Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Algebra

Write $n=\dim V$. If $Rv=\lambda v$ with $v\ne 0$, then $0=R^{a}v={\lambda }^{a}v$, hence $\lambda =0$. The characteristic polynomial splits over $\mathbb{C}$ and has degree $n$, soThere is indeed an eigenvector for the eigenvalue zero: a nonzero vector in the last nonzero image of $R$ lies in its kernel of a linear map. For the bound, choose $v$ with $R^{a-1}v\ne 0$, using minimality. The vectors $v,Rv,\dots ,R^{a-1}v$ are linearly independent. Otherwise choose the least $j$ with nonzero coefficient in a relation and apply $R^{a-1-j}$; only its nonzero multiple of $R^{a-1}v$ survives, a contradiction. Therefore $\boxed{a\le n}$.

Suppose $S^b=0$. Use the increasing chain of vector subspacesChoose a basis successively, extending a basis of each kernel of a linear map to one of the next kernel of a linear map. If a basis vector first enters at level $j$, its image under $S$ belongs to $\ker S^{j-1}$, spanned entirely by earlier basis vectors. Thus every matrix column has nonzero entries only above its diagonal: the matrix is strictly upper triangular.

Conversely, for such a matrix and its ordered basis $e_1,\dots ,e_n$, one has $S⟨e_1,\dots ,e_j⟩\subseteq ⟨e_1,\dots ,e_{j-1}⟩$. Iterating $n$ times kills every vector. Hence $\boxed{S^n=0}$, establishing both directions without assuming a Jordan form.

The polynomials $t^c$ and $(t-1{)}^d$ are coprime. Bézout's identity gives polynomials $u,v$ with $u(t)t^c+v(t)(t-1{)}^d=1$. SetFor every $x$, the decompositionhas first summand in $V_0$ and second in $V_1$, since $T^c(T-I{)}^d=0$ and polynomial expressions in $T$ commute. If $x$ belongs to both kernels, the same Bézout identity gives $x=0$. Thus $V=V_0\oplus V_1$. Both vector subspaces are invariant under $T$.

On $V_0$, the restriction of $T$ is nilpotent; on $V_1$, the restriction of $T-I$ is nilpotent. Apply (b) separately, then concatenate the two bases. This yieldswhere $A,N$ are strictly upper triangular. Empty blocks are allowed if one subspace is zero. In particular $C=I+N$ has precisely the required diagonal.

Let $D=T^{\ast }T-T{T}^{\ast }$. This operator is self-adjoint, andIf condition (iii) holds, this quadratic form vanishes. For a real inner product, polarization givesTherefore $D=0$, which is condition (i). This supplies the nontrivial reverse implication needed for the equivalence.

By the definition of the adjoint,If (i) holds, the expression is zero for all $x,y$, giving (ii). Conversely, if this expression vanishes for every $x$, nondegeneracy of the inner product gives $(T^{\ast }T-T{T}^{\ast })y=0$ for every $y$, so (ii) also implies (i).

Taking $y=x$ in (ii) gives (iii). Together with (iii)$\Rightarrow$(i) from part (i) and (i)$\Rightarrow$(ii) from part (ii), this completes

For a normal operator, (a) gives $\Vert Tx\Vert =\Vert T^{\ast }x\Vert$, so one side is zero exactly when the other is. Hence $\boxed{\ker T=\ker T^{\ast }}$.

For real $\lambda$, the adjoint of $T-\lambda I$ is $T^{\ast }-\lambda I$. Expanding both products shows that their difference is $T^{\ast }T-T{T}^{\ast }=0$. Thus $T-\lambda I$ is normal, and applying the kernel of a linear map conclusion to it givesThis proves equality of each real eigenspace, hence the same eigenvalues and eigenvectors on the stated real vector space. If neither has real eigenvalues, the conclusion is vacuous. Upon complexification one instead pairs $\lambda$ with $\overline{\lambda }$ for the adjoint; no claim of the same complex eigenvalues is needed here.

In fact the adjoint invariance follows from the orthogonal decomposition and the invariance under $T$, without using the normal operator property. For $x\in V_i$ and $y\in V_j$ with $j\ne i$,since $Ty\in V_j$. Hence $T^{\ast }x$ is orthogonal to every other summand and therefore lies in $V_i$. Moreover, for $x,y\in V_i$, the adjoint identity restricted to that subspace shows

Restricting $T^{\ast }T=T{T}^{\ast }$ to $V_i$, using (i), gives $T_i^{\ast }{T}_i=T_i{T}_i^{\ast }$, so the global normal operator property implies the normal operator property of every restriction. Conversely, if the restrictions are normal and the adjoint preserves each summand, then for $x={\sum }_i{x}_i$,Thus $\boxed{T\text{ is normal}⟺\text{every }{T}_i\text{ is normal}}$, with condition (i) automatic under the stated decomposition.

Interpret membership as having a representation with denominator coprime to $30$, equivalently having this property in lowest terms. Clearly $0,1\in R$. If $a/b,c/d\in R$ with admissible denominators, thenThe product $bd$ is still coprime to $30$. These are elements of $R$, proving that $\boxed{R\text{ is a subring of }\mathbb{Q}}$.

Take a reduced fraction $a/b\in R$. If $a\ne 0$, its only possible inverse in $\mathbb{Q}$ is $b/a$. Since $\gcd (a,b)=1$, this is in lowest terms up to sign and belongs to $R$ exactly when $\gcd (a,30)=1$. Zero is not invertible. ThereforeEquivalently, these are the rational numbers whose valuations at $2,3,5$ are all zero.

The intersection $J=I\cap \mathbb{Z}$ is an additive subgroup of $\mathbb{Z}$, and $nj\in I\cap \mathbb{Z}$ for $n\in \mathbb{Z}$, $j\in J$. Thus it is an ideal of $\mathbb{Z}$, necessarily $d\mathbb{Z}$ for a unique $d\ge 0$.

Because $d\in I$, one has $dR\subseteq I$. Conversely, if $a/b\in I$ with $b$ coprime to $30$, multiplication by $b\in R$ gives $a\in I\cap \mathbb{Z}=d\mathbb{Z}$. Write $a=dk$; then $a/b=d(k/b)\in dR$. Thus $I=dR$.

Every positive integer factors as $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }u$ with $\gcd (u,30)=1$. The factor $u$ is a unit, so it does not change the ideal. The complete distinct list isThey are distinct because the intersection of the displayed nonzero ideal with $\mathbb{Z}$ is $2^{\alpha }{3}^{\beta }{5}^{\gamma }\mathbb{Z}$: admissible denominators cannot cancel any of these prime powers. The triple $(0,0,0)$ gives $R$ itself.

The zero ideal is a prime ideal because $R$ is an integral domain. Each of $(2),(3),(5)$ is a prime ideal: for $p$ among these three, reduction $a/b\mapsto \overline{a}{\overline{b}}^{-1}$ is a surjective ring homomorphism $R\to {\mathbb{F}}_p$ with kernel of a ring homomorphism $(p)$, so its quotient ring is a field.

Any other nonzero proper ideal in (c) has generator $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }$ with total exponent at least two. Choose one prime factor $p$ and put $e=d/p$. Then $pe=d\in (d)$ but neither $p$ nor $e$ belongs to $(d)$, as seen from the required prime exponents. Such an ideal is not prime. Thus with the standard proper-ideal convention the answer isThe definition printed in this paper omits properness. Literally under that definition $R$ itself also satisfies the implication, so its literal complete list is the preceding four ideals together with $\boxed{R}$. Both interpretations are thereby accounted for.