I can not explain a difference between to calculations regarding linear functionals

Abuattallah

Hello,
Problem, let [itex]B=[/itex]{[itex]a_1,a_2,a_3[/itex]} be a basis for [itex]C^3[/itex] defined by [itex]a_1=(1,0,-1)[/itex] [itex]a_2=(1,1,1)[/itex] [itex]a_3=(2,2,0)[/itex]
Find the dual basis of [itex]B[/itex].

My Solution. Let [itex]W_1[/itex] be the subspace generated by [itex]a_2=(1,1,1)[/itex] [itex]a_3=(2,2,0)[/itex], lets find [itex][/itex] [itex]W*[/itex], where [itex]W*[/itex] is the set of linear anihilator of [itex]W_1[/itex]. Consider the system:

\begin{align*}
\begin{bmatrix}1 & 1 & 1 \\ 2 & 2 & 0 \\\end{bmatrix} &\begin{bmatrix} x_1 \\x_2\\x_3 \\\end{bmatrix}
=0
\end{align*}
solving this will yeild [itex]x_2=-x_1[/itex] and [itex]x_3=0[/itex]
so from this, by assinging [itex]x_1=1[/itex] ,we find out that the map [itex]f_1=x_1-x_2[/itex].
Similarly,




Let [itex]W_2[/itex] be the subspace generated by [itex]a_2=(1,0,-1)[/itex] [itex]a_3=(2,2,0)[/itex], lets find [itex][/itex] [itex]W*[/itex], where [itex]W*[/itex] is the set of linear anihilator of [itex]W_2[/itex]. Consider the system:

\begin{align*}
\begin{bmatrix}1 & 0 & -1 \\ 2 & 2 & 0 \\\end{bmatrix} &\begin{bmatrix} x_1 \\x_2\\x_3 \\\end{bmatrix}
=0
\end{align*}
solving this will yeild [itex]x_2=-x_1[/itex] and [itex]x_3=x_1[/itex]
so from this, by assinging [itex]x_1=1[/itex] ,we find out that the map [itex]f_2=x_1-x_2+x_3[/itex].

Finally,

Let [itex]W_3[/itex] be the subspace generated by [itex]a_2=(1,1,1)[/itex] [itex]a_3=(1,0,-1)[/itex], lets find [itex][/itex] [itex]W*[/itex], where [itex]W*[/itex] is the set of linear anihilator of [itex]W_3[/itex]. Consider the system:

\begin{align*}
\begin{bmatrix}1 & 1 & 1 \\ 1 & 0 & -1 \\\end{bmatrix} &\begin{bmatrix} x_1 \\x_2\\x_3 \\\end{bmatrix}
=0
\end{align*}
solving this will yeild [itex]x_2=-2x_1[/itex] and [itex]x_3=x_1[/itex]
so from this, by assinging [itex]x_1=1[/itex] ,we find out that the map [itex]f_3=x_1-2x_2+x_3[/itex].
So far so Good. The issue is that the last map [itex]f_3[/itex]does not agree with the solution I have, but every other thing is the same.Note that I have the final solution with a diffrenet method of solution than mine. In the solution, I got that [itex]f_3=-1/2x_1+x_2-1/2x_3[/itex], I know that I assigned [itex]x_1=1[/itex] when I was trying to find [itex]f_3[/itex], and I could get the same answer if I put [itex]x_1=-1/2[/itex].
This significantly affect the nature of the map. i.e.
let [itex](5,4,2)\in C^3[/itex], then
[itex](5,4,2)=1(1,0,-1)+3(1,1,1)+1/2(2,2,0)[/itex]
so we have [itex]c_1=1, c_2=3, c_3=1/2[/itex]. while
[itex]f_1(5,4,2)=1[/itex],
[itex]f_2(5,4,2)=3[/itex],
[itex]f_3(5,4,2)=-1[/itex]
so note that [itex]c_3 \ \ does \ \ not \ \ equal \ \ f_3(5,4,2)[/itex]
While if I used the map [itex]f_3=-1/2x_1+x_2-1/2x_3[/itex] we get that [itex]f_3(5,4,2)=1/2[/itex]. The equality should ocuur since [itex]f_3[/itex]is a vector in the dual which determine the scalar [itex]c_3[/itex].
So my explanation is that I should consider also another equation to each system saying that [itex]c_1x_1+ c_2x_2 + c_2x_3=1[/itex] for each system depending on our choice of the vectors to determine [itex]c_1,c_2,c_3[/itex].
My question, Is this method always work when we find the Dual basis to a given Basis?.
Because I figured this method by myself using a similar method to find anihilator space along with the fact that [itex]dim \ \ W + dim \ \ of \ \ annihilator \ \ space= dim \ \ V[/itex]
, where [itex]W\subset V[/itex]