Finding $B^*$, the dual basis

Question

Find a basis $B$ for $$V = \left\{ \left[ \begin{array}{cc} x\\ y\\ z \end{array} \right] \in \mathbb{R}^3 \vert x+y+z = 0\right\}$$ and then find $B^*$, the dual basis for $B$.

The way we learned it was that given a basis, we build a matrix A whose columns is the vectors, find $A^{-1}$, and those build linear functionals that are the rows of the inverse matrix.

For example, if $V=\mathbb{R}^2$ and $ B = $$\left\{ \left[ \begin{array}{cc} 3\\ 4 \end{array} \right], \left[ \begin{array}{c} 5\\ 7 \end{array} \right]\right\} $, then $ A = $$ \left[ \begin{array}{cc} 3&5\\ 4&7 \end{array} \right]$, then $A^{-1} = $$ \left[ \begin{array}{cc} 7&-5\\ -4&3 \end{array} \right]$, and the dual basis $B^* = (l_1, l_2)$ when :

$ l_1= \left( \left[ \begin{array}{cc} x\\ y\\ \end{array} \right]\right) = 7x_1 - 5x_2 $

$ l_2= \left(\left[ \begin{array}{cc} x\\ y\\ \end{array} \right]\right) = -4x_1 + 3x_2 $

However, when finding a basis for $V$, I get to the following solution vector $$\left[\begin{array}{cc} -y-z\\ y\\ z \end{array} \right] = y \left[\begin{array}{cc} -1\\ 1\\ 0 \end{array} \right] + z \left[\begin{array}{cc} -1\\ 0\\ 1 \end{array} \right]$$

If I build a matrix out of the basis$ \left\{ \left[\begin{array}{cc} -1\\ 1\\ 0 \end{array} \right] , \left[\begin{array}{cc} -1\\ 0\\ 1 \end{array} \right] \right\}$ I will have a matrix that is $3\times2$ and I cannot inverse this. How to proceed from here?

Answer 1

By definition, the dual basis functionals $f_1, f_2$ are given on your basis $\{v_1, v_2\}$ as $$f_1(\alpha_1v_1 + \alpha_2v_2) = \alpha_1, \quad f_2(\alpha_1v_1 + \alpha_2v_2) = \alpha_2$$

Now $$f_1\left(\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right) = f_1\left(\begin{bmatrix} -y-z \\ y \\ z\end{bmatrix}\right) = f_1\left(y\begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix} + z\begin{bmatrix} -1 \\ 0 \\ -1\end{bmatrix}\right) = f_1(yv_1 + zv_2) = y$$ $$f_2\left(\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right) = f_2\left(\begin{bmatrix} -y-z \\ y \\ z\end{bmatrix}\right) = f_2\left(y\begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix} + z\begin{bmatrix} -1 \\ 0 \\ -1\end{bmatrix}\right) = f_2(yv_1 + zv_2) = z$$

Answer 2

Be \begin{equation} V = \left\lbrace \ \left[\begin{array}{c} x \\ y \\ z \\ \end{array}\right] \in \mathbb{R}^3 : x+y+z = 0 \right\rbrace \end{equation} Then $z=-x-y$, thus basis $B$ is: \begin{equation} B = \left\lbrace\ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ -1 \\ \end{array}\right] \ \right\rbrace \end{equation} Where $B=\{v_1,v_2\}$. The dual basis is given by $f_i(v_j)=\delta_{ij}$ where $f_i\in B^{*}$, $v_j\in B$ and $\delta_{ij}$ is the Kronecker delta since: \begin{equation} \delta_{ij}=\begin{cases} 1 \quad\textrm{ if } i=j \\ 0 \quad\textrm{ if } i\neq j \\ \end{cases} \end{equation} Then for $f_1$: \begin{eqnarray} f_1\left(\ \left[\begin{array}{c} x \\ y \\ z \\ \end{array}\right]\ \right) = f_1\left(\ \left[\begin{array}{c} x \\ y \\ -x-y \\ \end{array}\right]\ \right) &=& xf_1\left(\ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ \end{array}\right]\ \right) + yf_1\left(\ \left[\begin{array}{c} 0 \\ 1 \\ -1 \\ \end{array}\right]\ \right) \\ f_1\left(\ \left[\begin{array}{c} x \\ y \\ -x-y \\ \end{array}\right]\ \right) &=& x \end{eqnarray} And $f_2$: \begin{eqnarray} f_2\left(\ \left[\begin{array}{c} x \\ y \\ z \\ \end{array}\right]\ \right) = f_2\left(\ \left[\begin{array}{c} x \\ y \\ -x-y \\ \end{array}\right]\ \right) &=& xf_2\left(\ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ \end{array}\right]\ \right) + yf_2\left(\ \left[\begin{array}{c} 0 \\ 1 \\ -1 \\ \end{array}\right]\ \right) \\ f_2\left(\ \left[\begin{array}{c} x \\ y \\ -x-y \\ \end{array}\right]\ \right) &=& y \end{eqnarray} Then $B^{*}=\{f_1,f_2\}$