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Math Help - Eigen value , eigen vector

  1. #1
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    Eigen value , eigen vector

    Find the eigen value and eigen vector of \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix}

    Solution :

    |A- \lambda I | = \begin{bmatrix} (1 - \lambda) & 2 & 3 \\ 3 & (1 - \lambda) & 0 \\ -2 & 0 & (1 - \lambda) \end{bmatrix} =

    (1-\lambda)[(1 - \lambda)(1-\lambda) - 0]-2[3(1 - \lambda)-0]+3[0-(-2)(1 - \lambda)]
    = (1-\lambda)(1 - 2 \lambda + 2)- 6 + 6 \lambda+ 6 - 6 \lambda
    = 1 -2\lambda+\lambda^2-\lambda+2\lambda^2-\lambda^3
    =(1-\lambda)(1-2\lambda+\lambda^2)
    (1-\lambda)(\lambda-1)(\lambda-1)

    Eigen value \lambda =1

    putting \lambda=1 in the matrix eq

    \begin{bmatrix} 0 & 2 & 3 \\ 3 & 0 & 0 \\ -2 & 0 & 0 \end{bmatrix} . \begin{bmatrix}x \\ y \\z \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix} \Rightarrow \begin{cases}2y + 3z = 0 \\ 3x = 0 \\ -2x = 0 \end{cases}

    Stuck here !!!!!!!
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  2. #2
    Senior Member Shanks's Avatar
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    x=0,y=c,z=\frac{-2c}{3}
    the solution set is a subspace generated by (0,1,\frac{-2}{3}).
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  3. #3
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    thanks
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  4. #4
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    Quote Originally Posted by flintstone View Post
    Find the eigen value and eigen vector of \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix}

    Solution :

    |A- \lambda I | = \begin{bmatrix} (1 - \lambda) & 2 & 3 \\ 3 & (1 - \lambda) & 0 \\ -2 & 0 & (1 - \lambda) \end{bmatrix} =

    (1-\lambda)[(1 - \lambda)(1-\lambda) - 0]-2[3(1 - \lambda)-0]+3[0-(-2)(1 - \lambda)]
    = (1-\lambda)(1 - 2 \lambda + 2)- 6 + 6 \lambda+ 6 - 6 \lambda
    = 1 -2\lambda+\lambda^2-\lambda+2\lambda^2-\lambda^3
    =(1-\lambda)(1-2\lambda+\lambda^2)
    (1-\lambda)(\lambda-1)(\lambda-1)

    Eigen value \lambda =1

    putting \lambda=1 in the matrix eq

    \begin{bmatrix} 0 & 2 & 3 \\ 3 & 0 & 0 \\ -2 & 0 & 0 \end{bmatrix} . \begin{bmatrix}x \\ y \\z \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix} \Rightarrow \begin{cases}2y + 3z = 0 \\ 3x = 0 \\ -2x = 0 \end{cases}

    Stuck here !!!!!!!
    Both second and third equations just tell you that x= 0. The first equation tells you that 2y= -3z so y= -(3/2)z. Any eigenvector is of the form <0, -(3/2)z, z>= z<0, -3/2, 1>. If z= 2, that is <0, -3 , 2> so the space of eigenvectors of this matrix is spanned by that single vector (which is, of course, a multiple of the one Shanks gave).
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