Finding basis for orthogonal subspace Find a basis for $S^\perp$ for the subspace $$ S = span\left\{\left[\begin{matrix}1\\1\\-2\end{matrix}\right]\right\}
$$
How do I start this question? 
 A: You have to find a basis of the null space of the matrix
$$
\begin{bmatrix}1 & 1 & -2\end{bmatrix}
$$
or, equivalently, to solve the system (with just one equation)
$$
x+y-2z=0
$$
The second and third variables are free, so you get two linearly independent solution with $y=1$, $z=0$ and with $y=0$, $z=1$. So the requested vectors are
$$
\begin{bmatrix}-1\\1\\0\end{bmatrix}
\qquad\text{and}\qquad
\begin{bmatrix}2\\0\\1\end{bmatrix}
$$
A: The orthogonal subspace of $S$ is $S^\perp = \{x:x\cdot v=0 \text{ for all } v\in S\}$.  So the first step is to find the vectors that are orthogonal to S.  Let $x$ be one such vector.  Then $x\cdot v$ is 0.  We know one such $v$; let $$v=\left[\begin{array}{c}1\\1\\-2\end{array}\right].$$
Then $x_1 + x_2 - 2 x_3=0$.  The set of vectors that satisfy this expression is a two dimensional subspace.  It's the set
$ \left\{\begin{array}{c}x_1\\x_2\\ \frac{1}{2} x_1 + \frac{1}{2} x_2\end{array}\right\}$.
So the orthogonal subspace is
$$S^\perp = \text{span}\left\{
  \left[\begin{array}{c}1\\0\\\frac{1}{2}\end{array}\right],
  \left[\begin{array}{c}0\\1\\\frac{1}{2}\end{array}\right]
\right\},$$
and a basis is
$$\left\{
  \left[\begin{array}{c}1\\0\\\frac{1}{2}\end{array}\right],
  \left[\begin{array}{c}0\\1\\\frac{1}{2}\end{array}\right]
\right\}.$$
