Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Digital systems, linear algebra and introduction to formal proof / 5

Take $V={\mathbb{R}}^2$ and the linear map $h(x,y)=(x,0)$. ThenThese two vector subspaces have zero intersection and together span ${\mathbb{R}}^2$, so $V=\mathcal{R}(h)\oplus \mathcal{N}(h)$ as a direct sum of vector spaces.

Take $V={\mathbb{R}}^2$ and the linear map $h(x,y)=(y,0)$. ThenTheir intersection is nonzero and their sum is only one-dimensional. Hence $V\ne \mathcal{R}(h)\oplus \mathcal{N}(h)$. This nilpotent linear map has $h^2=0$, illustrating why taking sufficiently high powers changes the situation.

For $v\in \mathcal{R}(h^{m+k})$, choose $u$ with $v=h^{m+k}u=h^m(h^{k}u)$. Thus $v\in \mathcal{R}(h^m)$, provingThis is the decreasing-image part of stabilization of images of powers of a linear map. The convention $h^0={\mathrm{id}}_V$ includes $m=0$ and $k=0$.

Apply $h$ to both sides of $\mathcal{R}(h^m)=\mathcal{R}(h^{m+1})$. Because $h(\mathcal{R}(h^j))=\mathcal{R}(h^{j+1})$, we obtain $\mathcal{R}(h^{m+1})=\mathcal{R}(h^{m+2})$. Repeating this argument proves equality at every later step. More formally, an induction on $k$ givesThe base case is $k=0$; the induction step follows from applying the linear map $h$ and then the assumed first equality. Thus one equality permanently stabilizes the descending sequence of images of a linear map.

The dimensions of the nested images of a linear mapcan strictly decrease at most $n$ times. If all the first $n$ inclusions are strict, $\mathcal{R}(h^n)=0$ and is stable. Otherwise an earlier equality occurs and (d) makes every subsequent image equal. Thus in all casesBy the rank-nullity theorem and inclusion of the null spaces, this also gives $\ker h^n=\ker h^{2n}$. If $v\in \mathcal{R}(h^n)\cap \ker h^n$, write $v=h^{n}u$. Then $0=h^{n}v=h^{2n}u$, so $u\in \ker h^{2n}=\ker h^n$ and $v=0$. Hence the intersection is zero. Finally the rank-nullity theorem givesZero intersection and this dimension sum proveThis is the finite-dimensional Fitting decomposition. It includes the zero-dimensional case, for which $V=0$ and the assertion is immediate.