- REGLAS DE VUELO Y OPERACIÓN GENERAL SUBPARTE B - REGLAS GENERALES DE VUELO

since = for k . This proves (4.1) with = and = a,, .

Now suppose independent elements . . . , and scalars . . . , exist satisfying (4.1). Since . . . , are independent they form a basis for V. If we define a,, and

= 0 for i then the matrix A = is a diagonal matrix which represents T relative to the basis . . . ,

Thus the of finding a diagonal matrix representation of a linear transformation has been transformed to another problem, that of finding independent elements . . . , and scalars . . . , 1, to satisfy (4.1). Elements and scalars satisfying (4.1) are called eigenvectors and eigenvalues of respectively. In the next section we study eigenvectors and eigenvaluesf in a more general setting.

4.2 Eigenvectors and eigenvalues of a linear transformation

In this discussion V denotes a linear space and S denotes a of V. The spaces and V are not required to be finite dimensional.

DEFINITION. Let S V be a linear transformation of S into V. A scalar is called an eigenvalue T is a element in S such that

T(x) =

The element is called an eigenvector of Tbelonging to 1. The scalar is called an eigenvalue corresponding to

There is exactly one eigenvalue corresponding to a given eigenvector x. In fact, if we have = and T(x) = for some then ^q = so = .

Note: Although Equation (4.2) always holds for x = 0 and any scalar I, the definition excludes 0 as an eigenvector. One reason for this prejudice against 0 is to have exactly one eigenvalue associated with a given eigenvector x.

The following examples illustrate the meaning of these concepts.

EXAMPLE 1. Multiplication by a fixed scalar. Let T: S V be the linear transformation defined by the equation T(x) = cx for each in S, where c is a fixed scalar. In this example every element of S is an eigenvector belonging to the scalar c.

The words eigenvector and eigenvalue are partial translations of the German words and respectively. Some authors use the terms characteristic vector, or proper vector as synonyms for eigenvector. are also called characteristic values, proper values, or latent roots.

98 Eigenvalues and eigenvectors

EXAMPLE 2. The eigenspace E(A) consisting of all x such that T(x) = Let T: V be a linear transformation having an eigenvalue Let be the set of all elements x in such that T(x) = This set contains the zero element 0 and all eigenvectors belonging to il. It is easy to prove that E(I) is a of S, because if x and are in we have

+ = + +

for all scalars a and Hence (ax + by) E(1) so is a subspace. The space E(1) is called the eigenspace corresponding to It may be finite- or infinite-dimensional. If is finite-dimensional then dim E(I) 1 , since E(1) contains at least one element x corresponding to 1.

EXAMPLE 3. Existence of zero eigenvalues. If an eigenvector exists it cannot be zero, by definition. However, the zero scalar can be an eigenvalue. In fact, if 0 is an eigenvalue for x then T(x) = Ox = 0, so x is in the null space of T. Conversely, if the null space of T contains any elements then each of these is an eigenvector with eigenvalue 0. In general, is the null space of T

EXAMPLE 5. Rotation of through This example is of special interest because it shows that the existence of eigenvectors may depend on the underlying field of scalars. The plane can be regarded as a linear space in two different ways: (1) As a dimensional real linear space, V = V,(R), with two basis elements (1, 0) and (0, and with real numbers as scalars; or (2) as a l-dimensional complex linear space, V =

with one basis element 1, and complex numbers as scalars.

Consider the second interpretation first. Each element z 0 of can be expressed in polar form, z = If T rotates z through an angle then T(z) = = Thus, each z 0 is an eigenvector with eigenvalue = Note that the eigenvalue is not real unless is an integer multiple of

Now consider the plane as a real linear space, V,(R). Since the scalars of V,(R) are real numbers the rotation T has real eigenvalues only if is an integer multiple of In other words, if is not an integer multiple of then T has no real eigenvalues and hence no eigenvectors. Thus the existence of eigenvectors and eigenvalues may depend on the choice of scalars for V.

EXAMPLE 6. The gperator. Let V be the linear space of all real functions having derivatives of every order on a given open interval. Let D be the linear transfor-mation which maps each onto its derivative, = The eigenvectors of D are those

functions satisfying an equation of the form

Eigenvectors and eigenvalues of a linear transformation 99

=

for some real This is a first order linear differential equation. All its solutions are given by the formula

f(x) =

where c is an arbitrary real constant. Therefore the eigenvectors of D are all exponential functions f(x) with c 0 . The eigenvalue corresponding to f(x) = is In examples this one where is a function space the eigenvectors are called eigenfunctions.

EXAMPLE 7. The integration operator. Let V be the linear space of all real functions continuous on a finite interval [a, b]. define = T(f) to be that function given by

The eigenfunctions of T (if any exist) are those nonzerofsatisfying an equation of the form

(4.3)

for some real 4. If an eigenfunction exists we may differentiate this equation to obtain the = from which we = provided 0. In other words, the only candidates for eigenfunctions are those exponential functions of the

with c 0 and 0. However, if we put = a in (4.3) we obtain

0 = = Ice”‘“.

Since is never zero we see that the equation T(f) = be satisfied with a zero so Thas no eigenfunctions and no eigenvalues.

EXAMPLE 8. The spanned by an eigenvector. Let T: S V be a linear trans-formation having an eigenvalue 1. Let x be an eigenvector belonging to and let be the spanned by x. That is, is the set of all scalar multiples of x. It is easy to show that maps into itself. In fact, if y = cx we

T(y) = = = = =

If c 0 then 0 so every element of is also an eigenvector belonging to

A of S is called invariant under Tif Tmaps each element of onto an element of We have j ust shown that the spanned by an eigenvector is invariant under T.

100 Eigenvalues and eigenvectors