Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2012 / 1

The dual space consists of the linear functionals on the original vector space, with addition and scalar multiplication defined pointwise:The dual map acts by function composition with $T$:For $u\in U$, its value is $(T^{\prime }\lambda )(u)=\lambda (Tu)$. The composite is a linear functional because both $T$ and $\lambda$ are linear maps. Moreover $T^{\prime }(a\lambda +b\mu )=a{T}^{\prime }\lambda +b{T}^{\prime }\mu$, so the dual map is itself a linear map. Its direction is reversed because a measurement on the output of $T$ becomes a measurement on its input.

The annihilator of a vector subspace consists of the linear functionals that vanish on every element of that vector subspace:For the dimension identity, choose a basis $w_1,\dots ,w_k$ of $W$ and use basis extension to obtain $w_1,\dots ,w_n$ for $V$. A linear functional is determined by its values on these $n$ basis vectors. Belonging to the annihilator of a vector subspace sets its first $k$ values to zero and leaves its last $n-k$ values arbitrary. Equivalently the last $n-k$ elements of the dual basis form a basis of $W^0$.

For a linear functional $\lambda \in V^{\prime }$, the definition of the dual map givesThus the kernel of a linear map is precisely the annihilator of a vector subspaceWrite $n=\dim U$, $m=\dim V$ and $r=\dim \mathrm{im}T$. The rank-nullity theorem, the dimension formula for an annihilator of a vector subspace, and $\dim V^{\prime }=m$ giveApplying the same dimension formula in $U$ givesFinally, if $\alpha =T^{\prime }\lambda$ and $u\in \ker T$, then $\alpha (u)=\lambda (Tu)=\lambda (0)=0$. Therefore $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^0$. These are finite-dimensional vector subspaces of equal dimension, so inclusion is equality:This is the dual image and kernel annihilator identity.