Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2013 / Algebra

The dual space is the vector space of all linear functionals on $V$:Addition and scalar multiplication are defined pointwise. The annihilator of a vector subspace isIt is a vector subspace of $V^{\prime }$: linear combinations of functionals vanishing on $U$ also vanish on $U$.

Choose a basis $u_1,\dots ,u_m$ of $U$. Because $v\notin U$, the list $u_1,\dots ,u_m,v$ is linearly independent. Extend it to a basis of $V$. A linear functional is specified uniquely by its values on a basis, so define $f$ to be zero on the $u_j$ and on every added basis vector other than $v$, and set $f(v)=1$.

Then $f$ vanishes on every linear combination of the $u_j$, so $f\in U^0$, whileThis proves that the annihilator of a vector subspace separates every vector outside the subspace from that subspace; no inner product or preferred coordinates are needed.

Every $g_i\in U^0$ vanishes on $U$, so $U\subseteq {\bigcap }_i\ker g_i$. Conversely, suppose $v\in {\bigcap }_i\ker g_i$. Every $f\in U^0$ is a linear combination ${\sum }_i{a}_i{g}_i$, giving $f(v)=0$. If $v\notin U$, part (b) supplies a functional in $U^0$ with $f(v)\ne 0$, a contradiction. ThereforeFor the dimension statement, take a basis $u_1,\dots ,u_m$ of $U$ and extend it to a basis $u_1,\dots ,u_n$ of $V$. Relative to the corresponding dual basis, a functional belongs to $U^0$ precisely when its first $m$ coefficients are zero. ThusIf $U=V$, this says $k=0$ and the empty intersection is understood to be $V$.

In the coordinate example choose the linear functionals $g_1(x,y,z)=y$ and $g_2(x,y,z)=z$. Their kernels are the two coordinate planesTheir intersection is the stated axis, and $k=2=3-1$.

The adjoint operator $T^{\ast }$ is characterized byThis identity works with either convention for which argument of the complex inner product is linear, provided the convention is used consistently.

For uniqueness, subtract the two given identities to get $⟨v,(S-T^{\ast })w⟩=0$ for every $v,w$. Fix $w$ and choose $v=(S-T^{\ast })w$. Then positive definiteness givesso $(S-T^{\ast })w=0$. As $w$ was arbitrary, $\boxed{S=T^{\ast }}$. The decisive fact is that a vector orthogonal to every vector, including itself, must be zero.

First suppose $E=E^{\ast }$. For $u\in U=\mathrm{im}E$, choose $v$ with $Ev=u$. For $w\in W=\ker E$,Thus $U\subseteq W^{\perp }$. The direct sum of vector spaces gives $\dim U=\dim V-\dim W$, while the orthogonal complement has $\dim W^{\perp }=\dim V-\dim W$. Equal finite dimensions imply $U=W^{\perp }$.

Conversely, suppose $U=W^{\perp }$. Decompose arbitrary vectors as $v=u+w$ and $v^{\prime }=u^{\prime }+w^{\prime }$, with $u,u^{\prime }\in U$ and $w,w^{\prime }\in W$. Since $E$ is the indicated projection, $Ev=u$ and $E{v}^{\prime }=u^{\prime }$. Orthogonality makes the cross terms vanish, soTherefore $E$ satisfies the defining identity of its own adjoint operator. By uniqueness from part (a), $E=E^{\ast }$. HenceThis is the characterization of an orthogonal projection among projections along a complementary subspace.

A Euclidean domain is an integral domain $R$ admitting a Euclidean function $\delta :R\setminus \{0\}\to {\mathbb{Z}}_{\ge 0}$ such that division with a smaller remainder is possible: for every $a\in R$ and nonzero $b\in R$, there are $q,r\in R$ withThus the Euclidean function measures progress in repeated division. Definitions which additionally normalize it to be nondecreasing under multiplication give the same class of Euclidean domains; the division property is the one needed below.

For the zero ideal, choose $d=0$. Otherwise choose a nonzero $d\in I$ for which the Euclidean function $\delta (d)$ is minimal; the well-ordering principle guarantees such a choice. For any $a\in I$, perform Euclidean division:Because $I$ is an ideal, $r=a-qd\in I$. A nonzero remainder would contradict the minimal choice of $d$, so $r=0$ and $a\in Rd$. Thus $I\subseteq Rd$. Conversely, $d\in I$ implies $Rd\subseteq I$, provingEvery ideal is therefore a principal ideal; in particular, a Euclidean domain is a principal ideal domain.

For the Gaussian integers, use the Gaussian normTo identify the ideal, note that $a^2+b^2$ is even if and only if $a,b$ have the same parity. AlsoThis quotient is a Gaussian integer precisely when $a,b$ have the same parity. Therefore the prescribed set consists exactly of the multiples of $1+i$, givingAny associate of $1+i$ would generate the same principal ideal.