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

Identify each linear functional on \(\mathbb{R}^3\) with a row vector \((p,q,r)\) acting on a column vector \((x,y,z)^T\) by \[f(x,y,z) = px + qy + rz\]

The subspace \(S = \operatorname{span}\{e_1,e_2\}\) consists of all vectors of the form \((x,y,0)\), that is vectors with free first and second coordinates and a zero third coordinate.

A functional \(f\) annihilates all of \(S\) exactly when \(f(x,y,0) = 0\) for every choice of \(x\) and \(y\): \[px + qy + r\cdot 0 = px + qy = 0 \quad \text{for all } x,y \in \mathbb{R}\]

Since \(x\) and \(y\) can be chosen independently and freely (take \(x=1,y=0\) then \(x=0,y=1\)), this forces \(p=0\) and \(q=0\), while \(r\) is left completely unrestricted.

So every functional in \(S^0\) has the form \[f(x,y,z) = r\cdot z, \qquad r \in \mathbb{R}\] This is exactly the set of scalar multiples of the third coordinate functional, which is precisely \(f_3\) (the dual basis vector satisfying \(f_3(e_1)=0, f_3(e_2)=0, f_3(e_3)=1\)). Hence \[S^0 = \{r f_3 : r \in \mathbb{R}\} = \operatorname{span}\{f_3\}\]

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