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Math Help - Eigenvector

  1. #1
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    Eigenvector

    Hello everyone, this is the following:

    Given the matrix
    \begin{bmatrix}{a}&{1}\\{0}&{3}\end{bmatrix} to determines the values of a such that the vector (7, 8) is an eigenvector of the matrix

    I know the value found should be a multiple of the vector given, I have done this:

    Multiplying

    \begin{bmatrix}{a}&{1}\\{0}&{3}\end{bmatrix}*  (7, 8)
    get

    (7a + 8, 21)


    (7a + 8, 21) = k(7, 8) , where k is the value to establish multiple.


    I do not know how to finish the exercise.

    I appreciate your help,

    or if someone tells me if there is another way to do it.
    Last edited by Dogod11; December 7th 2009 at 09:26 AM.
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  2. #2
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    You've started correctly. One math error, it should be
    (7a+8,24)=k(7,8).

    Then:
    7a+8 = 7k
    24 = 8k.

    Solve for k (2nd eqn), then solve for a(1st eqn).
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  3. #3
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    Quote Originally Posted by Dogod11 View Post
    Hello everyone, this is the following:

    Given the matrix
    \begin{bmatrix}{a}&{1}\\{0}&{3}\end{bmatrix} to determines the values of a such that the vector (7, 8) is an eigenvector of the matrix

    I know the value found should be a multiple of the vector given, I have done this:

    Multiplying

    \begin{bmatrix}{a}&{1}\\{0}&{3}\end{bmatrix}*  (7, 8)
    get

    (7a + 8, 21)


    (7a + 8, 21) = k(7, 8) , where k is the value to establish multiple.


    I do not know how to finish the exercise.

    I appreciate your help,

    or if someone tells me if there is another way to do it.


    You did great! Now just end the task: from (7a+8,21)=k(7,8) it follows that 21=8k\,\Longrightarrow\,k=\frac{21}{8} , and then 7a+8=7k=\frac{147}{8} you get an easy equation from where you get the value of a

    Tonio

    Ps. BTW, it is (7a+8,24) and not (7a+8,21) ...but the development is the same.
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  4. #4
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    O.K. thank you very much for the two

    The value found is 13 / 7.

    Find then the eigenvectors of the matrix


    \begin{bmatrix}{13/7}&{1}\\{0}&{3}\end{bmatrix}

    And I find the eigenvectors are

    (1, 0), and (1, 8/7)

    Any idea why not give me exactly 7, 8?

    Thank you very much


    Greetings
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  5. #5
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    Quote Originally Posted by Dogod11 View Post
    O.K. thank you very much for the two

    The value found is 13 / 7.

    Find then the eigenvectors of the matrix


    \begin{bmatrix}{13/7}&{1}\\{0}&{3}\end{bmatrix}

    And I find the eigenvectors are

    (1, 0), and (1, 8/7)

    Any idea why not give me exactly 7, 8?

    Thank you very much


    Greetings

    First, if  u is an eigenvector of a linear operator/matrix, then also tu is an eigenvector, for any scalar k\ne 0...so, can you see something interesting now?
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  6. #6
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    Yes, I understand,

    Thank you very much again.

    Greetings
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