Eigenvalue question

**Danneedshelp** · August 2nd 2009, 12:07 PM

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

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

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

$B=$ $ \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 $T$ at each vector in the basis...

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

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?

**Opalg** · August 2nd 2009, 12:35 PM

Remember that the basis $B$ consists of the four matrices $ \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 $ 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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