<s>[INST] Finding a basis of the annihilator of a subspace Let $V\subset\Bbb R^4$ be the subspace spanned by $e_1+e_2+e_3+e_4$ and $e_1+2e_2+3e_3+4e_4$. Find a basis of the subspace $V^\circ$ in the dual space $(\Bbb R)^*$.
My attempt: Let $a = e_1+e_2+e_3+e_4$ and $b = e_1+2e_2+3e_3+4e_4$, then $a$ and $b$ are lin. independent by looking at the row reducing matrix $ \color{white}{a}^{\begin{pmatrix} 
  1 & 1 & 1 & 1\\
  1 & 2 & 3 & 4 
\end{pmatrix}= \begin{pmatrix} 
  1 & 1 & 1 & 1\\
  0 & 1 & 2 & 3 
\end{pmatrix}}$. Thus we can extend the set $\{a,b\}$ to form the basis of $\Bbb R^4$. Example: $\{a,b,e_3,e_4\}$ forms a basis for $\Bbb R^4$ as $e_3,e_4$ are lin. independent of $a,b$. Let $\{f_1,f_2,f_3,f_4\}$ be the dual basis and we can say that $\{f_3,f_4\}$ forms the basis of $V^{\circ}$ because $f_3(a) = f_3(b) = f_4(a) = f_4(b) = 0$? Appreciate a hint.
 [/INST] Did it the long way here. To find the basis of $V^{\circ}$, we want linear functionals $f\in V^{*}$ s.t. $f(v)=0$, where $v = a_1v_1+a_2v_2$ ;  $a_1,a_2\in \mathbb{R}$. 
$$v_1 = e_1+e_2+e_3+e_4 = \begin{pmatrix}
  1 \\
  1 \\
  1 \\
  1 \\
\end{pmatrix} \& \text{ }v_2 = e_1+2e_2+3e_3+4e_4 = \begin{pmatrix}
  1 \\
  2 \\
  3 \\
  4 \\
\end{pmatrix}
$$
$$ f(v) = 0 \iff f(a_1v_1+a_2v_2)= a_1f(v_1)+a_2f(v_2) = 0 \iff f(v_1)=f(v_2)=0$$ 
For any $f\in V^{*}, \exists w\in R^{4}$, s.t. $f(v)= w.v$
$$f(v_1) = w.v_1 = 0\iff w_1+w_2+w_3+w_4 = 0$$ Similarly,
$$f(v_2) = w.v_2 = 0\iff w_1+2w_2+3w_3+4w_4 = 0 $$
$$ w = \begin{pmatrix}
  w_1 \\
  w_2 \\
  w_3 \\
  w_4 \\
\end{pmatrix} = \begin{pmatrix}
  t+2p \\
  -2t-3p \\
  t \\
  p \\
\end{pmatrix} = t\begin{pmatrix}
  1 \\
  -2 \\
  1 \\
  0 \\
\end{pmatrix} + p\begin{pmatrix}
  2 \\
  -3 \\
  0 \\
  1 \\
\end{pmatrix}$$
So $V^{\circ} = \{tf_1+pf_2; t,p\in \mathbb{R}\}$, where $f_1(x,y,z,w) = x-2y+z$ and $f_2(x,y,z,w) = 2x-3y+w$. Thus the basis of $V^{\circ}$ is $\{f_1,f_2\}$
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