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Math Help - 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

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

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

    hence

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

    v2 = -2v1

    eigenvalues are \lambda = 4  \lambda = 0

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

    however answers says when \lamba =4

    matrix is \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 \begin{bmatrix}0 & -\frac{1}{2} \\ 0 & 0 \end{bmatrix} leads to v2= -\frac{1}{2}v2. This really the reduced form of the "augmented matrix" \begin{bmatrix} 1 & -\frac{1}{2} & 0 \\ 0 & 0 & 0\end{bmatrix}
    for solving the matrix equation \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 v1- \frac{1}{2}v2= 0 so that you get v1= \frac{1}{2}v2, not v1= -\frac{1}{2}v2.

    Your second error is in thinking you can set v2= 1 to get v_1= \frac{1}{2} (using the correct v1= \frac{1}{2}v2) and then set v1= 1to get 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 \begin{bmatrix}\frac{1}{2} \\ 1\end{bmatrix}. If you take v1= 1, you get v2= 2 and so have the eigenvector \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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