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

The minimal polynomial $m_T$ is the monic polynomial of least polynomial degree such that $m_T(T)=0$. Here a polynomial $p(x)={\sum }_j{c}_j{x}^j$ is evaluated at the linear operator by $p(T)={\sum }_j{c}_j{T}^j$, with the constant term acting as $c_{0}I$.

An annihilating polynomial exists because $I,T,T^2,\dots$ lie in the finite-dimensional vector space ${\mathrm{End}}_F(V)$ and are therefore eventually linearly dependent. Uniqueness follows by subtracting two monic annihilators of the same least degree: their difference has smaller degree and still annihilates $T$.

Equivalently,For the forward implication evaluate the factorization at $T$. For the reverse implication divide $p$ by $m_T$ using polynomial division; the remainder annihilates $T$, so minimality makes it zero. This also identifies the annihilating polynomials as the principal ideal generated by $m_T$.

Suppose $E^2=E$. Every vector has the decompositionIf $w=Ez$ also belongs to $\ker E$, then $w=E^{2}z=Ew=0$. HenceOn its image of a linear map, $E$ acts as the identity; on its kernel of a linear map, it acts as zero. Concatenate a basis of the image with a basis of the kernel. The resulting matrix is $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}E$.

Conversely, this block matrix squares to itself, so any linear map having that matrix in a basis satisfies $E^2=E$. Thus projection operators are precisely the linear maps with this block form. This is an algebraic projection; no inner product or orthogonality is required.

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.