Past exam of the linear algebra course of the University of Oxford / 2017 / 2

The dual space isthe vector space of linear functionals, with pointwise operationsGiven the basis $e_1,\dots ,e_n$, its dual basis consists of the coordinate linear functionalsIn particular, $\dim V^{\prime }=\dim V=n$.

The dual map acts by function composition:The composition is a linear functional on $V$, and pointwise operations give $T^{\prime }(a\varphi +b\psi )=a{T}^{\prime }\varphi +b{T}^{\prime }\psi$. Thus this really is a linear transformation; its direction reverses that of $T$.

For $\varphi \in V^{\prime }$ and $v\in V$,so $I^{\prime }$ is the identity map on the dual space. If $S:U\to V$ and $T:V\to W$, then, for $\psi \in W^{\prime }$ and $u\in U$,Equality on all vectors provesApply this to $R{R}^{-1}=R^{-1}R=I$. It gives both $R^{\prime }(R^{-1}{)}^{\prime }=I^{\prime }$ and $(R^{-1}{)}^{\prime }{R}^{\prime }=I^{\prime }$. Consequently the dual map is invertible and

The annihilator of a vector subspace isChoose a basis $e_1,\dots ,e_r$ of $U$ and extend it to a basis $e_1,\dots ,e_d$ of $V$. In the dual basis, precisely $f_{r+1},\dots ,f_d$ span $U^{\circ }$. ThereforeThe canonical restriction linear transformationhas kernel of a linear map $U^{\circ }$. It is surjective: extend any linear functional on $U$ by assigning arbitrary values, for instance zero, to the additional basis vectors. The First isomorphism theorem for vector spaces gives the explicit linear isomorphismIt is natural because the displayed map itself makes no basis choice; a basis was used only to prove surjectivity. In particular, $U^{\prime }$ is canonically a quotient vector space of $V^{\prime }$, rather than canonically a subspace of $V^{\prime }$.

If $\varphi =T^{\prime }\psi$ and $v\in \ker T$, then $\varphi (v)=\psi (Tv)=0$. Thus $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^{\circ }$, using the annihilator of a vector subspace.

Conversely, let $\varphi$ vanish on the kernel of a linear map $\ker T$. Define a linear functional on $\mathrm{im}T$ by ${\psi }_0(Tv)=\varphi (v)$. This is well-defined: $Tv=Tw$ implies $v-w\in \ker T$, so $\varphi (v)=\varphi (w)$. Extend ${\psi }_0$ to a linear functional $\psi$ on $W$, using the extension argument from part (i). Then $T^{\prime }\psi =\varphi$, provingThe dimension formula from part (i) and the rank-nullity theorem now giveTo deduce the eigenvalue statement, apply this equality of rank of a linear map to $T-\lambda I$. Its dual map is $T^{\prime }-\lambda I$, so their kernel of a linear map dimensions agree. Thus the same $\lambda$ are eigenvalues, with the same geometric multiplicities.

For algebraic multiplicities as well, part (a)(iii) givesso equality of rank applies to every power. HenceAfter extending to a field where the characteristic polynomials split, take $r\ge \dim V$. These kernel of a linear map spaces are the generalized eigenspaces; the primary decomposition theorem for linear maps identifies their dimensions with the algebraic multiplicities. The same proof works over the original field for every eigenvalue in it. We obtainIn fact the argument also includes the zero eigenvalue.