Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra

Choose a basis and let $A$ represent $T$. Define the characteristic polynomial byUnder a change of basis with invertible change matrix $S$, the new matrix is $S^{-1}AS$. HenceThus the polynomial is independent of the chosen coordinates. It is monic of degree $\dim V$.

To prove the sum is direct, suppose ${\sum }_{j=1}^r{v}_j=0$ with $v_j\in V_{{\lambda }_j}$. Fix $i$ and apply the polynomial in the linear operatorEvery $v_j$ with $j\ne i$ is killed by one factor. On $v_i$, each factor is multiplication by ${\lambda }_i-{\lambda }_j$, givingThe scalar coefficient is nonzero because the eigenvalues are distinct. Thus $v_i=0$ for every $i$, provingIn particular, choosing a basis in each eigenspace yields a linearly independent combined list.

If ${\chi }_T(\lambda )=0$, then $\det (\lambda I-A)=0$, so $\lambda I-A$ is singular. By the rank-nullity theorem it has a nonzero kernel vector $v$. ThusThis is exactly the definition of an eigenvalue. Here $\lambda$ lies in the scalar field: for the stated real vector space it is real. A nonreal root instead gives an eigenvalue of the complexified operator, by the same argument over $\mathbb{C}$.

Expanding the determinant along the second column givesIf $b\notin \{-3,2\}$, there are three distinct real eigenvalues. Part (b)(ii) gives a nonzero eigenvector for each, and part (b)(i) makes them independent. They form a basis, so the matrix is diagonalisable.

When $b=-3$,has rank one for every $a$: the third row is nonzero, and the second row is $a/5$ times the third. Therefore the eigenspace for $-3$ has dimension two. The eigenvalue $2$ provides at least one further independent eigenvector by part (b), so again there is an eigenbasis.

When $b=2$,Its rank is one if $a=0$ and two otherwise. Thus the eigenspace for $2$ has dimension two exactly when $a=0$. For $a\ne 0$ it has dimension one, while the eigenspace for $-3$ also has dimension one, since $X+3I$ then has rank two. There are no other eigenvalues by the characteristic polynomial. A diagonal matrix in an eigenbasis would need three independent eigenvectors, whereas these two eigenspaces supply only two; diagonalisation is impossible. Conversely, at $a=0$ their dimensions add to three and part (b) makes their sum direct. ThereforeThe argument uses only the determinant, rank-nullity, the established direct-sum result, and the definition of diagonalisation.

Use the convention that the inner product is linear in its first argument, as required by the displayed representation of a linear functional. Choose an orthonormal basis $e_1,\dots ,e_n$, for instance by Gram-Schmidt orthogonalization. DefineFor $u={\sum }_j{u}_j{e}_j$, conjugate linearity in the second argument givesThusThe representation is unique as well: if two vectors represent the same functional, their difference is orthogonal to every vector and therefore to itself, so it is zero.

The adjoint operator is characterized byFor each fixed $v$, the map $u\mapsto ⟨Tu,v⟩$ is a linear functional. Part (i) gives a representing vector, which we define to be $T^{\ast }v$. Hence such a map exists on every vector of $V$. Part (b) proves that this constructed map is linear.

For scalars $\alpha ,\beta$ and vectors $v,w,u$, use the adjoint identity and conjugate linearity in the second argument:The difference of the two second-argument vectors is orthogonal to every $u$. Choosing $u$ to be that difference shows its squared norm is zero. ThereforeThe two conjugations are why the adjoint is a linear map, despite the conjugate-linear dependence of the representing functional on its second argument.

The finite-dimensional spectral theorem says that a self-adjoint operator on a complex inner-product space has an orthonormal basis of eigenvectors and all its eigenvalues are real. Equivalently, its matrix is unitarily diagonalisable with real diagonal entries.

For $a\in \mathbb{R}$, the matrix is equal to its conjugate transpose:Thus it represents a self-adjoint operator for the standard inner product on ${\mathbb{C}}^3$. The theorem applies directly, proving

Set $N=M_i-I$. Direct multiplication giveswith $N\ne 0$. If $M_i$ were diagonalisable, then subtracting the identity in the same basis would diagonalise $N$. A diagonal matrix whose cube is zero has every diagonal entry zero and is itself zero. Similarity would then imply $N=0$, a contradiction. HenceThis proves failure directly; merely observing that it is not self-adjoint would not suffice, because non-self-adjoint matrices can still be diagonalisable.

An ideal $I$ of a commutative ring is an additive subgroup such that $rx\in I$ for every $r\in R$ and $x\in I$. A prime ideal is a proper ideal $A\ne R$ such that $xy\in A$ implies $x\in A$ or $y\in A$.

If $(x+A)(y+A)=0+A$ in the quotient ring, then $xy\in A$. Primality implies $x+A=0+A$ or $y+A=0+A$. Therefore there are no two nonzero cosets whose product is zero:Properness also ensures its identity coset is not zero, so this quotient is an integral domain.

The set $\sqrt{I}$ contains zero. If $x^m\in I$, then $(-x{)}^m=(-1{)}^m{x}^m\in I$, giving closure under additive inverses; also $(rx{)}^m=r^m{x}^m\in I$ for every $r\in R$.

To prove closure under addition, let $x^m,y^n\in I$. Expand $(x+y{)}^{m+n-1}$ by the binomial theorem. In every monomial $x^j{y}^{m+n-1-j}$, either $j\ge m$ or $m+n-1-j\ge n$. Thus each term contains a factor $x^m$ or $y^n$ and belongs to $I$. Hence $(x+y{)}^{m+n-1}\in I$, so $x+y\in \sqrt{I}$. The additive subgroup and absorption properties are proved, givingThis ideal is the radical of an ideal. The finite exponent for a sum uses commutativity of the ring.

A coset is a nilpotent element of the quotient exactly when some positive power is zero:If $J=\sqrt{J}$, this forces every nilpotent coset to equal zero. Conversely, if there is no nonzero nilpotent coset, every $x\in \sqrt{J}$ has $x+J=0$, hence $x\in J$. Since $J\subseteq \sqrt{J}$ always, equality follows. ThereforeIn standard terms, an ideal is radical exactly when its quotient is a reduced ring.

Certainly $A\subseteq \sqrt{A}$, by taking exponent one. If $x\in \sqrt{A}$, choose $n\ge 1$ with $x^n\in A$. Primality applied to $x\cdot x^{n-1}$ gives $x\in A$ or $x^{n-1}\in A$. In the latter case repeat; induction reduces to $x\in A$. Thus $\sqrt{A}\subseteq A$ as well, provingEvery prime ideal is therefore a radical ideal.