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Math Help - Find the characteristic roots and characteristic vector

  1. #1
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    Find the characteristic roots and characteristic vector

    Find the characteristic roots and characteristic vectors of the matrix A given by

    <br />
A= \begin{bmatrix}<br />
1 & 2 & 3 \\<br />
0 & 2 & 3 \\<br />
0 & 0 & 2<br />
\end{bmatrix}<br />
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  2. #2
    Super Member Gamma's Avatar
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    This matrix is already in diagonal form. You find the roots of the characteristic polynomial det(A-Ix). In this case, they are just the things on the diagonal.

    To find the "characteristic vectors" I assume you are talking about the eigenvectors. Simply put the eigenvalue in for x in A-Ix (just subtract the value from the diagonal). Then you set that equal to 0 and find a basis for this solution space (the null space). This is the basis for the kernel of (A-I\lambda)x.
    Last edited by Gamma; May 4th 2009 at 02:11 AM. Reason: accidently wrote eigenvalues instead of vector
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by Gamma View Post
    This matrix is already in diagonal form. [snip]
    I'm sure you meant to say upper triangular form.

    (By the way, welcome and thanks for all your help so far at MHF )
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  4. #4
    Super Member Gamma's Avatar
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    Whoa! whoops. Yeah exactly. it is 4:42 am here, I think it is bedtime, lol. Take care.
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  5. #5
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    I am stuck here

    Quote Originally Posted by mr fantastic View Post
    I'm sure you meant to say upper triangular form.

    (By the way, welcome and thanks for all your help so far at MHF )

    I have got \lambda = 1,2

    Taking \lambda = 1

    \begin{bmatrix} 0 & 2 & 3\\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0  \end{bmatrix} \Rightarrow \begin{cases} 2y + 3z = 0 \\ y + 3z =0 \\z = 0 \end{cases}

    I am stuck here what should i do know from here onwards ?????
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  6. #6
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    Quote Originally Posted by zorro View Post
    I have got \lambda = 1,2

    Taking \lambda = 1

    \begin{bmatrix} 0 & 2 & 3\\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0  \end{bmatrix} \Rightarrow \begin{cases} 2y + 3z = 0 \\ y + 3z =0 \\z = 0 \end{cases}

    I am stuck here what should i do know from here onwards ?????
    That should tell you that y=0, z=0 and x is arbitrary. So take:

    \bold{x}_{(\lambda=1)}=\left[\begin{array}{c}1\\0\\0 \end{array}\right]

    as the eigenvector corresponding to eigen value \lambda=1

    When you work this for the other eigen value (which has multiplicity 2) you will find two solutions for the eigen vector (defined up to an arbitrary constant).

    CB
    Last edited by CaptainBlack; December 20th 2009 at 12:06 AM.
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  7. #7
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    Is this what u are talking

    Quote Originally Posted by CaptainBlack View Post
    That should tell you that y=0, z=0 and x is arbitrary. So take:

    \bold{x}_{(\lambda=1)}=\left[\begin{array}{c}1\\0\\0 \end{array}\right]

    as the eigenvector corresponding to eigen value \lambda=1

    When you work this for the other eigen value (which has multiplicity 2) you will find two solutions for the eigen vector (defined up to an arbitrary constant).

    CB

    \lambda = 2

    \begin{bmatrix} -1 & 2 & 3 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \begin{cases} -x + 2y + 3z = 0 \\ 3z = 0 \\ \end{cases}


    \therefore z = 0

    and - x + 2y = 0

    U are correct so What should i do now ?????
    Last edited by mr fantastic; December 20th 2009 at 04:00 AM. Reason: Added tags to a bit of latex code.
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  8. #8
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    Smile

    Quote Originally Posted by zorro View Post
    \lambda = 2

    \begin{bmatrix} -1 & 2 & 3 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \begin{cases} -x + 2y + 3z = 0 \\ 3z = 0 \\ \end{cases}


    \therefore z = 0

    and - x + 2y = 0

    U are correct so What should i do now ?????
    let y=t, t \in \mathbb{R}, then x=2t, and the solution is (x,y,z) = (2t, t, 0) = t(2, 1, 0)

    so, the eigen vector is
    \bold{x}_{(\lambda=2)}=\left[\begin{array}{c}2\\1\\0 \end{array}\right]
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