9.4. Eigenvalues and eigenvectors

We saw in the previous section that the analysis of isometrics of became possible under a special assumption. That assumption motivates our next definition.

9.4.1 Definition:

Let V be a vector space and T: V→V be a linear transformation. A scalar is called an eigen-value for T if there exists a non-zero vector vV such that Tv = λv. The non-zero vector v is called an
eigenvector for the eigenvalue λ .

9.4.2. Example:

(i)   Suppose V is one-dimensional vector space. Then, the only possible linear transformations
      T: V→V are the following : for fixed, T(v) = αv for every vV . Thus, α is an eigenvalue
       of T and every v ≠ 0 is an eigenvector for this eigenvalue.
(ii)   Let T:

→

be defined by T( x, y ) = ( 5x, 2y ) , " ( x, y) ∈

.
       Then α = 5 and α = 2 are eigenvalues of T, since T(1,0) = 5(1,0) and T(0,1) = 2(0,1).
(iii)   Let V be a finite dimensional inner-product space over F and S a non trivial proper subspace of V.
        Let

be the orthogonal complement of S. Let T: V→V be the orthogonal projection of T onto S.
Since, both S and

are nontrivial subspace of V and
                       T(v) = v   for every v∈S
      and
                       T(v) = 0   for every v∈

,
      it follows that 0 and 1 are both eigenvalues of T. We show that these are the only eigenvalues of T.
        Let 0 ≠ λ∈F and v∈V be such that v≠ 0 and P(v) =λ(v).
        Then P(P(v)) = P(λv) = λ(P(v)).             --------- (*)
        Since P is orthogonal projection ,

        Thus, (*) gives
                        λP(v) = P(v), i.e. (1-λ)P(v) = 0.
         Since P(v) = λv ≠ 0, we get λ=1.
         Thus, if λ ≠ 0 is an eigenvector, then λ = 1.
(iv)    Consider T :

→

given by
T(x, y) = ( y, -x )∈

.
Then, for any λ∈ℝ and ( x, y )∈

, T( x, y ) = λ( x, y ) iff -x =

x ,which is never possible.
         Hence, T has no eigenvalues.
(v)     Let T :V→V be invertible .Then λ =0 is not an an eigenvalue of T. for otherwise T(v) = 0 for some
          v ≠ 0.
                 In general, given a linear transformation T :V→V , it is not obvious how to analyze: Whether
         T has eigenvalues or not ? And if it has, how to find them. The following theorem is useful in this
          regards.

9.4.3. Theorem:

Let V be a n-dimensional vector space over F and B be any ordered basis of V. Let T :V→V
be a linear transformation .
(i)   λ∈F is an eigenvalue for T with an eigenvector v∈V if and only if

(ii) λ∈F is an eigenvalue for T if and only if

       The function is a polynomial over F of degree n with leading
       coefficient .

(iii) T has at most n distinct eigenvalues.
(iv) If λ∈F is an eigenvalue for T, then for every nonzero solution X of the homogenous system
is an eigenvector for the eigenvalue λ.

Proof

Above theorem motivates the following:

9.4.4. Definition:

Let A be a n×n matrix over F . A scalar λ∈F is called an eigenvalue for A if there exists a nonzero
such that Av = λv . And, any such nonzero vector is called as eigenvector for the eigenvalue λ.
Theorem 9.4.3 tells us that a scalar λ∈F is an eigenvalue for T iff it is an eigenvalue for the matrix representation of T with respect to every ordered basis of T. Further, eigenvalues of a matrix A are
the solution of the equation det(A-λI).

9.4.5 Examples:

(i) Let T :

→

be given by T( x, y ) = ( x + 3y, 4x + 2y ) , ( x, y )∈

.
With respect to basis B = { ( 1, 0 ), ( 0, 1 ) } of

Thus, λ∈ℝ is an eigenvalue of T if and only if det( ) = 0. since
0 = det( ) =

                                        = ( 1-λ )( 2-λ )-12

                                         = ( λ + 2 ( λ - 5 ),
       T has eigenvalues λ = -2 and λ = -5.
       To find corresponding eigenvectors, for λ = -2,we have to solve the equation

         This, gives x = 1, y = -1. Thus T has eigenvalue -2 with eigenvector ( 1, -1 ). Similarly for the
         eigenvalue λ = -5,

          gives x = 3, y = 4. Thus, ( 3, 4 ) is an eigenvector for T for the eigenvalue λ = -5.

9.4.6. Definition:

Let A be a n×n matrix with entries F . The polynomial

is called the characteristic polynomial of A .

9.4.7. Note:

Note that for the matrix A- λI , λ∈F,the entries are not scalars, they are infact polynomials over F
of degree at most one.

9.4.8 Examples:

(i) Let

Thus the characteristic polynomial of A is

     If we treat A as a matrix over ℝ, then A has no eigenvalues. If we treat A as matrix over C, the complex
      numbers, then A has two distinct eigenvalues : λ ± i.
(ii) Let A be any diagonal matrix,

Then the characteristic polynomial of

Hence, each diagonal entry of A is an eigenvalue for A.

As a consequence of theorem 2.5.3, we have the following

9.4.9 Corollary:

Let V be a vector space of dimension n and T : V→V be a linear map. If n is odd, then T will
have at least one real eigenvalue.

Proof

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Click here to take a Quiz: Quiz 9.4