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Math Help - Eigenvalue question

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

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

    T\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right) =<br />
\begin{bmatrix}a-c+d&b+d\\-2a+2c-2d&2b+2d\end{bmatrix} .

    Find the eigenvaulues and eigenvectors of T relative to the standard basis

    B= <br />
\begin{bmatrix}1&0\\0&0\end{bmatrix},<br />
\begin{bmatrix}0&1\\0&0\end{bmatrix},<br />
\begin{bmatrix}0&0\\1&0\end{bmatrix},<br />
\begin{bmatrix}0&0\\0&1\end{bmatrix}<br />
.

    A:

    I evaluated T at each vector in the basis...

    T\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right) =<br />
a-c+d
    T\left(\begin{bmatrix}0&1\\0&0\end{bmatrix}\right) =<br />
b+d
    T\left(\begin{bmatrix}0&0\\1&0\end{bmatrix}\right) =<br />
 -2a+2c-2d
    T\left(\begin{bmatrix}0&0\\0&1\end{bmatrix}\right) =<br />
 2b+2d

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

    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 \lambda=0 and \lambda=3 which is in the back of the book.

    But, when I evaluate \lambda{I}-A for either value, I end up with 4\times{1} column vector's. Both eigenvectors in the book are 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 B consists of the four matrices <br />
\begin{bmatrix}1&0\\0&0\end{bmatrix},<br />
\begin{bmatrix}0&1\\0&0\end{bmatrix},<br />
\begin{bmatrix}0&0\\1&0\end{bmatrix},<br />
\begin{bmatrix}0&0\\0&1\end{bmatrix}<br />
. 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 <br />
p\begin{bmatrix}1&0\\0&0\end{bmatrix}<br />
+q\begin{bmatrix}0&1\\0&0\end{bmatrix}<br />
+r\begin{bmatrix}0&0\\1&0\end{bmatrix}<br />
+s\begin{bmatrix}0&0\\0&1\end{bmatrix} =\begin{bmatrix}p&q\\r&s\end{bmatrix}<br />
 .
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