Gram-Schmidt of 2 by 2 matrix Given $A=\begin{bmatrix}
2 &-1 \\ 
-1 & 2
\end{bmatrix}$, I take the first column of $A$, and divide it by its norm to find $q_{1}=\begin{bmatrix}\frac{2}{\sqrt{5}} \\ \frac{-1}{\sqrt{5}}\end{bmatrix}$
Now if I denote the second column of $A$ as $b$, then $b-(q_{1}^{T}b)q_{1}$ divided by $||b-(q_{1}^{T}b)q_{1}||$ will give me $q_{2}$, correct? 
$\begin{bmatrix}
-1 \\ 
2
\end{bmatrix}-(\begin{bmatrix}\frac{2}{\sqrt{5}} \frac{-1}{\sqrt{5}} \end{bmatrix}\begin{bmatrix}
-1 \\ 
2
\end{bmatrix})\begin{bmatrix}\frac{2}{\sqrt{5}} \\ \frac{-1}{\sqrt{5}}\end{bmatrix}$
$=\begin{bmatrix}
-1 \\ 
2
\end{bmatrix}-(\frac{-4}{\sqrt{5}}\begin{bmatrix}\frac{2}{\sqrt{5}} \\ \frac{-1}{\sqrt{5}}\end{bmatrix})$
$=\begin{bmatrix}
-1 \\ 
2
\end{bmatrix}-\begin{bmatrix}
-8/5 \\ 
1/5
\end{bmatrix}$
$=\begin{bmatrix}
13/5 \\ 
9/5
\end{bmatrix}$
And its norm is $\sqrt{10}$, so $q_{2}=\begin{bmatrix}
13/(5\sqrt{10}) \\ 
9/(5\sqrt{10})
\end{bmatrix}$
However this is not the right answer in the book. Where did I go wrong?

The error was: $\displaystyle \left(\frac{-4}{\sqrt{5}}\right)\left(\frac{-1}{\sqrt{5}}\right)\neq \frac{1}{5}$
