# Finding the Dual Basis

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.

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Perhaps in this case the following method is viable. Let $g_{ij}=v_i \bullet v_j$ and define the matrix $(g^{ij})$ to be $(g_{ij})^{-1}$. Then the dual basis $v^1, v^2, v^3, v^4$ is given by the formula $$v^i= \sum_{j=1}^4 g^{i j}v_j.$$ (This is called raising the index $i$). Usually this procedure is heavy on computations because of that inverse matrix. – Giuseppe Negro May 27 '12 at 23:17

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}

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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 $i,j$.

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.

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