Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 1 / b

Evaluate Bézout's identity for polynomials at $T$:All these linear operators commute because they are polynomials in the same operator. Thus every $v\in V$ can be writtenThe first summand belongs to $\ker p(T)$ and the second to $\ker q(T)$, since applying the corresponding operator introduces the product $p(T)q(T)=0$.

If $v$ lies in both kernels, the same identity gives $v=0+0=0$. Therefore the sum is direct:This is the kernel decomposition for coprime polynomials. It uses both annihilation by $pq$ and the coprimality expressed by the Bézout identity.

Write $U=\ker p(T)$, $W=\ker q(T)$. The linear operator $E=a(T)p(T)$ vanishes on $U$. On $W$, the identity $E=I-b(T)q(T)$ gives $Ew=w$.

By the direct sum in part (i), every vector has a unique expression $u+w$, andConsequentlyThe complementary projection operator is $I-E=b(T)q(T)$, onto $U$ along $W$. These are polynomial projections for a coprime decomposition. In particular the projection does not depend on which Bézout coefficients are chosen: its action on the two summands uniquely determines it.

Let $P=b(T)q(T)$ and $Q=a(T)p(T)$. By part (ii), these are complementary projection operators with images $U=\ker (T-\lambda I{)}^{\alpha }$ and $W=\ker (T-\mu I{)}^{\beta }$. Thus $D=\lambda P+\mu Q$ acts as $\lambda I$ on $U$ and $\mu I$ on $W$.

The linear operator $T$ preserves both summands because it commutes with its own polynomials. Therefore $N=T-D$ preserves them too, and its restrictions satisfyPut $k=\max (\alpha ,\beta )$. Both restrictions of $N^k$ vanish, so the direct sum givesHence $T-D$ is a nilpotent linear map. The operator $D$ is diagonalizable, since its two summands are its eigenspaces, and it commutes with $T$. This illustrates the Jordan–Chevalley decomposition in the case of two distinct eigenvalues.