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?
2 Answers
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 $$ \begin{bmatrix}-1\\1\\0\end{bmatrix} \qquad\text{and}\qquad \begin{bmatrix}2\\0\\1\end{bmatrix} $$
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 $$\left\{ \left[\begin{array}{c}1\\0\\\frac{1}{2}\end{array}\right], \left[\begin{array}{c}0\\1\\\frac{1}{2}\end{array}\right] \right\}.$$
