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Math Help - Linear Algebra, eigenvectors problem

  1. #1
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    Linear Algebra, eigenvectors problem

    If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

    (1) v is also an eigenvector of 2A
    (2) v is also an eigenvector of A^2
    (3) v is also an eigenvector of A^-1

    A) 1 only
    B) 2 only
    C) 3 only
    D) 1 and 3 only
    E) 1,2 and 3

    I am pretty sure 2 is true because if we look at Av = (lamba)v.
    A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v

    So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.

    2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v

    Can someone help show 3? If Av = lambda v
    then (A^-1)Av =A^-1(lambda)v
    I am not sure how to show it
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  2. #2
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    Quote Originally Posted by ksheehan62 View Post
    If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

    (1) v is also an eigenvector of 2A
    (2) v is also an eigenvector of A^2
    (3) v is also an eigenvector of A^-1

    A) 1 only
    B) 2 only
    C) 3 only
    D) 1 and 3 only
    E) 1,2 and 3

    I am pretty sure 2 is true because if we look at Av = (lamba)v.
    A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v
    So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.
    Very good!


    2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v
    I presume you meant (1) here.

    Can someone help show 3? If Av = lambda v
    then (A^-1)Av =A^-1(lambda)v
    I am not sure how to show it
    Well, what is (A^-1)Av? And, of course, because A and A^{-1} are linear operators, A^{-1}(\lambda v)= \lambda A^{-1}v.
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  3. #3
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    Quote Originally Posted by ksheehan62 View Post
    If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

    (1) v is also an eigenvector of 2A
    (2) v is also an eigenvector of A^2
    (3) v is also an eigenvector of A^-1

    A) 1 only
    B) 2 only
    C) 3 only
    D) 1 and 3 only
    E) 1,2 and 3

    I am pretty sure 2 is true because if we look at Av = (lamba)v.
    A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v
    So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.
    Very good!


    2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v
    I presume you meant (1) here.

    Can someone help show 3? If Av = lambda v
    then (A^-1)Av =A^-1(lambda)v
    I am not sure how to show it
    Well, what is (A^-1)Av? And, of course, because A and A^{-1} are linear operators, A^{-1}(\lambda v)= \lambda A^{-1}v.

    As long as \lambda\ne 0, you can divide both sides by \lambda. Of course, if \lambda= 0, then A does not have an inverse.
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  4. #4
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    Quote Originally Posted by ksheehan62 View Post
    If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

    (1) v is also an eigenvector of 2A
    (2) v is also an eigenvector of A^2
    (3) v is also an eigenvector of A^-1

    A) 1 only
    B) 2 only
    C) 3 only
    D) 1 and 3 only
    E) 1,2 and 3

    I am pretty sure 2 is true because if we look at Av = (lamba)v.
    A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v

    So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.

    2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v

    Can someone help show 3? If Av = lambda v
    then (A^-1)Av =A^-1(lambda)v
    I am not sure how to show it


    Av=\lambda v\Longrightarrow v=Iv=A^{-1}Av=A^{-1}(Av)=A^{-1}(\lambda v)=\lambda A^{-1}v\Longrightarrow multiplying now by \lambda^{-1} in both extremes of the above equality (why

    can we do such a thing?) you get that v is also an eigenvector of A^{-1} .

    Tonio
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