Past exam of the algebra course of the University of Oxford / 2007 / A / 4

An $R$-module isomorphism $T:V_{\alpha }\to V_{\beta }$ is a $K$-linear isomorphism satisfying $T\alpha =\beta T$, by applying it to multiplication by $X$. Conversely that identity implies $Tf(\alpha )=f(\beta )T$ for every polynomial, so a similarity is an $R$-module isomorphism.

The cyclic decomposition theorem over the principal ideal domain $R=K[X]$ states that any finitely generated module iswith nonunit monic invariant factors, uniquely determined, and free rank $r$ uniquely determined. A finite-dimensional $K$-space has $r=0$. In each cyclic summand the basis $1,X,\dots ,X^{\mathrm{deg}{d}_i-1}$ makes multiplication by $X$ the companion matrix of $d_i$. Their block diagonal sum is the rational canonical form.

Here the two polynomials are already in divisibility order. With the companion convention having ones below the diagonal,Integer coefficients are interpreted in $K$, so this remains valid in characteristics two and three.

In the cyclic case there is one companion block $C(d)$. Its defining polynomial is both characteristic and minimal: the vector represented by $1$ is cyclic and a polynomial annihilates it exactly when divisible by $d$. Thus the minimal polynomial is read directly from that block.

The eigenvalues in $K$ are precisely the roots in $K$ of the companion polynomial $d$; over an algebraic closure take all its roots. Thus the form determines their root set, but an explicit list generally requires factoring or solving $d$ and cannot be read off by inspection. This is the relevant distinction in the qualification “with no further calculations”. Over a general field the polynomial may have no roots in $K$ at all, and its diagonal entries are not its eigenvalues. If “determinable” means only uniquely specified rather than immediately readable, all the information is of course fixed by the matrix, including this root set.

For $d(X)=X^n+a_{n-1}{X}^{n-1}+\cdots +a_0$, the determinant of its companion matrix is $(-1{)}^n{a}_0$. ThereforeThus invertibility is readable from the constant coefficient, with no factorization or elimination.

In the cyclic quotient $R/(d)$, the kernel is formed by residue classes of polynomials $h$ such that $d\mid Xh$. If $X\nmid d$, coprimality forces $d\mid h$, so the kernel is zero. If $d=Xe$, divisibility is equivalent to $e\mid h$, and the residue classes of multiples of $e$ modulo $Xe$ form a one-dimensional $K$-space. HenceThis too is immediately readable in the cyclic case, even if zero is a repeated root. The zero-dimensional module has kernel dimension zero separately.