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Math Help - Finding the eigenvalues

  1. #1
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    Finding the eigenvalues

    Consider the matrix B =

    <br />
B = \begin{bmatrix} 1&-3\\1&1\end{bmatrix}<br />



    Find the eigenvalues 1 and 2 of B. Write eugenvalues in the form a+bi, where a; b E R.



    When I solve for the determinant i get (1-/|(eigenvalue) )+3 and I'm stuck after because when i substitute a +bi instead of the eigenvalue I get a^2 + 2abi +b^2 +4 and it doesn't really help me.

    If you know how to solve this I would really appreciate your help.Thanks!
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  2. #2
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    When it says write eigenvalues in the form a + bi, it means that you are going to get complex numbers as your values.

    To find the eigen values we need to solve this following equation, (which I think you have already stated above, kind of.)

    (1- \lambda)(1- \lambda) - 1(-3) = 0

    Expanding out the brackets above and rearranging a little gives you the following equation:

    \lambda^2 - 2\lambda + 4 = 0.

    Solve this using the quadratic formula to get your eigenvalues.

    Hope this helps.
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  3. #3
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    Quote Originally Posted by anna123456 View Post
    Consider the matrix B =

    <br />
B = \begin{bmatrix} 1&-3\\1&1\end{bmatrix}<br />



    Find the eigenvalues 1 and 2 of B. Write eugenvalues in the form a+bi, where a; b E R.



    When I solve for the determinant i get (1-/|(eigenvalue) )+3 and I'm stuck after because when i substitute a +bi instead of the eigenvalue I get a^2 + 2abi +b^2 +4 and it doesn't really help me.

    If you know how to solve this I would really appreciate your help.Thanks!
     det (B-\lambda I) = det \begin{bmatrix} 1-\lambda &-3\\1&1- \lambda \end{bmatrix} = (1-\lambda )^2 + 3 = 0 \Leftrightarrow (1-\lambda )^2 = -3 ...

    Can you solve now?
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  4. #4
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    Quote Originally Posted by Defunkt View Post
     det (B-\lambda I) = det \begin{bmatrix} 1-\lambda &-3\\1&1- \lambda \end{bmatrix} = (1-\lambda )^2 + 3 = 0 \Leftrightarrow (1-\lambda )^2 = -3 ...

    Can you solve now?
    The problem is it doesn't have a solution in this form because when something is squared it's always going to be positve or equal to 0.
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  5. #5
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    Quote Originally Posted by anna123456 View Post
    The problem is it doesn't have a solution in this form because when something is squared it's always going to be positve or equal to 0.
    Over \mathbb{R}, yes, but not over the field of complex numbers \mathbb{C}. The question even says that you are going to have a complex solution.

    (1-\lambda )^2 = -3 \Leftrightarrow 1 - \lambda = \pm \sqrt{-3} = \pm \sqrt{3}i \Leftrightarrow \lambda = 1 \pm \sqrt{3}{i}
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