1. ## Eigenvalue question

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

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

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?

2. 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}
$
.