Finding Dual Basis of Linear Functionals for a Given Basis in C^3

[Abuattallah]

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

My Solution. Let W_1 be the subspace generated by a_2=(1,1,1) a_3=(2,2,0), let's find W*, where W* is the set of linear anihilator of W_1. 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 x_2=-x_1 and x_3=0
so from this, by assinging x_1=1 ,we find out that the map f_1=x_1-x_2.
Similarly,

Let W_2 be the subspace generated by a_2=(1,0,-1) a_3=(2,2,0), let's find W*, where W* is the set of linear anihilator of W_2. 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 x_2=-x_1 and x_3=x_1
so from this, by assinging x_1=1 ,we find out that the map f_2=x_1-x_2+x_3.

Finally,

Let W_3 be the subspace generated by a_2=(1,1,1) a_3=(1,0,-1), let's find W*, where W* is the set of linear anihilator of W_3. 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 x_2=-2x_1 and x_3=x_1
so from this, by assinging x_1=1 ,we find out that the map f_3=x_1-2x_2+x_3.
So far so Good. The issue is that the last map f_3does 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 f_3=-1/2x_1+x_2-1/2x_3, I know that I assigned x_1=1 when I was trying to find f_3, and I could get the same answer if I put x_1=-1/2.
This significantly affect the nature of the map. i.e.
let (5,4,2)\in C^3, then
(5,4,2)=1(1,0,-1)+3(1,1,1)+1/2(2,2,0)
so we have c_1=1, c_2=3, c_3=1/2. while
f_1(5,4,2)=1,
f_2(5,4,2)=3,
f_3(5,4,2)=-1
so note that c_3 \ \ does \ \ not \ \ equal \ \ f_3(5,4,2)
While if I used the map f_3=-1/2x_1+x_2-1/2x_3 we get that f_3(5,4,2)=1/2. The equality should ocuur since f_3is a vector in the dual which determine the scalar c_3.
So my explanation is that I should consider also another equation to each system saying that c_1x_1+ c_2x_2 + c_2x_3=1 for each system depending on our choice of the vectors to determine c_1,c_2,c_3.
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 dim \ \ W + dim \ \ of \ \ annihilator \ \ space= dim \ \ V
, where W\subset V

 

[fresh_42]

You could just have taken the suitably normed perpendicular vectors of your basis. Should have been easier to find, and in the end to cross check by a matrix multiplication.