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Thread: How to find nonsingular matrix?

  1. #1
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    How to find nonsingular matrix?

    Find, if possible, a nonsingular matrix P such that $\displaystyle P^{-1}AP$ is diagonal.

    $\displaystyle
    \begin{pmatrix}1 & 1 & 2\\0 & 1 & 0\\0 & 1 & 3\end{pmatrix}
    $
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  2. #2
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    Quote Originally Posted by jennifer1004 View Post
    Find, if possible, a nonsingular matrix P such that $\displaystyle P^{-1}AP$ is diagonal.

    $\displaystyle
    \begin{pmatrix}1 & 1 & 2\\0 & 1 & 0\\0 & 1 & 3\end{pmatrix}
    $
    First you need to find the eigenvalues so you can find the eigen Basis for the matrix (if it exists)


    $\displaystyle
    \begin{pmatrix}1-\lambda & 1 & 2\\0 & 1-\lambda & 0\\0 & 1 & 3 -\lambda \end{pmatrix}
    $

    Expanding along the first column we get

    $\displaystyle (1-\lambda)[(1-\lambda)(3-\lambda)-(1)(0)]=(1-\lambda)^2(3-\lambda)$

    So the two eigenvalues are $\displaystyle \lambda =1,\lambda=3 $

    Using $\displaystyle \lambda =1 $ we get the matrix

    $\displaystyle
    \begin{pmatrix}0 & 1 & 2\\0 & 0 & 0\\0 & 1 & 2 \end{pmatrix}
    $

    Reducing we get

    $\displaystyle
    \begin{pmatrix}0 & 1 & 2\\0 & 0 & 0\\0 & 0 & 0 \end{pmatrix}
    $

    This implies that there are two parameters in the solution so let

    $\displaystyle x=s, z=t$ so the last equation is

    $\displaystyle y+2z=0 \iff y+2t=0 \implies y =-2t$

    so the solution is

    $\displaystyle \begin{pmatrix}
    s \\
    -2t \\
    t
    \end{pmatrix} =
    s \cdot \begin {pmatrix}
    1 \\ 0 \\ 0
    \end {pmatrix}
    +t \cdot \begin {pmatrix}
    0 \\ -2 \\ 1
    \end{pmatrix}
    $

    So these are the first two eigenvectors

    Now using the eigen values $\displaystyle \lambda =3$ we get

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

    Reducing we get

    $\displaystyle
    \begin{pmatrix}1 & 0 & -1\\0 & 1 & 0\\0 & 0 & 0 \end{pmatrix}
    $

    This has 1 parameter in its solution so let $\displaystyle z=t$

    We can see that $\displaystyle y=0$ and $\displaystyle x=z \iff x=t$

    So the solution is

    $\displaystyle \begin{pmatrix}
    t \\ 0 \\ t
    \end{pmatrix}=t \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end {pmatrix}$

    The eigenvectors are the columns of the matrix P.
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  3. #3
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    so my answer is simply $\displaystyle
    \begin{pmatrix}1 & 0 & 1\\0 & -2 & 0\\0 & 1 & 1\end{pmatrix}
    $?

    Thank you for showing your work. It has helped me so much. I can now use your example to help me figure out problems on my own.
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    Quote Originally Posted by jennifer1004 View Post
    so my answer is simply $\displaystyle
    \begin{pmatrix}1 & 0 & 1\\0 & -2 & 0\\0 & 1 & 1\end{pmatrix}
    $?

    Thank you for showing your work. It has helped me so much. I can now use your example to help me figure out problems on my own.
    Yes that is the solution
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  5. #5
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    Out of curiosity, you wouldn't multiply the eigenvector for $\displaystyle \lambda=3$ by 3 for the final matrix, which would make it $\displaystyle
    \begin{pmatrix}1 & 0 & 3\\0 & -2 & 0\\0 & 1 & 3\end{pmatrix}$?
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