Finding $B^*$, the dual basis 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]$$
If I build a matrix out of the basis$ \left\{ \left[\begin{array}{cc}
  -1\\
  1\\
0 
\end{array}
\right] , \left[\begin{array}{cc}
  -1\\
  0\\
1 
\end{array}
\right] \right\}$ I will have a matrix that is $3\times2$ and I cannot inverse this.
How to proceed from here?
 A: By definition, the dual basis functionals $f_1, f_2$ are given on your basis $\{v_1, v_2\}$ as $$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$$
A: 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 $B^{*}=\{f_1,f_2\}$
