Past exam of the algebra and analysis I for mathematics and philosophy course of the University of Oxford

Source: Mathematics Part A release, member FOI-20210708-1 - Data/4307.pdf, dated 17 June 2013. All 22 source leaf parts are answered below.

The department's solution-release policy excludes ordinary odd years; bounded exact-paper checks found no matching advertisement of full or selected worked answers. No parts are skipped.

The six question bodies coincide exactly with questions 1–6 of paper 4301; their solutions are reused verbatim while retaining this paper's distinct cover identity and section tree.

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.

Under the assumed continuous real differentiability, the necessary and sufficient conditions are the Cauchy-Riemann equations at every point of $U$:Necessity follows by computing the complex derivative along real and imaginary increments. For sufficiency, real differentiability gives the first-order expansionThe Cauchy-Riemann equations turn the linear part into $(u_x+i{v}_x)h$, so the complex derivative exists and equals $u_x+i{v}_x$. Thus $f$ is a holomorphic function on $U$.

At any $z\ne 0$, take sufficiently small $h$ so $z+h\ne 0$. The complex difference quotient isHence the complex derivative exists at every point of the punctured plane, andThe excluded point is essential to the domain statement: the reciprocal has a simple pole there.

Write $z=x+iy$. Both moduli are nonnegative, so squaring preserves their strict inequality:ThusThe boundary is the perpendicular bisector of $0$ and $i$ in the complex plane; the strict inequality selects the half-plane containing $0$.

Use the reciprocal Möbius transformationIt is a holomorphic function on $U_1$ by part (b). For $z\ne 0$,Part (c) now gives $z\in U_1$ if and only if $|F(z)+i|>1$, so $F(U_1)\subseteq U_2$. Conversely, $0\notin U_2$ because $|0+i|=1$. For any $w\in U_2$, put $z=1/w$; the same identity gives $|z-i|>|z|$, so $z\in U_1$ and $F(z)=w$. Finally, reciprocation is injective on nonzero complex numbers. This proves bijectivity, with inverseThe inverse is also holomorphic, so this is a biholomorphism.

For a piecewise continuously differentiable contour $\gamma :[a,b]\to U$, the contour integral is defined byIf the contour has corners, integrate on its finitely many differentiable pieces and sum the results. The integral on the right is taken componentwise as an ordinary real-parameter integral of a complex-valued function. The factor ${\gamma }^{\prime }(t)$ encodes both the speed and orientation of the contour.

Choose the continuous argument with values in $(-\pi /2,3\pi /2)$ and defineIts branch cut is the nonpositive imaginary axis, including zero, exactly the removed ray. Locally away from that ray, the Cauchy-Riemann equations or the local inverse of the exponential show that it is a holomorphic logarithm, with derivative $1/z$ and $e^{\mathrm{Log}z}=z$. The chosen interval fixes the additive ambiguity by integer multiples of $2\pi i$.

Use the branch from part (b) and close the real interval $[-R,R]$ with the upper semicircle, oriented counterclockwise. For $\mathrm{Im}z\ge 0$, the number $z+i\alpha$ has positive imaginary part, so the numerator is a holomorphic logarithm throughout a neighbourhood of the contour and its interior. The branch point $-i\alpha$ lies strictly below the contour. The only enclosed pole ofis the simple pole at $i\beta$, with residueTo justify discarding the arc, take $R>\max \{\beta ,\alpha +1\}$. On it, $|z+i\alpha |$ lies between $R-\alpha$ and $R+\alpha$, and the argument has bounded absolute value. HenceThe real-axis integral is absolutely convergent, since its integrand is $O(\log |x|/x^2)$ at infinity and has no real singularities. The residue theorem therefore givesFinally, $\mathrm{Re}\mathrm{Log}(x+i\alpha )=\frac{1}{2}\log (x^2+{\alpha }^2)$. Taking twice the real part yieldsThe branch choice affects the first integral's imaginary part; the second integral is real and independent of that choice.

An isolated singularity at $a$ means that $f$ is holomorphic in some punctured disk $0<|z-a|<r$, while $a$ itself is an excluded point or a point where the given function is not holomorphic. There are no other singularities arbitrarily close to $a$. By Laurent's theorem, the function has a unique Laurent series on that punctured disk:The negative-power terms form its principal part of a Laurent series.

A removable singularity is an isolated singularity for which the function on the punctured disk extends to a holomorphic function on the full disk, by assigning or correcting its value at $a$. In Laurent series terms,The remaining power series defines the extension with value $c_0$ at $a$. Conversely, a holomorphic extension has a Taylor series there, so uniqueness of the Laurent series forces all negative coefficients to vanish.

A pole of order $m\ge 1$ at $a$ has finitely many negative powers in its Laurent series, the lowest being exactly $-m$:Equivalently,Thus $(z-a{)}^{m}f(z)$ has a holomorphic extension with nonzero value at $a$, whereas multiplying by any smaller power does not remove the singularity.

A zero of order $n\ge 1$ at $a$ means that $f$ is holomorphic near $a$ andEquivalently its Taylor series starts with the nonzero coefficient of $(z-a{)}^n$, soUnlike a pole, a zero is a regular point of the function itself; its reciprocal has a pole of the corresponding order.

By the finite-order zero factorizations,with $F,G$ holomorphic near $a$. Shrinking the disk if necessary, continuity ensures $G$ never vanishes there, so $F/G$ is holomorphic and nonzero at $a$. ConsequentlyIf $n>m$, the exponent is $-(n-m)$, with a holomorphic nonvanishing factor. Hence

With the same factors, if $n\le m$, thenhas no negative power and extends holomorphically across $a$. Thus its apparent singularity is a removable singularity, with extension valueFor $m>n$ the extended quotient has a zero of order $m-n$; for $m=n$ the extension is nonzero.

The complex sine has simple zeros exactly at $z=k\pi$, $k\in \mathbb{Z}$, because $\sin (k\pi )=0$ and $\cos (k\pi )=(-1{)}^k\ne 0$. Hence the denominator has a zero of order three at each of these points and nowhere else.

At $0$, the numerator has order two, so the quotient has a simple pole. Indeed its leading term is $-{\pi }^4/z$. At $\pi$, the numerator has order one, so there is a pole of order two. At $-\pi$, the numerator has order three, cancelling the denominator's order-three zero, so the singularity is removable. Its extension value isAt every remaining $k\pi$, the numerator is nonzero, giving a pole of order three. Thus the complete finite-plane classification isAway from this discrete set the denominator is nonzero, so the quotient is holomorphic.