Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2011 / 1

The annihilator of a vector subspace is the vector subspace of the dual space consisting of the linear functionals that vanish on $U$:It is a vector subspace because every linear combination of functionals vanishing on $U$ still vanishes there.

For the basis $e_1,\dots ,e_n$, define the dual basis byUniqueness of the coordinates makes each $e_i^{\prime }$ well-defined and linear. If ${\sum }_i{a}_i{e}_i^{\prime }=0$, evaluation at $e_j$ gives $a_j=0$, so these linear functionals are linearly independent. For any $\lambda \in V^{\prime }$, evaluation on each basis vector givesBoth sides agree on a basis, hence on every linear combination. Thus they also span the dual space and are a basis of it.

Let ${\epsilon }_i$ denote the coordinate linear functional dual to the $i$th vector of the standard basis. Writing a vector in the supplied basis givesSuccessive differences recover the coefficients. Hence the dual basis isThese formulas directly give $v_i^{\prime }(v_j)={\delta }_{ij}$ and exemplify the dual basis of cumulative coordinate sums.

The span $U$ is the coordinate plane $\{(x_1,x_2,0,0{)}^T\}$. Consequently its annihilator of a vector subspace is $U^{\circ }=\mathrm{span}\{{\epsilon }_3,{\epsilon }_4\}$. The dual map is composition with $T$, soIt follows thatEquivalently, $T^{\prime }(a{\epsilon }_3+b{\epsilon }_4)$ sends $x$ to $a{x}_3$.