ϕ_{k}
(λ)
^{mk}.
Corollary 9.2.4Let the minimal polynomial of A be p
(λ)
= ∏_{k=1}^{q}ϕ_{k}
(λ )
^{mk}where each ϕ_{
k}isirreducible. Let V_{k} = ker
(ϕ (A )mk)
. Then
V1 ⊕ ⋅⋅⋅⊕ Vq = V
and letting A_{k}denote the restriction of A to V_{k}, it follows the minimal polynomial of A_{k}isϕ_{k}
(λ)
^{mk}.
Proof:Recall the direct sum, V_{1}⊕
⋅⋅⋅
⊕ V_{q} = V where V_{k} = ker
mk
(ϕk(A) )
for
p
(λ)
= ∏_{k=1}^{q}ϕ_{k}
(λ)
^{mk} the minimal polynomial for A where the ϕ_{
k}
(λ)
are all irreducible. Thus
each V_{k} is invariant with respect to A. What is the minimal polynomial of A_{k}, the restriction
of A to V_{k}? First note that ϕ_{k}
(Ak )
^{mk}
(Vk)
=
{0}
by definition. Thus if η
(λ)
is the
minimal polynomial for A_{k} then it must divide ϕ_{k}
Theorem 9.2.5Suppose V is a vector space with field of scalars F and A ∈ℒ
(V,V )
. Supposealso
V = V1 ⊕ ⋅⋅⋅⊕ Vq
where each V_{k}is A invariant. (AV_{k}⊆ V_{k}) Also let β_{k}be an ordered basis for V_{k}andlet A_{k}denotethe restriction of A to V_{k}. Letting M^{k}denote the matrix of A_{k}withrespect to this basis, it follows the matrix of A with respect to the basis
{β1,⋅⋅⋅,βq}
is
( )
M 1 0
|| .. ||
( . )
0 M q
Proof: Let β denote the ordered basis
{β1,⋅⋅⋅,βq}
,
|βk|
being the number of vectors in β_{k}.
Let q_{k} : F^{|βk|
}→ V_{
k} be the usual map such that the following diagram commutes.
Ak
Vk → Vk
qk ↑ ∘ ↑ qk
F |βk| → F|βk|
M k
Thus A_{k}q_{k} = q_{k}M^{k}. Then if q is the map from F^{n} to V corresponding to the ordered basis β just
described,
( )
q 0 ⋅⋅⋅ x ⋅⋅⋅ 0 T = q x,
k
where x occupies the positions between ∑_{i=1}^{k−1}
|βi|
+ 1 and ∑_{i=1}^{k}
|βi|
. Then M will be the
matrix 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 the invariance of each
V_{k},
( ) ( 1 ) ( )
| 0 | | M 0 | | 0 |
|| ... || || ... || || ... ||
Aq|| x || = A q x = q M kx = q || M k || || x ||
|| . || kk k || . || || . ||
( .. ) ( .. ) ( .. )
0 0 M q 0
It follows that the above block diagonal matrix is the matrix of A with respect to the given
ordered basis. ■
An examination of the proof of the above theorem yields the following corollary.
Corollary 9.2.6If any β_{k}in the above consists of eigenvectors, then M^{k}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 in the
direct sum. This leads to the Jordan form or more generally other canonical forms such as the
rational canonical form.