Finding the orthogonal complement where a single subspace is given

Question

For this question, am I doing it right so far?

Find the orthogonal complement of the given subspace $sp([1,2,-1]) in \Bbb R^3$

$a[1,2,-1].[x,y,z]$

= $ a+x, 2a+2y, -a+z$

$A = [1,1][2,2][-1,1]$

$A^T = [1,2,-1][1,2,1]$

After row reducing I get $[1,2,0][0,0,1]$

and then I let $x_2 = s$

$x_1 = -2s$

$x_3 = 0$

so I get $s[-2,1,0]$ as my final answer.

Answer 1

Let $W$ be the subspace of $\mathbb{R}^3$ given by all the vectors orthogonal to $u=[1,2,-1]$

Finding the orthogonal compliment is finding a basis of unit vectors of $W$.

$$x+2y-z=0$$

$$x=-2y+z$$ Now, $$(x,y,z)^T=(-2y+z,y,z)^T=(-2y,y,0)^T+(z,0,z)^T$$ $$=y(-2,1,0)^T+z(1,0,1)^T$$ So, the required vectors are $[-2,1,0]^T$ and $[1,0,1]^T$

Answer 2

The correct way to solve this is find the general vector $v=(x,y,z)$ satisfying $$\left<v,(1,2,-1)\right>=0$$ Using the usual inner product for $\mathbb{R^{3}}$, $$\left<v,(1,2,-1)\right>=0\Rightarrow x+2y-z=0$$