How to find dual basis for basis B* Let $\varepsilon^1$, $\varepsilon^2$ and $\varepsilon^3$ be elements of the dual canonical basis $K^*$. How to find dual basis to basis $B^*=(\varepsilon^1+\varepsilon^2-\varepsilon^3,\varepsilon^1+2\varepsilon^3,\varepsilon^1+\varepsilon^2)$
 A: Name 
$$\beta^1=\varepsilon^1+\varepsilon^2-\varepsilon^3,$$
$$\beta^2=\varepsilon^1+2\varepsilon^3,$$
$$\beta^3=\varepsilon^1+\varepsilon^2.$$
You are seeking $b_1,b_2,b_3$ three vectors in $K$ such that
$\beta^i(b_j)=\delta^i_j$.
Let us take the matrix of your basis change as
$$[B^*]
=\left[\begin{array}{ccc}
1&1&1\\1&0&1\\-1&2&0\end{array}\right].$$
So its inverse is 
$$[B^*]^{-1}
=\left[\begin{array}{ccc}
2&-2&-1\\1&-1&0\\-2&3&1\end{array}\right].$$
Now observing that $[B^*]^{-1}[B^*]=1\!\!1$, we have
$$\left[\begin{array}{ccc}
1&1&1\\1&0&1\\-1&2&0\end{array}\right]
\left[\begin{array}{ccc}
2&-2&-1\\1&-1&0\\-2&3&1\end{array}\right]=1\!\!1,$$
which suggests that the rows of $[B^*]^{-1}$ paired with the columns of 
$[B^*]$ gives delta-Kronecker.
Hence, by taking
$$b_1=e_1+e_2+e_3,$$
$$b_2=e_1+e_3,$$
$$b_3=-e_1+2e_2,$$
you will see $\beta^i(b_j)=\delta^i_j$. 
All this is a not so well known trick but pretty effective.
A: See this related post. As I state in my answer there, we can identify dual vectors with row-vectors. In particular, 
$$
w_1 = (1,1,-1)^T, \quad w_2 = (1,0,2)^T, \quad w_3 = (1,1,0)^T
$$
are the vectors of the elements of $B^*$ relative to the standard dual basis $K^*$. These vectors are such that if $\phi_i$ is the $i$th element of $B^*$ and $v = x_1 \varepsilon_1 + x_2 \varepsilon_2 + x_3 \varepsilon_3$ is a vector, then $\phi_i(v) = w_i^Tx$, where $x = (x_1,x_2,x_3)^T$.
If $v_i = (v_{i,1},v_{i,2},v_{i,3})^T$ is the coordinate vector relative to the canonical basis $K$ of the $i$th element of the basis $B$ (which is dual to $B^*$), then we must have $G^TF = I$, where $G$ is the matrix whose columns are $w_1,w_2,w_3$ and $F$ is the matrix whose columns are $v_1,v_2,v_3$.
