Vtyjkyuk

Find the dual basis $beta^*$ of basis $beta =v_1,v_2,v_3$in vector space $mathbb C^3$ .
where $v_1 = (1, 0, −1),v_2 = (1, 1, 1), v_3 = (2, 2, 0)$.

I know that dual basis is the set of linearly independent functionals which satisfy $f_i(c_j)=delta_ij$. Where $delta_ij$ is Kronecker-Delta. Also represented as,

$beta^*=f_1,f_2,f_3$ .

We know that any linear functional can be represented as

$f(x_1,...,x_n)=c_1x_1+...+c_nx_n$.

To find the dual basis(by the method I am trying to use), what we do is consider $f_1$ and write equations $f_1(c_1)=1$

$f_1(c_2)=0$

$f_1(c_3)=0$

to find the coefficients $c_1,c_2,c_3$ for $f_1$. Similarly for $f_2&f_3$.

Now, what is the dual basis? Is it the set of tuples of $c_i$ for each $f$ ? or the set of functions given by $f$ ie. the linear equation given by $f$?
Also what is the dual space?

The dual space $V^*$ of a $K$-vectorspace $V$ is defined as all linear maps from $V$ to $K$
$$V^*=lambda colon V to K mid lambda text linear.$$ Since we can identify every linear map with a matrix, we denote the elements of the dual space with $1times n$-matrices where $n$ is the dimension of $V$. Now let $beta^* = f_1,f_2,f_3$ be the dual basis of $beta = v_1,v_2,v_3$. By definition the dual basis must satisfy $f_i(v_j) = delta_ij$. We can convert this into a linear equation system:
$$beginpmatrix
f_1^1 & f_1^2 & f_1^3 \
f_2^1 & f_2^2 & f_2^3 \
f_3^1 & f_3^2 & f_3^3
endpmatrix cdot
beginpmatrix
v_1^1 & v_2^1 & v_3^1 \
v_1^2 & v_2^2 & v_3^2 \
v_1^3 & v_2^3 & v_3^3
endpmatrix =
beginpmatrix
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
endpmatrix.$$
The first matrix on the left-hand side is obtained by writing the dual space basis vectors (the $1times n$-transformation matrices) row-wise. Let’s call it $A$. The second matrix on the left-hand side is obtained by writing the basis vectors column-wise. We will call it $B$. If we multiply the above equation with $B^-1$ from right, we get
$$A= mathbb1_mathrm n cdot B^-1 = B^-1.$$ This means the rows of $B^-1$ are the dual basis vectors.

In summary you have to write $v_1,v_2,v_3$ column-wise in a matrix, invert it and then read the dual basis vectors from the rows.

The solution would be $$f_1=(1,-1,0), f_2=(1,-1,1), f_3=(-1/2,1,-1/2)$$