Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2004 / 2

The minimal polynomial of $T$ is the monic nonzero polynomial of smallest degree annihilating $T$. Such a polynomial exists, since sufficiently many powers of $T$ are linearly dependent in the finite-dimensional space $\mathrm{End}(V)$. If two monic annihilators had the same least degree, their difference would be an annihilator of smaller degree and hence zero. Thus it is unique. Division with remainder also shows that every annihilating polynomial is a multiple of it.

The primary decomposition theorem states that if $m={\prod }_{j=1}^r{f}_j^{e_j}$ with distinct monic irreducible polynomials, thenThese subspaces are $T$-invariant, and the restriction to $V_j$ has minimal polynomial $f_j^{e_j}$.

Let $r=\gcd (q,m)$, chosen monic. The Bézout's identity for polynomials gives $r=aq+bm$. If $q(T)v=0$, then $r(T)v=a(T)q(T)v+b(T)m(T)v=0$. Conversely, $q=rh$, so $r(T)v=0$ implies $q(T)v=0$. ThereforeThe following Roman scopes correspond to the three labelled conditions, with their implications proved from the indicated condition.

If a basis consists of eigenvectors, let ${\lambda }_1,\dots ,{\lambda }_r\in K$ be the distinct eigenvalues. A polynomial annihilates the diagonal action exactly when it vanishes at each ${\lambda }_j$. Consequently the minimal polynomial is ${\prod }_j(X-{\lambda }_j)$, a product of distinct linear factors. It therefore has no repeated irreducible factor, proving (i) implies (ii).

Suppose $m$ has no repeated irreducible factor. For every polynomial $p$ and integer $n\ge 1$, its monic greatest common divisors satisfy $\gcd (p^n,m)=\gcd (p,m)$, since raising $p$ to a power changes no membership of an irreducible factor of the square-free $m$. This is the square-free minimal polynomial criterion for polynomial kernel stabilization. The kernel identity proved above givesThus (ii) implies (iii) over every field.

Over $\mathbb{C}$, the distinct irreducible factors of $m$ are linear. The primary decomposition theorem therefore reads $V={⨁}_j\ker (T-{\lambda }_{j}I)$, a direct sum of eigenspaces. Taking a basis in each summand gives an eigenbasis, proving (ii) implies (i) over the complex numbers.

Over $\mathbb{R}$, take $T=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Its minimal polynomial is $X^2+1$, which is an irreducible polynomial over $\mathbb{R}$ and has no repeated factor. But it has no real eigenvalues, so no real eigenbasis. This is the required counterexample. Over a general field, the phrase “no repeated factors” here refers to irreducible factors in $K[X]$, rather than a prior assumption that those factors split.

Assume condition (iii), and factor $m={\prod }_j{f}_j^{e_j}$. If some $e_j\ge 2$, the primary decomposition theorem gives the nonzero summand $V_j=\ker f_j(T{)}^{e_j}$ on which the minimal polynomial is exactly $f_j^{e_j}$. The operator $f_j(T)$ cannot vanish on this whole summand, or its restriction would have minimal polynomial dividing $f_j$. Choose $v\in V_j$ with $f_j(T)v\ne 0$. Thencontradicting (iii) with $p=f_j$ and $n=e_j$. Hence every $e_j=1$, proving (iii) implies (ii).