Let $S$ be the Subspace of $R^3$ Spanned by x $=(1,-1,1)^T)$ 
Let $S$ be the Subspace of $R^3$ Spanned by x $=(1,-1,1)^T)$ 
Find a Basis for $S^⊥$

Solution: 

I know that vectors of $S^⊥$ are in the form $(y_2-y_3,y_1+y_3,y_2-y_1)^T$, but I'm not sure how to find the basis for $S^⊥$. I'm unsure why the solution assigns scalars to $y_2$ and $y_3$. What is a better method?
 A: From the answer: since $y_1=y_2-y_3$, one has that the solution may be written as
$$
\left( \begin{array}{c} y_1\\y_2\\y_3 \end{array}\right)=
\left( \begin{array}{ccc} y_2-y_3\\y_2\\y_3 \end{array}\right)=
\left( \begin{array}{ccc} y_2 \\ y_2 \\ 0 \end{array}\right)+\left( \begin{array}{ccc} -y_3 \\ 0 \\ y_3 \end{array}\right) = y_2\left(\begin{array}{c}1 \\ 1 \\ 0 \end{array}\right)+y_3\left( \begin{array}{ccc} -1 \\ 0 \\ 1 \end{array}\right)
$$
that is, it is a linear combination of the last two column vectors. Since it's a linear combination, it need scalars, which will be $y_2$ and $y_3$. Now, just give those scalars names: $y_2=\alpha$ and $y_3=\beta$.
Note, however that, since you wrote $(y_2-y_3,y_1+y_3,y_2-y_1)^T$, it is not wrong, but you will not be able to find a basis for the vector space writing it in this form (you'll end up finding three vectors, but in true you need only two). Best way is to let one of the scalars (in this case, $y_1$) be written as dependent of the last two (in this case, $y_2$ and $y_3$) and separate them into a linear combination as I did above. 
A: Your basis is fine. 
Any other set or two linearly independent vectors in $ S^\perp $ is also a basis for$ S^\perp$. 
