Find a Basis $B$ of $R^2$ so that $B$ matrix of $T$ is diagonal $T([1,1]^t) = [3,7]^t$
$T([1,-1]^t) = [1,1]^t$
Here's what I get:
$T=
\left(\begin{array}{cc}3 & 1 \\7 & 1\end{array}\right)
$
The eigenvectors of $T$ is $E = \left(\begin{array}{cc} .4798 & -.2527 \\.8774 & .9675\end{array}\right)$.
$E^{-1}TE$ gives us a diagonal matrix. Am I doing this right? So my answer is that the columns of $E$ are the bases.
The answer I was given was $v_1 = (\frac{\sqrt(33)-3}{6},1)^t$ and $v_2 = (\frac{-(\sqrt(33) +3)}{6},1)^t$. Is this right? If so, how do I arrive at this solution?
 A: It is convenient to work with the standard basis $\left( \begin{matrix} 1 \\ 0  \end{matrix} \right), \left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$ to write the answers directly as vectors in the standard basis.
So we must solve, with $T= \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$
$$ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) =  \left( \begin{matrix} a+b  \\ c+d \end{matrix} \right) =  \left( \begin{matrix} 3 \\ 7 \end{matrix} \right) \textrm{ and }  \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) =  \left( \begin{matrix} a-b  \\ c-d \end{matrix} \right) =  \left( \begin{matrix} 1 \\ 1 \end{matrix} \right)$$
So we have two sets equations, namely $a+b =3$ and $ a -b =1$ which yields $a=2$ and $b=1$, and $c+d =7$ and $c-d=1$ which yields $c=4$ and $d=3$. 
So we find $T= \left( \begin{matrix} 2 & 1 \\ 4 & 3 \end{matrix} \right)$.
Then we can indeed find $E$ such that $E^{-1} T E = \left( \begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix} \right)$ is diagonal, which gives us
$$ E^{-1} T E \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \lambda_1\left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \textrm{ which gives } TE \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \lambda_1 E \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) $$
So if we take our basis the column vectors of $E$ we have that $E \left( \begin{matrix} 1 \\ 0 \end{matrix} \right)$ is the first column of $E$ and $E \left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$ is the second column of $E$.
Then we indeed have that this basis makes $T$ a diagonal matrix as we saw with $E \left( \begin{matrix} 1 \\ 0 \end{matrix} \right)$, and $E \left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$  can be shown at the same way.
This yields in the standard basis the vectors $v_1 =\left( \begin{matrix} \frac{1}{8} \left( \sqrt{17} -1 \right) \\ 1 \end{matrix} \right), v_2 =\left( \begin{matrix} \frac{-1}{8} \left( \sqrt{17} +1 \right) \\ 1 \end{matrix} \right)$. And if we write it in the basis $\left( \begin{matrix} 1 \\ 1 \end{matrix} \right), \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$ we find the vectors $v_1 =\left( \begin{matrix} \frac{-1}{4} \left( \sqrt{17} +5 \right) \\ 1 \end{matrix} \right),v_2 =\left( \begin{matrix} \frac{1}{4} \left( \sqrt{17} -5 \right) \\ 1 \end{matrix} \right)$. So the answer given to you is not correct, whereas your approach is the right approach. I would only recommend to write $T$ in the standard basis.
