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Math Help - Eigenvectors/values

  1. #1
    Junior Member
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    Eigenvectors/values

    Let A = \begin{bmatrix}1&-1&0\\-1&1&0\\0&0&1\end{bmatrix} and T: R^3 \rightarrow R^3 by T(X) = AX.
    Find eigenpairs for A, find a basis for Ker(T) using the eigenvectors previously found, find a basis for Im(T) using eigenvectors previously found.

    The eigenpairs I got were...
    (0, \begin{bmatrix}1\\1\\0\end{bmatrix}), (1, \begin{bmatrix}0\\0\\1\end{bmatrix}), and (2, \begin{bmatrix}-1\\1\\0\end{bmatrix}

    I wasn't sure about the Kernel or Image parts but for the kernel I got
    t \begin{bmatrix}1\\1\\0\end{bmatrix}
    And for the image I got
    [r\begin{bmatrix}0\\0\\1\end{bmatrix}, s\begin{bmatrix}-1\\1\\0\end{bmatrix}]

    Could someone please verify if this is correct or not? I assume the kernel and image are not so could someone also please help me with that?

    Thanks
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  2. #2
    MHF Contributor

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    Yes, that is correct. Your matrix is 3 by 3 so the base space is 3 dimensional and you have three independent eigenvectors. Since there are 3 and they are independent (eigevectors corresponding to distinct eigenvalues are always independent), they form a basis for the space. That means that any vector v, can be written as a\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ c\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}.

    Applying A to that, Av= A(a\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ c\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}) = aA\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ bA\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ cA\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix} = a(0)\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ b(1)\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ c(2)\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}.

    That is, Av= 0 if and only if v= a\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} so \{\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}\} is a basis for the kernel of A and Av= b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ 2c\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix} so \{\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}, \begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}\} is a basis for the image.
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