Determine Orthonormal basis in $R^4$ without calculator
Matthew Martinez 
In $R^4$ with the standard inner product, the linear subspace $U$ is given by:
$$U = <(0,1,2,2), (2,1,0,-1), (0,0,0,3)>$$
The linear subspace $V$ of $U$ consists of those elements in $U$ satisfying the equation $x_1 + x_2 + x_3 = 0$.
Determine an orthonormal basis $a_1, a_2, a_3, a_4$ of $R^4$ with $V = <a_1, a_2>$ and $U = <a_1, a_2, a_3>$.
I know how to determine an orthonormal basis using Gram-Schmidt. However, in the course I follow, we have to do this exercise without using a calculator, and the numbers I encounter are not nice to work with. So, am I missing something or making a mistake?
$$ u_1 = \left[
\begin{array}{c} 0\\ 1\\ 2\\ 2
\end{array}
\right] * \frac{1}{3} =
\left[ \begin{array}{c} 0\\ \frac{1}{3}\\ \frac{2}{3}\\ \frac{2}{3}
\end{array} \right]$$
$$u'_2 = \left[
\begin{array}{c} 2\\ 1\\ 0\\ -1
\end{array}
\right] - (u_1 \cdot \left[
\begin{array}{c} 0\\ 1\\ 2\\ 2
\end{array}
\right]) * u_1 = \left[
\begin{array}{c} 2\\ 1\\ 0\\ -1
\end{array}
\right] + \frac{1}{3}*\left[
\begin{array}{c} 0\\ \frac{1}{3}\\ \frac{2}{3}\\ \frac{2}{3}
\end{array}
\right] = \left[
\begin{array}{c} 2\\ \frac{10}{9}\\ \frac{2}{9}\\ \frac{-7}{9}
\end{array}
\right]$$
$$u_2 = \left[
\begin{array}{c} 2\\ \frac{10}{9}\\ \frac{2}{9}\\ \frac{-7}{9}
\end{array}
\right] * \frac{1}{\sqrt{2^2 + \frac{10}{9}^2+\frac{2}{9}^2+\frac{-7}{9}^2}} = \left[
\begin{array}{c} 2\\ \frac{10}{9}\\ \frac{2}{9}\\ \frac{-7}{9}
\end{array}
\right] * \frac{1}{\sqrt{\frac{477}{81}}}$$
From this point on, it is very difficult to calculate the other basis vectors.
$\endgroup$1 Answer
$\begingroup$Start with the vectors $u_1=(0,1,2,2)-(2,1,0,-1)=(-2,0,2,3)$, $u_2=(0,0,0,3)$, and $u_3=(0,1,2,2)$. Then $\{u_1,u_2\}$ is an orthogonal basis of $V$ and therefore it is very easy to apply Gramm-Shmidt to it. And, if $u_3=(2,1,0,-1)$, it's not hard to apply it to $\{u_1,u_2,u_3\}$ in order to get an orthonormal basis of $U$.
$\endgroup$ 3
