Question

Show that f(1, 1, 1), (1, – 1, 0), (1, 1, 0)1 is a
basis of R3 . Find a dual basis to this basis.

Answer 1

Let's consider $e_1=(1,1,1), e_2=(1,-1,0),e_3=(1,1,0)$ .

 Let's show that $e_1,e_2,e_3$  is a basis of $R^3$ :

$\alpha_1 e_1+\alpha_2 e_2+\alpha_3 e_3=0$ ,

where $\alpha_1 =\alpha_2=\alpha_3=0$ .

 Really

$\alpha_1 +\alpha_2+\alpha_3=0\\ \alpha_1 -\alpha_2+\alpha_3=0\\ \alpha_1 =0$

Then $\alpha_1 =\alpha_2=\alpha_3=0$ .

So $e_1,e_2,e_3$  is a basis of $R^3$.

 Find a dual basis to this basis.

We can find the biorthogonal (dual) basis $\{e^1,e^2,e^3\}$ by formulas below:

$e^1=(\frac{e_2\times e_3}{V})^T,e^2=(\frac{e_3\times e_1}{V})^T,e^3=(\frac{e_1\times e_2}{V})^T,$

where $^T$ denotes the transpose and $V=(e_1,e_2,e_3)=e_1\cdot(e_2\times e_3)$

is the volume of the parallelepiped formed by the basis vectors $e_1,e_2,e_3$ .

In our case

$V=\begin{vmatrix} 1 & 1&1 \\ 1 & -1&0\\ 1&1&0 \end{vmatrix}=1\cdot(-1)\cdot0+1\cdot0\cdot1+\\ +1\cdot1\cdot1-1\cdot(-1)\cdot1-1\cdot1\cdot0-1\cdot1\cdot0=2\\ e_2\times e_3=(\begin{vmatrix} -1& 0 \\ 1 & 0 \end{vmatrix},-\begin{vmatrix} 1& 0 \\ 1 & 0 \end{vmatrix},\begin{vmatrix} 1& -1 \\ 1 & 1 \end{vmatrix})=\\ =(0,0,2)\\ e_3\times e_1=(\begin{vmatrix} 1& 0 \\ 1 & 1 \end{vmatrix},-\begin{vmatrix} 1& 0 \\ 1 & 1 \end{vmatrix},\begin{vmatrix} 1& 1 \\ 1 & 1 \end{vmatrix})=\\ =(1,-1,0)\\ e_1\times e_2=(\begin{vmatrix} 1& 1 \\ -1 & 0 \end{vmatrix},-\begin{vmatrix} 1& 1 \\ 1 & 0 \end{vmatrix},\begin{vmatrix} 1& 1 \\ 1 & -1 \end{vmatrix})=\\ =(1,1,-2)$

Then

$e^1=(\frac{(0,0,2)}{2})^T=\begin{pmatrix} 0 \\ 0\\1 \end{pmatrix},\\ e^2=(\frac{(1,-1,0)}{2})^T=\begin{pmatrix} \frac{1}{2} \\ -\frac{1}{2}\\0 \end{pmatrix},\\ e^3=(\frac{(1,1,-2)}{2})^T=\begin{pmatrix} \frac{1}{2}\\ \frac{1}{2} \\-1 \end{pmatrix}.$

Denoting the indexed vector sets as ${\displaystyle B=\{v_{i}\}_{i\in I}}$  and ${\displaystyle B^{*}=\{v^{i}\}_{i\in I}}$ , being biorthogonal means that the elements pair to have an inner product equal to 1 if the indexes are equal, and equal to 0 otherwise. Symbolically, evaluating a dual vector in V^∗ on a vector in the original space V:

${\displaystyle v^{i}\cdot v_{j}=\delta _{j}^{i}={\begin{cases}1&{\text{if }}i=j\\0&{\text{if }}i\neq j{\text{,}}\end{cases}}}$

where ${\displaystyle \delta _{j}^{i}}$  is the Kronecker delta symbol.

In our case

$e^1\cdot e_1=0+0+1=1\\ e^1\cdot e_2=0+0+0=0\\ e^1\cdot e_3=0+0+0=0\\ e^2\cdot e_1=\frac{1}{2}-\frac{1}{2}+0=0\\ e^2\cdot e_2=\frac{1}{2}+\frac{1}{2}+0=1\\ e^2\cdot e_3=\frac{1}{2}-\frac{1}{2}+0=0\\ e^3\cdot e_1=\frac{1}{2}+\frac{1}{2}-1=0\\ e^3\cdot e_2=\frac{1}{2}-\frac{1}{2}+0=0\\ e^3\cdot e_3=\frac{1}{2}+\frac{1}{2}+0=1$

https://en.wikipedia.org/wiki/Dual_basis