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Math Help - 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 P^{-1}AP is diagonal.

     <br />
\begin{pmatrix}1 & 1 & 2\\0 & 1 & 0\\0 & 1 & 3\end{pmatrix}<br />
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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 P^{-1}AP is diagonal.

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


     <br />
\begin{pmatrix}1-\lambda & 1 & 2\\0 & 1-\lambda & 0\\0 & 1 & 3 -\lambda \end{pmatrix}<br />

    Expanding along the first column we get

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

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

    Using \lambda =1 we get the matrix

     <br />
\begin{pmatrix}0 & 1 & 2\\0 & 0 & 0\\0 & 1 & 2 \end{pmatrix}<br />

    Reducing we get

     <br />
\begin{pmatrix}0 & 1 & 2\\0 & 0 & 0\\0 & 0 & 0 \end{pmatrix}<br />

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

    x=s, z=t so the last equation is

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

    so the solution is

    \begin{pmatrix}<br />
s \\<br />
-2t \\<br />
t <br />
\end{pmatrix} = <br />
s \cdot \begin {pmatrix}<br />
1 \\ 0 \\ 0 <br />
\end {pmatrix}<br />
+t \cdot \begin {pmatrix}<br />
0 \\ -2 \\ 1 <br />
\end{pmatrix}<br />

    So these are the first two eigenvectors

    Now using the eigen values \lambda =3 we get

     <br />
\begin{pmatrix}-2 & 1 & 2\\0 & -2 & 0\\0 & 1 & 0 \end{pmatrix}<br />

    Reducing we get

     <br />
\begin{pmatrix}1 & 0 & -1\\0 & 1 & 0\\0 & 0 & 0 \end{pmatrix}<br />

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

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

    So the solution is

    \begin{pmatrix}<br />
t \\ 0 \\ t<br />
\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  <br />
\begin{pmatrix}1 & 0 & 1\\0 & -2 & 0\\0 & 1 & 1\end{pmatrix}<br />
?

    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  <br />
\begin{pmatrix}1 & 0 & 1\\0 & -2 & 0\\0 & 1 & 1\end{pmatrix}<br />
?

    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 \lambda=3 by 3 for the final matrix, which would make it <br />
\begin{pmatrix}1 & 0 & 3\\0 & -2 & 0\\0 & 1 & 3\end{pmatrix}?
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