Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 8G (Linear Algebra)

The dual space isthe vector space of linear functionals on $V$. If $e_1,\dots ,e_n$ is a basis of a finite-dimensional $V$, define $e_i^{\ast }(e_j)={\delta }_{ij}$. Every $f\in V^{\ast }$ has the unique expansionso the dual basis $e_1^{\ast },\dots ,e_n^{\ast }$ proves

For $U\le V$, its annihilator of a vector subspace isIf $u_1,\dots ,u_k$ is a basis of $U$ and is extended to a basis $u_1,\dots ,u_n$ of $V$, thenConsequentlyIf $U\ne V$, this dimension is positive, giving a nonzero functional that vanishes on $U$.

For a linear map $\alpha :V\to W$, the dual map isNowsoEvery $g\circ \alpha$ vanishes on $\ker \alpha$, henceThe two spaces have the same dimension, sinceThus

Let $Q:V\to V/U$ be the quotient map. Since $Q$ is onto, $Q^{\ast }$ is injective, and the preceding identity givesThereforeFor the inclusion $J:U\hookrightarrow V$, the map $J^{\ast }:V^{\ast }\to U^{\ast }$ is restriction to $U$. It is onto and has kernelThe first isomorphism theorem now gives the other duals of a subspace and its quotient:

Suppose that $Q$ were not surjective. Its image would be a proper subspace of $F^n$, so there would be a nonzero functionalvanishing on $\mathrm{im}Q$. Thenfor every $x\in V$, contradicting the linear independence of the $q_j$. Hence the surjectivity of independent linear functionals gives

Let $K={\bigcap }_j\ker q_j=\ker Q$ and suppose $K\subseteq \ker f$. DefineThis is well-defined: if $Qx=Qy$, then $x-y\in K$, so $f(x)=f(y)$. It is linear, and surjectivity of $Q$ means it is defined on all of $F^n$. Thus there are scalars $a_1,\dots ,a_n$ such that $g(y)={\sum }_j{a}_j{y}_j$. Thereforeand hence