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Thread: characteristic polynomial

  1. #1
    Junior Member piglet's Avatar
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    characteristic polynomial

    What's the easiest way to calculate the characteristic polynomial of a 3x3 matrix?

    I'm aware this is the formula: det(A-lambda(I)), dont really know how to apply it though.

    Say matrix was: 1 2 1
    1 3 1
    1 2 1


    i dont want the solution, just the actual method.. or if someone could run through step by step it would be great
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  2. #2
    MHF Contributor
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    Suppose that

    $\displaystyle A\mathbf{x} = \lambda \mathbf{x}$

    for some matrix $\displaystyle A$, some vector $\displaystyle \mathbf{x}$ and some scalar $\displaystyle \lambda$.

    Then $\displaystyle \mathbf{x}$ is called an eigenvector and $\displaystyle \lambda$ is the corresponding Eigenvalue.

    The thing is, how do you find $\displaystyle \lambda$ and $\displaystyle \mathbf{x}$?


    Notice if $\displaystyle A\mathbf{x} = \lambda \mathbf{x}$

    It might make sense to move everything to one side.

    Unfortunately $\displaystyle A$ is a matrix and $\displaystyle \lambda$ is a scalar. So you instead write

    $\displaystyle A\mathbf{x} = \lambda I \mathbf{x}$.

    Now that the matrices have the same dimensions, you can subtract them


    $\displaystyle A\mathbf{x} - \lambda I \mathbf{x} = \mathbf{0}$

    $\displaystyle (A - \lambda I)\mathbf{x} = \mathbf{0}$.


    Clearly, this has a trivial solution if $\displaystyle \mathbf{x} = \left[\begin{matrix}0 \\ 0 \\ \vdots \\ 0\end{matrix}\right]$.

    For a nontrivial solution

    $\displaystyle A - \lambda I = \mathbf{0}$.

    Since this is a zero matrix, it is singular.

    So $\displaystyle |A - \lambda I| = 0$.


    We can use this information to find $\displaystyle \lambda$.



    In your case:

    $\displaystyle A = \left[\begin{matrix}1 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 1\end{matrix}\right]$.


    You need to work out

    $\displaystyle |A - \lambda I|$.


    $\displaystyle A - \lambda I = \left[\begin{matrix}1 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 1\end{matrix}\right] - \left[\begin{matrix}\lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda\end{matrix}\right]$

    $\displaystyle = \left[\begin{matrix}1 - \lambda & 2 & 1 \\ 1 & 3 - \lambda & 1 \\ 1 & 2 & 1 - \lambda\end{matrix}\right]$


    Therefore $\displaystyle |A - \lambda I| = \left|\begin{matrix}1 - \lambda & 2 & 1 \\ 1 & 3 - \lambda & 1 \\ 1 & 2 & 1 - \lambda\end{matrix}\right|$

    $\displaystyle = (1 - \lambda)\left|\begin{matrix}3 - \lambda & 1 \\ 2 & 1 - \lambda\end{matrix}\right| - 2\left|\begin{matrix}1 & 1 \\ 1 & 1 - \lambda\end{matrix}\right| + 1\left|\begin{matrix}1 & 3-\lambda \\ 1 & 2\end{matrix}\right|$

    $\displaystyle = (1 - \lambda)[(3 - \lambda)(1 - \lambda) - 1\cdot 2] - 2[1(1 - \lambda) - 1\cdot 1] + 1[1 \cdot 2 - (3 - \lambda)1]$

    $\displaystyle = (1 - \lambda)(3 - 3\lambda - \lambda + \lambda^2 - 2) - 2(1 - \lambda - 1) + 2 - 3 + \lambda$

    $\displaystyle = (1 - \lambda)(1 - 4\lambda + \lambda^2) + 2\lambda -1 + \lambda$

    $\displaystyle = 1 - 4\lambda + \lambda^2 - \lambda + 4\lambda^2 - \lambda^3 + 3\lambda - 1$

    $\displaystyle = -\lambda^3 + 5\lambda^2 - 2\lambda$.


    Since this determinant is zero...

    $\displaystyle -\lambda^3 + 5\lambda^2 - 2\lambda = 0$

    $\displaystyle -\lambda(\lambda^2 - 5\lambda + 2) = 0$


    You should be able to solve for $\displaystyle \lambda$ now.
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  3. #3
    Junior Member piglet's Avatar
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    thanks so much, especially for the theory and worked through problem
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