Find the dual space and dual basis of basis $\beta$ in vector space $\mathbb{C^3}$

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 $1\times n$-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: $$\begin{pmatrix} f_1^1 & f_1^2 & f_1^3 \\ f_2^1 & f_2^2 & f_2^3 \\ f_3^1 & f_3^2 & f_3^3 \end{pmatrix} \cdot \begin{pmatrix} v_1^1 & v_2^1 & v_3^1 \\ v_1^2 & v_2^2 & v_3^2 \\ v_1^3 & v_2^3 & v_3^3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$ The first matrix on the left-hand side is obtained by writing the dual space basis vectors (the $1\times n$-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 $$A= \mathbb{1}_\mathrm n \cdot B^{-1} = B^{-1}.$$ 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 $$f_1=(1,-1,0), f_2=(1,-1,1), f_3=(-1/2,1,-1/2)$$
• @krishna The dual space contains all linear maps from the vectorspace $V$ over a field $K$ to K. This means the dual basis is made of linear maps. But we can represent every linear map as a matrix (there exists an isomorphism). The matrix is called the transformation matrix. Let be $\phi$ a linear map and $A$ the transformation matrix to the standard basis. Then $\phi(x)=A\cdot x$. Hence it is easier to speak in terms of matrices instead of functions. Aug 17, 2018 at 20:13
• @krishna For the linear maps in the dual space we get $1\times n$-matrices as transformation matrices (i.e. a row-vector). Aug 17, 2018 at 20:42