Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2004 / Algebra

${\dim }_{\mathbb{R}}V=4$. For arbitrary matrices,This is bilinear and symmetric. Also $⟨A\mid A⟩={\sum }_{i,j}{a}_{ij}^2\ge 0$, with equality exactly when every entry vanishes. Thus it is an inner product, the Frobenius inner product.

If $S^t=S$ and $K^t=-K$, invariance of trace under transposition and cyclicity giveFurthermore $⟨I\mid A⟩=\mathrm{tr}A$, so every traceless matrix is orthogonal to the identity. One convenient orthonormal basis adapted to these observations isTheir entrywise inner products are zero in distinct pairs and one on the diagonal; four orthonormal vectors in this four-dimensional space form a basis.

Self-adjointness means $⟨Tv,w⟩=⟨v,Tw⟩$ for all $v,w\in V$, or $T=T^{\ast }$. Take the inner product linear in its first variable. If $Tv=\lambda v$ with $v\ne 0$, thenTherefore every eigenvalue is real.

If $U$ is invariant and $w\in U^{\perp }$, then for every $u\in U$, $Tu\in U$ and $⟨Tw,u⟩=⟨w,Tu⟩=0$. Hence $Tw\in U^{\perp }$, proving invariance of the orthogonal complement.

Induct on $d=\dim V$. The zero-dimensional case is empty. For $d>0$, choose an eigenvector and normalize it to a unit vector $e$. Its span is invariant, so its orthogonal complement is invariant as just proved. The restriction of $T$ to that complement is again self-adjoint. Its dimension is $d-1$, so induction supplies an orthonormal basis of eigenvectors there. Adjoining $e$ gives an orthonormal eigenbasis of $V$. This proves the finite-dimensional spectral theorem without assuming it in the argument.

A ring is a set with an abelian-group addition, associative multiplication and both distributive laws; here we use rings with multiplicative identity. A field is a nonzero commutative ring in which every nonzero element has a multiplicative inverse.

The positive integers form neither a ring nor a field. They do not contain an additive identity zero, and they also lack additive inverses.

$\mathbb{Z}/4\mathbb{Z}$ is a ring but not a field. Quotient operations inherit associativity and distributivity from the integers, and residues have additive inverses. However the nonzero residue $\overline{2}$ satisfies $\overline{2}\overline{2}=\overline{0}$, so it cannot be invertible. A field has no nonzero zero divisors.

$\mathbb{Z}/3\mathbb{Z}$ is a field, and consequently also a ring. Its nonzero elements are $\overline{1}$ and $\overline{2}$, with inverses $\overline{1}$ and $\overline{2}$ respectively. The quotient ring axioms are inherited as in the preceding part.

This is the Gaussian integer ring $\mathbb{Z}[i]$. It contains zero and one and is closed under addition, additive inverses and multiplication, sinceIts operations inherit the ring axioms from $\mathbb{C}$. It is a ring but not a field: the nonzero element 2 has complex inverse $1/2$, which is not a Gaussian integer.