Given linear functionals, in $\mathbb{R}^3$, find the annihilator Consider the following linear functionals in $\mathbb{R^3}$:
$\phi_1(x_1, x_2, x_3)=\frac{1}{2}x_1+\frac{1}{2}x_3$
$\phi_2(x_1, x_2, x_3)=x_3$
$\phi_3(x_1, x_2, x_3)=\frac{1}{2}x_1-\frac{1}{2}x_3$
$\phi_4(x_1, x_2, x_3)=-\frac{1}{2}x_1+x_2+\frac{1}{2}x_3$


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*Find a basis $\mathcal{B}=\{u_1, u_2, u_3\}$ of $\mathbb{R^3}$ such that $\mathcal{B^*}=\{\phi_1, \phi_2, \phi_4\}$

*Calculate the dimension of $span\{\phi_1, \phi_2, \phi_3\}$ and find a subspace $V$ of $\mathbb{R^3}$ such that $V^0=span\{\phi_1, \phi_2, \phi_3\}$


*

*To answer at the first question I can set a the matrix 



$$ \left[
\begin{array}{ccc|ccc}
  \frac{1}{2}&0&\frac{1}{2}&1&0&0\\
  0&0&1&0&1&0\\-\frac{1}{2}&1&\frac{1}{2}&0&0&1
\end{array}
\right] $$
then calculate the inverse, whose columns should be, respectively, $u_1 =A^1$, $u_2=A^2$, $u_3=A^3$. Is that right?


*

*How should I proceed for answering at the point 2?


I think that the dimension is 2: the number of indipendent columns of the matrix with the coefficients of the functionals as rows. Then the subspace $V$ is composed by those vectors given by the kernel of that matrix, $(0,1,0)$ in this case; $dim(V)=1=dim(\mathbb{R^3})-dim(V^0)$. Am I right? Thank you!

1) Here $ A=\left(
          \begin{array}{ccc}
            1/2 & 0&1/2 \\
            0&0 & 1 \\
            -1/2&1& 1/2
          \end{array}
        \right)
$ and $Au_i=e_i,\ 1\leq i\leq 3 $. That is, $u_i= A^{-1}e_i$
2) $\phi_1+\phi_3=\phi_2$ so that $(
\phi_1,\phi_2,\phi_3)=(\phi_1,\phi_2)$. It has a dimension 2.
Hence $(\phi_1,\phi_2)=V^0:=\{f| f:V\rightarrow \mathbb{R} \ {\rm is}\ 0 \ {\rm map} \}$. Note that $V=(u_3)$. 
Proof : For any $v=\sum_i\ c_iu_i\in V$, $\phi_1(v)=0$ so that $c_1=0$ And $\phi_2(v)=0$ so that $c_2=0$. Hence $V\subset (u_3)$. 
If $V$ is $0$-dimensional, then $V^0=(\phi_1,\phi_2,\phi_4)$ That is, $V=(u_3)$.
