Orthogonal complements really confuse me, I think its the notation? For example what do I do here, I know wha to do for part a but then...?

Let$$W=\operatorname{Span}\left\{\left(\begin{array}{c}1\\1\\0\\0\end{array}\right),\,\left(\begin{array}{c}0\\1\\1\\-1\end{array}\right)\right\}$$
(a) Show $v=\left(\begin{array}{c}1\\-1\\1\\0\end{array}\right)\in W^{\perp}$.
(b) Determine a basis for $W^{\perp}$.
(c) Determine a matrix $A$ such that $W$ is the row space of $W$ and $W^{\perp}$ is the null space of $A$.

 A: Let $S$ be some set of vectors. Then the 0rthogonal complement of $S$ is denoted by $S^{\perp}$ and is defined as the space of all those vectors in the vector space $V$ such that they are orthogonal to every vector in $S$. So
$$S^{\perp}=\{x \in V\, | \, x \cdot s=0 \,\, \forall s \in S\}.$$
In your question $W$ is a subspace generated by two given vectors (call them) $a,b$. So any vector $w \in W$ can be written as a linear combination of $a$ and $b$. This means $w=c_1a+c_2b$ for some $c_1,,c_2 \in \Bbb{R}$ (assuming the field is real numbers).
So if we want to test that the given $v \in W^{\perp}$, then all we have to do is to verify if $v$ is orthogonal to BOTH $a$ and $b$. Because once it is then it will be orthogonal to every $w \in W$. So try testing that.

For finding a basis, there are a couple of ways of doing that. Here is a (sort of) standard approach. First find $W^{\perp}$. Let $X=\begin{bmatrix}x &y&z&w\end{bmatrix}^T \in W^{\perp}$, then $X \cdot a=0$ and $X \cdot b=0$ implies
\begin{align*}
x+y&=0\\
y+z-w&=0
\end{align*}
Thus the solution set is
$$W^{\perp}=\left\{z\begin{bmatrix}1\\-1\\1\\0\end{bmatrix}+w\begin{bmatrix}-1\\1\\0\\1\end{bmatrix} \, \Big| \, z,w \in \Bbb{R}\right\}.$$
Thus a basis for $W^{\perp}$ is
$$\left\{\begin{bmatrix}1\\-1\\1\\0\end{bmatrix}, \,\begin{bmatrix}-1\\1\\0\\1\end{bmatrix} \right\}.$$
