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Thread: Basic probelm on 2x2 Eigenvector

  1. #1
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    Basic probelm on 2x2 Eigenvector

    Hi

    can someone explain to be how they get the eigenvectors for the following matrix

    $\displaystyle \begin{matrix}2 & -1 \\ -4 & 2 \end{matrix}$

    $\displaystyle \begin{matrix}1 & \frac{-1}{2} \\ 0 & 0 \end{matrix}$

    hence

    $\displaystyle v1 = \frac{-1}{2}v2$

    $\displaystyle v2 = -2v1$

    eigenvalues are $\displaystyle \lambda = 4 \lambda = 0$

    i get $\displaystyle \begin{matrix} \frac{-1}{2} \\ -2 \end{matrix}$

    however answers says when $\displaystyle \lamba =4 $

    matrix is $\displaystyle \begin{matrix} 1 \\ -2 \end{matrix}$

    i don't understand how they got this result, someone please explain this to me.

    P.S
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Basic probelm on 2x2 Eigenvector

    (2 - L)^2 - 4 = 0 is the characteristic polynomial.
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  3. #3
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    Re: Basic probelm on 2x2 Eigenvector

    You are making two errors. The first is in thinking that the row reduced matrix $\displaystyle \begin{bmatrix}0 & -\frac{1}{2} \\ 0 & 0 \end{bmatrix}$ leads to $\displaystyle v2= -\frac{1}{2}v2$. This really the reduced form of the "augmented matrix" $\displaystyle \begin{bmatrix} 1 & -\frac{1}{2} & 0 \\ 0 & 0 & 0\end{bmatrix}$
    for solving the matrix equation $\displaystyle \begin{bmatrix}2 & -1 \\ -4 & 2\end{bmatrix}\begin{bmatrix}v1 \\ v2\end{bmatrix}= \begin{bmatrix}4v1 \\ 4v2\end{bmatrix}$.

    That is, it corresponds to the equation $\displaystyle v1- \frac{1}{2}v2= 0$ so that you get $\displaystyle v1= \frac{1}{2}v2$, not $\displaystyle v1= -\frac{1}{2}v2$.

    Your second error is in thinking you can set v2= 1 to get $\displaystyle v_1= \frac{1}{2}$ (using the correct $\displaystyle v1= \frac{1}{2}v2$) and then set $\displaystyle v1= 1$to get $\displaystyle v_2= 2$. You can do one or other, not both! If you set v2= 1, you get v1= 1/2 and so have the eigenvector $\displaystyle \begin{bmatrix}\frac{1}{2} \\ 1\end{bmatrix}$. If you take v1= 1, you get v2= 2 and so have the eigenvector $\displaystyle \begin{bmatrix}1 \\ 2\end{bmatrix}$.

    Since one is a multiple of the other, both are eigenvectors corresponding to eigenvalue 4.
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  4. #4
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    Re: Basic probelm on 2x2 Eigenvector

    ok i understand now thanks for the explanation
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