Let $S=\{f_1,f_2,f_3\}$ Find a basis for $S^\perp$ Let $S=\{f_1,f_2,f_3\}$ where $f_1,f_2,f_3$ are linear forms defined by $f_1(x,y,z)=x+y-z, f_2(x,y,z)=2x-3y+z,f_3(x,y,z)=3x-2y$  Find a basis for $S^\perp$
Definition: $S^\perp=\{f\in V^*:<f,v>=0 \forall v\in V\}$
Definition: $V^*=\{f:V\rightarrow \mathbb{K}\}$ where $\mathbb{K}$ is a field.
My work:
$f\in S^\perp\leftrightarrow <f,f_1(x,y,z>+<f,f_2(x,y,z)>+<f,f_3(x,y,z)>=0\leftrightarrow f(f_1(x,y,z))+f(f_2(x,y,z)+f(f_3(x,y,z))=0\leftrightarrow f(x+y-z)+f(2x-3y+z)+f(3x-2y)=0$
In this step i'm stuck. Can someone help me?
 A: Let's assume that for $T\subseteq V$, the definition of $T^\perp$ is $\{f\in V^*\colon\langle f,v\rangle=0,\forall v\in S\}.$ Because if we instead use $\{f\in V^*\colon\langle f,v\rangle=0,\forall v\in V\},$ well the only linear form in $V^*$ that annihilates all vectors in $V$ is $0$. And the zero vector space has an empty basis.
In our case, $S$ is a subset of $V^*$, so $S^\perp$ will be a subset of $V^{**}.$
If $V$ is finite dimensional, then let $\{e_i\}_{1\leq i\leq n}$ be a basis for $V$. Define $\delta_i(e_j)=\begin{cases}1 & i=j\\0 & i\neq j\end{cases}.$ Then $\{\delta_i\}_{1\leq i\leq n}$ is a basis for $V^*.$
Let us also assume that these formulas, $f_1(x,y,z)=x+y-z, f_2(x,y,z)=2x-3y+z,f_3(x,y,z)=3x-2y$, are given with respect to this basis. In other words, we are assuming $f_1(xe_1+ye_2+ze_3)=x+y-z,$ etc. And $n=3,$ so $V$ is three dimensional.
There is an isomorphism $\tilde{}\colon V\cong V^{**}$ given by $\tilde x(f) = f(x).$ Thus the $e_i$ are also a basis for $V^{**}.$ A vector $\tilde v$ is in $S^\perp\subseteq V^{**}$ iff $\langle \tilde v,f_i\rangle = 0$ for $i=1,2,3.$ But $\langle \tilde v,f_i\rangle = \tilde v(f_i) = f_i(v).$ So we're looking for the vectors which all three $f_i$ vanish on. We're looking for the nullspace of 
$$\begin{pmatrix}
1 & 1 & -1\\
2 & -3 & 1\\
3 & -2 & 0
\end{pmatrix}.
$$
By doing column operations, we get
$$
\begin{pmatrix}
1 & 1 & -1\\
2 & -3 & 1\\
3 & -2 & 0\\
\hline
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 0\\
2 & -5 & 3\\
3 & -5 & 3\\
\hline
1 & -1 & 1\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 0\\
2 & -15 & 15\\
3 & -15 & 15\\
\hline
1 & -3 & 5\\
0 & 3 & 0\\
0 & 0 & 5
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 0\\
2 & -5 & 0\\
3 & -5 & 0\\
\hline
1 & -1 & 2\\
0 & 1 & 3\\
0 & 0 & 5
\end{pmatrix},
$$
And we stop when we reduce all the non-pivot columns to all zeros. The corresponding augmented columns give the basis for the nullspace. Or if you like you can work algebraically with the system of equations
$$
x+y-z=0\\
2x-3y+z=0\\
3x-2y=0$$
and use substitution or Gaussian elimination (which of course is exactly what the elementary column operations are, in different notation).
Anyway, however you get it, we have that a basis for the nullspace of this matrix is $\begin{pmatrix}2 \\3\\5\end{pmatrix}.$ Thus our basis for $S^\perp$ is $\{2\tilde e_1+3\tilde e_2+5\tilde e_3\}.$
