Given linear functionals, in $\mathbb{R}^3$, find the annihilator

Question

Consider the following linear functionals in $\mathbb{R^3}$:

$\phi_1(x_1, x_2, x_3)=\frac{1}{2}x_1+\frac{1}{2}x_3$

$\phi_2(x_1, x_2, x_3)=x_3$

$\phi_3(x_1, x_2, x_3)=\frac{1}{2}x_1-\frac{1}{2}x_3$

$\phi_4(x_1, x_2, x_3)=-\frac{1}{2}x_1+x_2+\frac{1}{2}x_3$

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

2. 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!

Answer 1

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, $V=(u_3)$.