Past exam of the rings and modules course of the University of Oxford / 2017 / 3

Write $d=\dim V$. A cyclic vector for $S$ is a vector $v$ whose successive iterates span $V$; equivalently, when $d>0$,are a basis. The equivalence follows because the first linear dependence in the iterates makes their preceding span invariant under $S$, so no later iterate can enlarge it.

In this basis there are unique coefficients $c_0,\dots ,c_{d-1}$ with $S^{d}v=-{\sum }_{j<d}{c}_j{S}^{j}v$. Put $f(t)=t^d+{\sum }_{j<d}{c}_j{t}^j$. Then $f(S)v=0$, and commutation with powers of $S$ gives $f(S)S^{j}v=S^{j}f(S)v=0$. Therefore $f(S)$ kills an entire basis and is the zero map. No nonzero polynomial of degree less than $d$ annihilates $v$, because that would be a linear dependence of the displayed basis. Thus $m_S=f$.

The matrix of $S$ in the cyclic basis is the companion matrix $C(f)$ with ones on its subdiagonal and last column $(-c_0,\dots ,-c_{d-1}{)}^{\mathsf{T}}$. Expanding $\det (tI-C(f))$ along its last column givesthe cofactors are $1,t,\dots ,t^{d-1}$. This proves the characteristic polynomial of a companion matrix formula here. HenceThis proves that a cyclic operator has equal minimal and characteristic polynomials. If $V=\{0\}$, both polynomials are one by convention, so the equality remains valid.

The rational canonical form of $T$ is a block diagonal matrixsimilar to a matrix of $T$, where the $f_j\in F[t]$ are monic nonconstant polynomials and the sum of their degrees is $\dim V$. They are the invariant factors of a linear operator. The block $C(f_j)$ is the companion matrix for $f_j$, representing multiplication by $t$ on $F[t]/(f_j)$. The invariant factors are uniquely determined by $T$.

Every companion block has a cyclic vector and therefore has both minimal polynomial and characteristic polynomial equal to $f_j$, by part (a). The determinant of a block diagonal matrix is the product of its block determinants, givingA polynomial $h$ annihilates the whole matrix exactly when it annihilates every block. On block $j$, dividing $h$ by $f_j$ leaves a remainder of smaller degree; if that remainder annihilated the block without being zero, it would contradict minimality of $f_j$. Thus the annihilation condition is exactly $f_j\mid h$ for every $j$. The monic polynomial of least degree with this property is their least common multiple. The divisibility chain makes this $f_r$, soThese are the minimal and characteristic polynomials from invariant factors formulae. On the zero vector space the product is empty and both polynomials are defined to be one.

Let $J=\left(\begin{array}{cc}0& -2\\ 1& 0\end{array}\right)$, so the displayed rational canonical form of $T$ is $J\oplus J$. Since $J^2=-2I_2$, we have $T^2=-2I_4$.

If $a=0$, then $U=b{I}_4$ is a scalar matrix. Its invariant factors of a linear operator are four copies of $t-b$, so its rational canonical form isHere $m_U=t-b$ and $c_U=(t-b{)}^4$, including when $b=0$.

Now let $a\ne 0$ and putOn each two-dimensional invariant block, $A=aJ+b{I}_2$ satisfies $q(A)=0$. The vectors $e_1,A{e}_1$ form a cyclic vector basis, since their coordinate matrix is $P=\left(\begin{array}{cc}1& b\\ 0& a\end{array}\right)$ with determinant $a\ne 0$. Directly,Applying this basis change to both blocks givesThus $m_U=q$ and $c_U=q^2$. In particular there are two invariant factors $q,q$, rather than a single block for $q^2$. Also $q(t)>0$ for every real $t$, so it is an irreducible polynomial over $\mathbb{R}$. This is an instance of rational canonical form under an affine change.

Finally $T^2=-2I_4$ implies $T^{-1}=-T/2$. Substituting $a=-1/2$, $b=0$ in the preceding result gives $q(t)=t^2+1/2$, and henceIts minimal polynomial is $t^2+1/2$ and its characteristic polynomial is $(t^2+1/2{)}^2$. The normalized reciprocal of $t^2+2$ is also $t^2+1/2$, agreeing with rational canonical form of an inverse.