Finding the Dual Basis

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.

So you need to find $u^m$ (using covariant notation) such that $$\boldsymbol{u^m}\cdot \boldsymbol{v_n}=\delta^m_n$$ Considering that $$v_1=e_1$$ $$v_2=e_1+e_2$$ $$v_3=e_1+e_2+e_3$$ $$v_4=e_1+e_2+e_3$$ where $e_n$ are the vectors of the orthonormal Euclidean basis and writing $$\boldsymbol{u^m}=u_{1m}e^1+u_{2m}e^2+u_{3m}e^3+u_{4m}e^3$$ It is quite easy to find the coefficients by consecutive calculations of the scalar products. Recall also that Euclidean orthonormal basis is dual to itself. It is also possible to write the result in a more concise form which is a generalisation of that for $\mathbb{R}$ and is essentially the application of Kramer's formula to the resulting linear system $$\boldsymbol{u_{i}}=\frac{1}{\Delta}\left|\begin{array}{cccc} e_{1} & e_{2} & e_{3} & e_{4}\\ v^{j1} & v^{j2} & v^{j3} & v^{j4}\\ v^{k1} & v^{k2} & v^{k3} & v^{k4}\\ v^{l1} & v^{l2} & v^{l3} & v^{l4} \end{array}\right|$$

where $$\Delta=\left|\begin{array}{cccc} v_{11} & v_{12} & v_{13} & v_{14}\\ v_{21} & v_{22} & v_{23} & v_{24}\\ v_{31} & v_{32} & v_{33} & v_{34}\\ v_{41} & v_{42} & v_{43} & v_{44} \end{array}\right|$$ is the volume of the paralleliped constructed on $\boldsymbol{u_n}$ Both methods should lead to the answer $$\begin{aligned}\boldsymbol{v}^{1}=\left(1,-1,0,0\right)\\ \boldsymbol{v}^{2}=\left(0,1,-1,0\right)\\ \boldsymbol{v}^{3}=\left(0,0,1,-1\right)\\ \boldsymbol{v}^{4}=\left(0,0,0,1\right) \end{aligned}$$