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

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 $a{x}_3$.