Finding a basis for a subspace Find a basis for the subspace of $\mathbb{R}^4$ consisting of all vectors that satisfy $x_1 + 2x_2 - x_3 = 0$
My general idea so far is:
$x_1 = -2x_2 +x_3 + 0x_4$
$x_2 = $ free
$x_3 = $ free
$x_4 = $ free
Where should I go from here? Am I even going about it correctly?
 A: What you have is an expression for every vector in the subspace in parametric form, with three parameters:
$$\begin{align*}
x_1 &= -2r + s\\
x_2 &=r\\
x_3 &=s\\
x_4 &=t
\end{align*}$$
with $r,s,t\in\mathbb{R}$, arbitrary.
To get a basis for the space, for each parameter, set that parameter equal to $1$ and the other parameters equal to $0$ to obtain a vector. Each parameter gives you a vector. So setting $r=1$ and $s=t=0$ gives you one vector; setting $s=1$ and $r=t=0$ gives you a second vector; setting $t=1$ and $r=s=0$ gives you a third.
Alternatively, you can try rewriting the parametric solution in vector form:
$$\left(\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{c}-2r-s\\r\\s\\t\end{array}\right) = \left(\begin{array}{r}-2\\1\\0\\0\end{array}\right)r + \cdots$$
(I'll let you finish it up).
A: Hint:  given that you have one restriction, you expect a three-dimensional subspace.  You need to find three independent vectors in that subspace.  $(0,0,0,1)$ is in the subspace, but $(0,1,0,0)$ is not.  How can you modify $(0,1,0,0)$ to satisfy $x_1 + 2x_2 - x_3 = 0$?
A: Hint: You just need to find the subspace/set of vectors orthogonal to the vector $(1,2,-1,0)^T$.
Edit
You can do orthogonal projection of the standard basis of $\mathbb{R}^4$ to $(1,2,-1,0)^{T {\perp}}$.
