Explore BrainMass

Explore BrainMass

    Finding eigen values and eigen space

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Please see the attached file for homework specifics.
    Thank-you so much for your expertise.

    Let A = . Find each of the following (by hand, showing your work).

    (a) The characteristic polynomial of A
    (b) The eigenvalues of A.
    (c) The algebraic multiplicity for each eigenvalue.
    (d) A basis for each eigenspace.
    (e) Is "A" diagonalizable? Why or why not?

    © BrainMass Inc. brainmass.com October 2, 2022, 11:47 am ad1c9bdddf
    https://brainmass.com/math/linear-algebra/finding-eigen-values-eigen-space-204062

    Attachments

    SOLUTION This solution is FREE courtesy of BrainMass!

    HI,

    Please find the explanations/solution attached herewith.

    Let A = . Find each of the following (by hand, showing your work).
    (a) The characteristic polynomial of A
    (b) The eigenvalues of A.
    (c) The algebraic multiplicity for each eigenvalue.
    (d) A basis for each eigenspace.
    (e) Is "A" diagonalizable? Why or why not?

    Solution:
    (a)

    The characteristic equation is

    (b)

    Eigen values are 3 and 2.

    (c)
    The multiplicity of eigen value 2 is 2.
    The multiplicity of eigen value 3 is 2.
    (d)
    The Eigen vector corresponding to =3, will be
    Substitute =3 into the gives

    Which gives y = z = w = 0
    Therefore, the eigenvectors of A associated with the eigenvalue λ = 3 are all vectors of the form ( x, 0, 0, 0)T = x(1, 0, 0, 0)T for x≠ 0. Removing the restriction that the scalar multiple be nonzero includes the zero vector and gives the full eigenspace:

    Basis for this eigenspace is .
    The Eigen vector corresponding to =2, will be
    Substitute =3 into the gives

    Which gives z = w = 0 , x = -2y
    Therefore, the eigenvectors of A associated with the eigenvalue λ = 2 are all vectors of the form ( -2y, y, 0, 0)T = y(-2, 1, 0, 0)T for y≠ 0. Removing the restriction that the scalar multiple be nonzero includes the zero vector and gives the full eigenspace:

    Basis for this Eigen space is .

    (e)
    A is not diagonalizable as the its all eigen values are not different.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 2, 2022, 11:47 am ad1c9bdddf>
    https://brainmass.com/math/linear-algebra/finding-eigen-values-eigen-space-204062

    Attachments

    ADVERTISEMENT