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Math Help - Upper triangular matrix

  1. #1
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    Upper triangular matrix

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    How do I form P and T? Is P just made up of 4 eigenvectors? If so, how do I find the 4th eigenvector? And how do I find it? Is upper triangular matrix just the eigenvalues along the main entry?

    edit: the 4 eigenvalues are 2,2,-2,1
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  2. #2
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    Quote Originally Posted by qwerty1234 View Post
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    How do I form P and T? Is P just made up of 4 eigenvectors? If so, how do I find the 4th eigenvector? And how do I find it? Is upper triangular matrix just the eigenvalues along the main entry?

    edit: the 4 eigenvalues are 2,2,-2,1
    Since two is a repeated eigenvalue that admits only one eigenvector you need to find a generalized eigenvalue vector

    You need a vector

    \mathbf{P}=\begin{pmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \end{pmatrix}

    (A-2I)\mathbf{P}=\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}

    Note that the right hand side is the eigenvector for lambda equals 2.

    Note 2: Generalized eigenvectors are not unique you just need to pick one vector that works.

    Once you have this vector p use the other 3 eigenvectors to make the transformation matrix.
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  3. #3
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    Thanks, and how do I find T? Is upper triangular matrix just the eigenvalues along the main entry?

    lambda 0 0 0
    0 lambda 0 0
    0 0 lambda 0
    0 0 0 lambda
    ?

    Thanks
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  4. #4
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    Quote Originally Posted by qwerty1234 View Post
    Thanks, and how do I find T? Is upper triangular matrix just the eigenvalues along the main entry?

    lambda 0 0 0
    0 lambda 0 0
    0 0 lambda 0
    0 0 0 lambda
    ?

    Thanks
    No the matrix is not diagonalizable! You can get it into a Jordan canonical form which is an upper triangular matrix.

    Since lambda = 2 is the repeated e-value its Jordan block will be a 2 by 2 block. The other blocks will be 1 by 1.

    So you will get

     \begin{bmatrix} -2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2\end{bmatrix}
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    Since two is a repeated eigenvalue that admits only one eigenvector you need to find a generalized eigenvalue vector

    You need a vector

    \mathbf{P}=\begin{pmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \end{pmatrix}

    (A-2I)\mathbf{P}=\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}

    Note that the right hand side is the eigenvector for lambda equals 2.
    Thanks, i tried solving (A-2I)P, and I got:
    1
    0
    -1
    1

    And i checked it with a online calculator.

    BTW, do i just add the 4 eigenvectors to get P?
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  6. #6
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    Quote Originally Posted by qwerty1234 View Post
    Thanks, i tried solving (A-2I)P, and I got:
    1
    0
    -1
    1

    And i checked it with a online calculator.

    BTW, do i just add the 4 eigenvectors to get P?
    Yes but What I typed in the previous post is wrong the vector on the right hand side is not the eigenvector for lambda =2. I wrote down the wrong vector.

    It should be

    (A-2I)\marhbf{P}=\begin{pmatrix}1 \\ 0 \\ -1 \\ 1 \end{pmatrix}

    and I get

    \mathbf{P} = \begin{pmatrix}-1 \\ -1 \\ 1 \\ 0 \end{pmatrix}

    Again sorry for the typo
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