EIGENVALUES, EIGENVECTORS, AND EIGENVALUES, EIGENVECTORS, AND DIAGONALIZATION Note: In these...

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Transcript of EIGENVALUES, EIGENVECTORS, AND EIGENVALUES, EIGENVECTORS, AND DIAGONALIZATION Note: In these...

  • CHAPTER 5 EIGENVALUES, EIGENVECTORS, AND DIAGONALIZATION

    Note: In these definitions v ∈ Rn and λ ∈ R, but sometimes it is necessary to extend the domain of T to allow v ∈ Cn and λ ∈ C.

    Example: reflection operator T about the line y = (1/2)x

    b1 is an eigenvector of T corresponding to the eigenvalue 1.

    b2 is an eigenvector of T corresponding to the eigenvalue -1.

    Definition

    x

    y

    L b2

    b1 = T (b1)

    T (b2) = �b2

  • Note: In these definitions v ∈ Rn and λ ∈ R, but sometimes it is necessary to allow v ∈ Cn and λ ∈ C.

    Example:

    non-eigenvectors eigenvectors

    Definition

    A =

     0.6 0.8 0.8 �0.6

    A

     �5 5

    � =

     � , A

     7 6

    � =

     � , A

     2 1

    � =

     � , A

     �1 2

    � =

     �

  • Example:

    An eigenvector of A corresponds to a unique eigenvalue. An eigenvalue of A has infinitely many eigenvectors.

    A =

    2

    4 5 2 1 �2 1 �1 2 2 4

    3

    5 v =

    2

    4 1 �1 1

    3

    5

    Proof T(v) = λv ⇔ Av = λv.

    The eigenvectors and corresponding eigenvalues of a linear operator are the same

    as those of its standard matrix.

    Property

    Av =

    2

    4 5 2 1 �2 1 �1 2 2 4

    3

    5

    2

    4 1 �1 1

    3

    5 =

    2

    4 4 �4 4

    3

    5 = 4

    2

    4 1 �1 1

    3

    5 = 4v

    A(cv) = c(Av) = c(4v) = 4(cv), 8c 2 R

  • Proof Av = λv ⇔ Av - λv = 0 ⇔ Av - λInv = 0 ⇔ (A - λIn)v = 0.

    Example: to check 3 and -2 are eigenvalues of the linear operator

    Definition

    Definition

    T

    ✓ x1

    x2

    �◆ =

     �2x2

    �3x1 + x2

    Property Let A be an n⇥nmatrix with eigenvalue �. The eigenvectors of A corresponding to � are the nonzero solutions of (A� �In)v = 0.

  • consider its standard matrix

    The rank of A - 3I2 is 1, so its null space is not the zero space, and every nonzero vector in the null space is an eigenvector of T corresponding to the eigenvalue 3.

    Similarly, the rank of A + 2I2 is 1, so its null space is not the zero space, and every nonzero vector in the null space is an eigenvector of T corresponding to the eigenvalue -2.

    A =

     0 �2 �3 1

  • Example: to check that 3 is an eigenvalue of B and find a basis for the corresponding eigenspace, where

    find the reduced row echelon form of B - 3I3 and the solution set of (B - 3I3)x = 0, respectively, to be

    Thus {[ 1 0 0 ]T, [ 0 1 1 ]T} is a basis of the eigenspace of B corresponding to the eigenvalue 3.

    Example: some square matrices and linear operators on Rn have no

    real eigenvalues, like the 90º-rotation matrix

    B =

    2

    4 3 0 0 0 1 2 0 2 1

    3

    5

    2

    4 0 1 �1 0 0 0 0 0 0

    3

    5

    2

    4 x1

    x2

    x3

    3

    5 =

    2

    4 x1

    x3

    x3

    3

    5 = x1

    2

    4 1 0 0

    3

    5+ x3

    2

    4 0 1 1

    3

    5

     0 �1 1 0

  • 7

    T (x) = Ax = �x An eig. vec. x ===> unique eig. val. λ.

    An eig. val. λ ===> multiple eig. vectors x. ===> The set of all eig. vectors corr. to λ is {x 2 Rn : (A� �In)x = 0} \ {0}

    eigenspace corr. to the eig. val. λ

  • Section 5.1: Problems 1, 3, 7, 9, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59

    Homework Set for Sections 5.1