8.4 Eigenvalues And Eigenvectors Of Linear Transformations
Let V be a finite dimensional vector space. For example, it could be a subspace of ℂ^{n}or ℝ^{n}. Also
suppose A ∈ℒ
(V,V)
.
Definition 8.4.1The characteristic polynomialof A is defined as q
(λ)
≡ det
(λI − A)
.The zeros of q
(λ)
in F are called the eigenvaluesof A.
Lemma 8.4.2When λ is an eigenvalue of A which is also in F, the field of scalars, thenthere exists v≠0 such that Av = λv.
Proof:This follows from Theorem 8.3.16. Since λ ∈ F,
λI − A ∈ ℒ (V,V)
and since it has zero determinant, it is not one to one. ■
The following lemma gives the existence of something called the minimal polynomial.
Lemma 8.4.3Let A ∈ℒ
(V,V )
where V is a finite dimensional vector space of dimension n witharbitrary field of scalars. Then there exists a unique 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.
Proof: Consider the linear transformations, I,A,A^{2},
⋅⋅⋅
,A^{n2
}. There are n^{2} + 1 of these
transformations and so by Theorem 8.2.3 the set is linearly dependent. Thus there exist constants,
c_{i}∈ F such that
n∑2
c0I + ckAk = 0.
k=1
This implies there exists a polynomial, q
(λ)
which has the property that q
(A)
= 0. In
fact, one example is q
(λ)
≡ c_{0} + ∑_{k=1}^{n2
}c_{k}λ^{k}. Dividing by the leading term, it can
be assumed this polynomial is of the form λ^{m} + c_{m−1}λ^{m−1} +
⋅⋅⋅
+ c_{1}λ + c_{0}, a monic
polynomial. Now consider all such monic polynomials, q such that q
(A)
= 0 and pick the one
which has the smallest degree m. This is called the minimal polynomial and will be
denoted here by p
(λ)
. If there were two minimal polynomials, the one just found and
another,
λm + dm −1λm−1 + ⋅⋅⋅+ d1λ +d0.
Then subtracting these would give the following polynomial,
= 0, this requires each d_{k} = c_{k} since otherwise you could divide by d_{k}− c_{k} where
k is the largest one which is nonzero. Thus the choice of m would be contradicted.
■
Theorem 8.4.4Let V be a nonzero finite dimensionalvector space of dimension n with thefield of scalars equal to F. Suppose A ∈ℒ
(V,V )
and for p
(λ)
the minimal polynomialdefined above, let μ ∈ F be a zero of this polynomial. Then there exists v≠0,v ∈ V suchthat
Av = μv.
IfF = ℂ, then A always has an eigenvectorand eigenvalue. Furthermore, if
{λ ,⋅⋅⋅,λ }
1 m
are thezeros of p
(λ)
in F, these are exactly the eigenvalues of A for which there exists an eigenvector inV.
Proof: Suppose first μ is a zero of p
(λ)
. Since p
(μ)
= 0, it follows
p(λ) = (λ − μ)k(λ)
where k
(λ)
is a polynomial having coefficients in F. Since p has minimal degree, k
(A)
≠0 and so
there exists a vector, u≠0 such that k
(A)
u ≡ v≠0. But then
(A − μI)v = (A − μI)k (A)(u) = 0.
The next claim about the existence of an eigenvalue follows from the fundamental theorem of
algebra and what was just shown.
It has been shown that every zero of p
(λ)
is an eigenvalue which has an eigenvector in V . Now
suppose μ is an eigenvalue which has an eigenvector in V so that Av = μv for some v ∈ V,v≠0.
Does it follow μ is a zero of p
(λ)
?
0 = p(A)v = p(μ)v
and so μ is indeed a zero of p
(λ)
. ■
In summary, the theorem says that the eigenvalues which have eigenvectors in V are exactly
the zeros of the minimal polynomial which are in the field of scalars F.
class=”left” align=”middle”(V,W) As A Vector Space8.5. EXERCISES