Question

On the vector space \(\mathbb{R}^3\) over the field \(\mathbb{R}\) with the standard basis \(\{e_1,e_2,e_3\}\), what is the annihilator \((S^0)\) of the subspace \(S=\text{span}\{e_1,e_2\}\)?

• A. \(S^0=\{0\}\), the zero functional on \(\mathbb{R}^3\)
• B. \(S^0=\text{span}\{f_1,f_2\}\), where \(f_1,f_2\) are the dual basis elements corresponding to the basis elements \(e_1,e_2\)
• C. \(S^0=\text{span}\{f_3\}\), where \(f_3\) is the dual basis element corresponding to the basis element \(e_3\)
• D. \(S^0=(\mathbb{R}^3)^*\), the entire dual space of \(\mathbb{R}^3\)

Hint:

A functional is in S^0 iff it kills e_1 and e_2, leaving only the coefficient of f_3 free.

Answer 1

The Correct Option is C

Step 1: Recall the definition of the annihilator. For a subspace \(S \subseteq V\), the annihilator is \[S^0 = \{f \in V^* : f(s) = 0 \text{ for all } s \in S\}\] where \(V^*\) is the dual space of linear functionals on \(V\).

Step 2: Set up a general functional in coordinates. Let \(\{f_1,f_2,f_3\}\) be the dual basis of \(\{e_1,e_2,e_3\}\), meaning \(f_i(e_j) = 1\) if \(i=j\) and \(0\) otherwise. A general element of \((\mathbb{R}^3)^*\) is \[f = a f_1 + b f_2 + c f_3\]

Step 3: Impose the condition that \(f\) vanishes on \(S = \operatorname{span}\{e_1,e_2\}\), which happens exactly when \(f(e_1)=0\) and \(f(e_2)=0\) (checking on a spanning set suffices by linearity). \[f(e_1) = a = 0, \qquad f(e_2) = b = 0\]

Step 4: With \(a=b=0\), the functional reduces to \(f = c f_3\) for arbitrary \(c \in \mathbb{R}\). So \[S^0 = \{c f_3 : c \in \mathbb{R}\} = \operatorname{span}\{f_3\}\] This matches the general dimension formula for annihilators, \(\dim(S^0) = \dim(V) - \dim(S) = 3 - 2 = 1\), and \(\operatorname{span}\{f_3\}\) is indeed 1-dimensional.

\[\boxed{S^0 = \operatorname{span}\{f_3\}}\]