I am trying to find the orthogonal complement $W^{\bot}$ of the subspace $W = \text{span}\{(1,-1,2,0), (-2,1,0,1)\}$ of $\mathbb{R}^4$. My lecturer said that if some vector $v$ is a member of $W^{\bot}$ , then $v$ is orthogonal to the vectors $(1,-1,2,0)$ and $(-2,1,0,1)$. From this point, I understand how to get linear equations for the components of $v$ and use matrix methods to solve those equations, which gives the orthogonal complement of $W$.
However, I am not sure how to prove that a vector $v \in W^\bot$ implies that $v$ is orthogonal to each of the spanning vectors of $W$. It seems obvious, and I tried to show this implication by considering a general element of $w$ and expressing it as a linear combination of the spanning vectors. Then taking some arbitrary element $v$ of $W^\bot$ which is orthogonal to this linear combination. From there, I am not sure how to show that this implies that $v$ is orthogonal to each of the spanning vectors of $W$.
How does one show that this is true?
As астон вілла олоф мэллбэрг suggests, you seem confused. By definition $W^\perp$ consists of all vectors that are orthogonal to every vector in $W$. This includes the spanning vectors $x = (1,-1,2,0), y = (-2,1,0,1)$. So if you've identified $W^\perp$ correctly, then every vector in it has to be orthogonal to $x$ and $y$.
To prove that any specific vector $v = (v_1, v_2, v_3, v_4)$ is orthogonal to them is easy. Just take the dot product. $v$ and $w$ are orthogonal if and only if $$v \cdot w = v_1w_1 + v_2w_2 + v_3w_3 + v_4w_4 = 0$$
So let's look at that process you say you understand. To find the vectors $v$, we look at the dot products with all vectors $w \in W$. They all need to be $0$. But since $x$ and $y$ span $W$, there are always scalars $a,b$ such that $w = ax + by$, and by properties of the dot product, we get $$v \cdot w = v \cdot (ax + by) = a(v \cdot x) + b(v\cdot y)$$ Therefore if $v \cdot x$ and $v \cdot y$ are both $0$, then so will be $v \cdot w$ for every $w \in W$. That is, we only need to look for $v$ that satisfy $v \cdot x = 0$ and $v \cdot y = 0$, instead of having to look at every vector in $W$. This gives us the two equations
$$v_1x_1 + v_2x_2 + v_3x_3 + v_4x_4 = 0\\v_1y_1 + v_2y_2 + v_3y_3 + v_4y_4 = 0$$ or plugging in the coefficient values: $$\begin{align}v_1 - v_2 + 2v_3 &= 0\\-2v_1 + v_2 + v_4 &= 0\end{align}$$
These are the equations you are solving. But notice what those equations are saying: they are exactly the conditions of being orthogonal to $x$ and to $y$. So any solution to them is automatically orthogonal to the spanning vectors of $W$.
If you have specific solutions that you want to test to make sure you've solved them correctly, just plug the values of $v_1, v_2, v_3, v_4$ into the two equations and see if you really do get $0$. If so, then they are actual solutions, and so are orthogonal to $x$ and $y$.