Basis for orthogonal complement 
Consider $\mathbb R^4$ together with the standard inner product and let $W=\{(x_1,
 x_2, x_3, x_4) \in \mathbb R^4 |\ x_1+x_2-x_4=0\}$. Find a basis for $W^\perp$.

How do you solve this question? 
All I can think of is that you have to put the elements in $W$ in terms of others.
 A: This is not the most elegant way, but here are some hints.
(1) First note that $W$ is the vector space given by
$$W = \{(t, s, u, t+s)\in \mathbb{R}^4 \lvert t,s,u \in \mathbb{R} \}.
$$
(2) Then for a vector $(a,b,c,d)$ to be in $W^{\perp}$ we need
$$(a,b,c,d)\cdot (t,s,u,t+s) = 0
$$
for all $t,s,u \in \mathbb{R}$.
(3) We get from this that
$$W^{\perp} = \{(a, a, 0, -a) \lvert a \in \mathbb{R}\}.
$$
(4) From here you should be able to write down a basis.
A: Since this is a hw, I will only give some hint!
First find a basis for $W$. To do so use the identity $x_4=x_1+x_2$ that defines $W$. You should find that any member $x=(x_1,x_2,x_3,x_4)$ of $W$ is a linear combination of 3 linearly independent vectors $w_1,w_2,w_3$, i.e. 
$w=\alpha_1 w_1+\alpha_2 w_2+\alpha_3w_3$ for some real numbers $\alpha_1,\alpha_2, \alpha_3$ (e.g. depending on $x_1,x_2,x_3$). Next, find a non-zero vector $v=(v_1,v_2,v_3,v_4)$ which is orthogonal to $w_1, \ w_2$, and $w_3$, meaning that
$$
v\cdot w_1=v\cdot w_2= v\cdot w_3=0.
$$
Any vector of the form $\alpha.v$, where $\alpha \in \mathbf{R}\setminus\{0\}$ is a basis for $W^\perp$.
A: Consider that $(1,1,0,-1)$ is orthogonal to all $(x_1,x_2,x_3,x_4)\in W$ because by definition $x_1+x_2-x_4=0$. Then $W^{\perp}=\langle(1,1,0,-1)\rangle$. 
A: Note that $W = \{(1,1,0,-1)\}^\perp =\langle(1,1,0,-1)\rangle^\perp$. So $W^\perp = \big(\langle(1,1,0,-1)\rangle^\perp\big)^\perp= \langle(1,1,0,-1)\rangle$, with basis $\{(1,1,0,-1)\}$.
A: I would first find a basis for the solution space to the homogenous system of equations given by $x_1 + x_2 - x_4 = 0$.  This turns out to be a 3 dimensional vector space.  To find a basis for the solution space, set $x_2 = 1, x_3 = 0$, and $x_4 = 0$, then solve for $x_1$.  This will provide you with your first vector in the basis.  For the second vector, repeat this process, again solving for $x_1$, but setting $x_3 = 1$ instead of $x_2$.  For the third and final vector, set $x_4 = 1$ and solve for $x_1$.  You can then set up a new system of equations where the vector $\langle y_1, y_2, y_3, y_4 \rangle$ inner-producted with each of the basis vectors you just found will be zero.  This will give you 3 equations in 4 variables.  The solution space to this new system will provide you with a basis for $W^{\perp}$.
