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Thread: Eigenvalue question

  1. #1
    Senior Member Danneedshelp's Avatar
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    Eigenvalue question

    Q1: Let $\displaystyle T:M_{2,2}\rightarrow{M_{2,2}}$ be represented by

    $\displaystyle T\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)$$\displaystyle =
    \begin{bmatrix}a-c+d&b+d\\-2a+2c-2d&2b+2d\end{bmatrix}$ .

    Find the eigenvaulues and eigenvectors of $\displaystyle T$ relative to the standard basis

    $\displaystyle B=$ $\displaystyle
    \begin{bmatrix}1&0\\0&0\end{bmatrix},
    \begin{bmatrix}0&1\\0&0\end{bmatrix},
    \begin{bmatrix}0&0\\1&0\end{bmatrix},
    \begin{bmatrix}0&0\\0&1\end{bmatrix}
    $ .

    A:

    I evaluated $\displaystyle T$ at each vector in the basis...

    $\displaystyle T\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)$$\displaystyle =
    a-c+d$
    $\displaystyle T\left(\begin{bmatrix}0&1\\0&0\end{bmatrix}\right)$$\displaystyle =
    b+d$
    $\displaystyle T\left(\begin{bmatrix}0&0\\1&0\end{bmatrix}\right)$$\displaystyle =
    -2a+2c-2d$
    $\displaystyle T\left(\begin{bmatrix}0&0\\0&1\end{bmatrix}\right)$$\displaystyle =
    2b+2d$

    Then, I created a matrix $\displaystyle A$ from the above information...

    $\displaystyle A=\begin{bmatrix}1&0&-1&1\\0&1&0&1\\-2&0&2&-2\\0&2&0&2\end{bmatrix}$

    I then found my eigenvauls to be $\displaystyle \lambda=0$ and $\displaystyle \lambda=3$ which is in the back of the book.

    But, when I evaluate $\displaystyle \lambda{I}-A$ for either value, I end up with $\displaystyle 4\times{1}$ column vector's. Both eigenvectors in the book are $\displaystyle 2\times{2}$ matrices.

    Where did I mess up?
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Remember that the basis $\displaystyle B$ consists of the four matrices $\displaystyle
    \begin{bmatrix}1&0\\0&0\end{bmatrix},
    \begin{bmatrix}0&1\\0&0\end{bmatrix},
    \begin{bmatrix}0&0\\1&0\end{bmatrix},
    \begin{bmatrix}0&0\\0&1\end{bmatrix}
    $. So if your eigenvector calculation leads to a column vector with entries p,q,r,s, then those four numbers are actually the coefficients in a linear combination of the basis vectors. So the eigenvector is the matrix $\displaystyle
    p\begin{bmatrix}1&0\\0&0\end{bmatrix}
    +q\begin{bmatrix}0&1\\0&0\end{bmatrix}
    +r\begin{bmatrix}0&0\\1&0\end{bmatrix}
    +s\begin{bmatrix}0&0\\0&1\end{bmatrix} =\begin{bmatrix}p&q\\r&s\end{bmatrix}
    $.
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