TITLE: Finding the Dual Basis
QUESTION [9 upvotes]: Define the four vectors in $\mathbb{R}^4$ by 
$$v_1=\left( \begin{array}{ccc}
1 \\
0 \\
0 \\
0 \end{array} \right),
v_2=\left( \begin{array}{ccc}
1 \\
1 \\
0 \\
0 \end{array} \right),
v_3=\left( \begin{array}{ccc}
1 \\
1 \\
1 \\
0 \end{array} \right),
v_4=\left( \begin{array}{ccc}
1 \\
1 \\
1 \\
1 \end{array} \right). $$
I'm now asked to find the basis dual to $\{v_1,v_2,v_3,v_4 \}$ in $\mathbb{R}^4$, wth each vector expressed as a linear combination of the standard basis in $\mathbb{R}^4$.
Now, this is one of those situations where I 'know' all of the bookwork regarding dual bases etc. however, what seems like a simple application presents quite a hurdle.
Any explanation of how to progress would be very appreciated.

REPLY [6 votes]: Let $\{u_1,\ldots,u_4\}$ be the dual basis for basis $\{v_1,\ldots, v_4\}$, to be written as co-ordinate (column) vectors relative to the standard basis of $\mathbb{R}^4$.  By definition of Dual basis, these sets are biorthogonal, that is, $u_i^T v_j = \delta_{ij}$, for all $i,j$ in $\{1,\ldots,4\}$. 
Let $$
U = \left[\begin{matrix}
u_1 &u_2& u_3& u_4
\end{matrix}\right]
$$
and
$$
V = \left[\begin{matrix}
v_1 &v_2& v_3& v_4
\end{matrix}\right].
$$
Now the biorthogonality equations can be expressed as
$$U^T V = \left[\begin{matrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{matrix}\right]
= I.
$$
So, $U^T=V^{-1}$, which you can easily compute, and the dual basis is formed by the columns of U.