Dual basis in linear algebra Let V1=(3,4,2) V2=(1,2,0) and V3=(1,-1,-2).Find the dual basis of B={V1,V2,V3}
I don't know how to proceed with this problem. Can someone please help me?
 A: Here is an example: suppose $\mathbf{x} = c_1\mathbf{e}_1 + \dots c_k\mathbf{e}_k$, where the vectors $\mathbf{e}_i$ are a (not necessarily standard) basis for a vector space $V$ (these could be your vectors $V_1, V_2, V_3$). The dual basis is a set of linear functionals that, given a vector $\mathbf{x}$ in your vector space, give you the $i^{th}$ coordinate $c_i$ of $\mathbf{x}$ for your chosen basis. For example, suppose that $\mathbf{x} = [2$ $3]^T \in \mathbb{R}^2$ with the basis $[1$ $1]^T, [1$ $-1]^T$. So
$$ \mathbf{x} = c_1\begin{bmatrix} 1 \\ 1\end{bmatrix} + c_2\begin{bmatrix} 1 \\ -1\end{bmatrix},$$
and you want to know $c_1$. The dual to basis vector $[1$ $1]^T$ can be used to give it to you: calculate the matrix with columns the dual basis as in the Wikipedia link above , which in this case is
$$ G = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix},$$
then take the dot product of the first column of $G$ with $\mathbf{x}: [1/2$ $1/2]^T [2$ $3] = 5/2 = c_1$.
A: The vectors $v_1,v_2,v_3$ form a basis of $\Bbb R^3$. By definition, a dual basis is a set of functionals (linear transformations from $\Bbb R^3$ to $\Bbb R$) $\phi_1,\phi_2,\phi_3$ for which $\phi_i(v_j)$ is $1$ when $i = j$ and $0$ otherwise.  The functionals $\phi_1,\phi_2,\phi_3$ can be identified with $3 \times 1$ matrices, i.e. "row-vectors".
So, our goal is ultimately to find $3$ vectors $w_1,w_2,w_3$ such that $w_i^Tx = \phi_i(x)$. In other words, we want $3$ vectors such that $w_i^Tv_j = 1$ when $i = j$ and $0$ otherwise.
Since each $w_i$ has $3$ entries, writing all this out gives us a system of $9$ equations on $9$ variables. However, as the wikipedia page points out, we can frame this problem nicely in terms of matrix multiplication. Let $F$ be the matrix whose columns are $v_1,v_2,v_3$, and let $G$ be the matrix whose columns are $w_1,w_2,w_3$. Note that if the vectors $w_i$ satisfy the equations that we want to solve, then we'll have
$$
G^TF = \pmatrix{w_1^T\\ w_2^T\\ w_3^T} \pmatrix{v_1 &v_2&v_3} = 
\pmatrix{w_1^Tv_1 & w_1^Tv_2 & w_1^Tv_3\\
w_2^Tv_1 & w_2^Tv_2 & w_2^Tv_3\\
w_3^Tv_1 & w_3^Tv_2 & w_3^Tv_3}
= \pmatrix{1&0&0\\0&1&0\\0&0&1}.
$$
In other words, $G^T$ needs to be the inverse of the matrix $F$. Once you find this matrix $G$, the answer you're looking for is the list of vectors made from the columns of $G$.
A: Hint:
Write an arbitrary $x \in \Bbb{R}^3$ as 
$$x = \alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3.$$
Since $\{v_1, v_2, v_3\}$ is a basis this is possible and the coefficients are unique. Now the dual basis vectors act on $x$ as
$$v_1^*(x) = \alpha_1, \quad v_2^*(x) = \alpha_2, \quad v_3^*(x) = \alpha_3.$$
A: By filtering the information of your basis into a matrix
$$
B=
\left[\begin{array}{ccc}
3&1&1\\4&2&-1\\2&0&-2
\end{array}\right]
$$
then its inverse is
$$
B^{-1}=
\left[\begin{array}{ccc}
2/5& -1/5& 3/10\\
-3/5&4/5&-7/10\\
2/5&-1/5&-1/5
\end{array}\right].$$
So, the product $B^{-1}B=1\!\!1$ is the identity matrix.
Reinterpreting this as 
$$
\left[\begin{array}{ccc}
2/5& -1/5& 3/10\\
-3/5&4/5&-7/10\\
2/5&-1/5&-1/5
\end{array}\right]
\left[\begin{array}{ccc}
3&1&1\\4&2&-1\\2&0&-2
\end{array}\right]
=
\left[\begin{array}{ccc}
1&0&0\\0&1&0\\0&0&1
\end{array}\right]
,$$
you can see that the rows of $B^{-1}$ multiplied by the columns of $B$
satisfy the condition for duality.
So, defining
$$\gamma^1=[2/5, -1/5, 3/10],$$
$$\gamma^2=[-3/5, 4/5, -7/10],$$
$$\gamma^3=[2/5, -1/5, -1/5],$$
you are going to verify duality as 
$$\gamma^i(V_j)=\delta^i_j.$$
Verifying this means that ought to use the rows $\gamma^i$ multiply the $V_j$ 
as columns.
