Triangularisation, Generalised Eigenspaces and Jordan Canonical Form

Suppose $v+W=v^{\mathrm{\prime }}+W$. By (W1) of [L2] we have $v=v+0_V\in v+W$, so $v\in v^{\mathrm{\prime }}+W$, giving $w\in W$ with $v=v^{\mathrm{\prime }}+w$ and hence $v-v^{\mathrm{\prime }}=w\in W$. Conversely suppose $v-v^{\mathrm{\prime }}\in W$. For $w\in W$, $v+w=v^{\mathrm{\prime }}+((v-v^{\mathrm{\prime }})+w)$ and $(v-v^{\mathrm{\prime }})+w\in W$ by (W2), so $v+W\subseteq v^{\mathrm{\prime }}+W$; since $v^{\mathrm{\prime }}-v=(-1)(v-v^{\mathrm{\prime }})\in W$ by (W3), the same argument gives $v^{\mathrm{\prime }}+W\subseteq v+W$. Hence $v+W=v^{\mathrm{\prime }}+W$ exactly when $v-v^{\mathrm{\prime }}\in W$.

Let $v+W=v^{\mathrm{\prime }}+W$ and $u+W=u^{\mathrm{\prime }}+W$. By step 1.1, $v-v^{\mathrm{\prime }}\in W$ and $u-u^{\mathrm{\prime }}\in W$, so $(v+u)-(v^{\mathrm{\prime }}+u^{\mathrm{\prime }})=(v-v^{\mathrm{\prime }})+(u-u^{\mathrm{\prime }})\in W$ by (W2) and $av-a{v}^{\mathrm{\prime }}=a(v-v^{\mathrm{\prime }})\in W$ by (W3). Applying step 1.1 in the converse direction gives $(v+u)+W=(v^{\mathrm{\prime }}+u^{\mathrm{\prime }})+W$ and $(av)+W=(a{v}^{\mathrm{\prime }})+W$, so both quotient operations are independent of representatives.

The vector-space identities in $V\mathrm{/}W$ follow by applying the corresponding identities in $V$ to representatives, with zero coset $W=0_V+W$ and inverse $-(v+W)=(-v)+W$; moreover $\pi (av+bu)=(av+bu)+W=a\pi (v)+b\pi (u)$ by [L1] and [L3], every coset is $\pi (v)$ by definition, and by step 1.1 $\pi (v)=0_V+W$ exactly when $v-0_V=v\in W$, so $\pi$ is linear and surjective with $\ker \pi =W$.

If ${\sum }_i{a}_i{w}_i+{\sum }_j{b}_j{v}_j=0$, applying $\pi$ gives ${\sum }_j{b}_j(v_j+W)=0$; independence of the quotient basis forces every $b_j=0$, and independence of the basis of $W$ then forces every $a_i=0$.

For $v\in V$, expand $\pi (v)={\sum }_j{b}_j(v_j+W)$; then $v-{\sum }_j{b}_j{v}_j\in \ker \pi =W$ and is a combination of the $w_i$, so the displayed independent family spans $V$ and is a basis; counting its $r+s$ members gives $\dim V=\dim W+\dim (V\mathrm{/}W)$, including $W=0$, $W=V$, and $V=0$.

Define $\bar{f}(v+W):=f(v)$; if $v+W=v^{\mathrm{\prime }}+W$, then $v-v^{\mathrm{\prime }}\in W\subseteq \ker f$ by [L1], so $f(v)-f(v^{\mathrm{\prime }})=f(v-v^{\mathrm{\prime }})=0$ by [L2] and the linearity of $f$, giving $f(v)=f(v^{\mathrm{\prime }})$, and $\bar{f}$ is well defined; the operation formulas of [L3] then give $\bar{f}(a(v+W)+b(u+W))=\bar{f}((av+bu)+W)=f(av+bu)=af(v)+bf(u)$, so $\bar{f}$ is linear.

The definition gives $\bar{f}(\pi (v))=f(v)$, hence $\bar{f}\pi =f$; if $g:V\mathrm{/}W\to U$ also satisfies $g\pi =f$, then every coset is $\pi (v)$ and $g(v+W)=f(v)=\bar{f}(v+W)$, so $g=\bar{f}$, including the cases $W=V$ and $V=0$.

Apply [L1] with $W=\ker T$ and codomain restricted to $\mathrm{im}T$ to obtain the linear map $\tilde{T}(v+\ker T)=T(v)$; it is surjective by the definition of $\mathrm{im}T$.

Its kernel consists of cosets $v+\ker T$ with $T(v)=0$, hence only the zero coset $\ker T$; [L2] makes $\tilde{T}$ injective, so it is an isomorphism, including the zero map and the zero-space case.

If $v+W=v^{\mathrm{\prime }}+W$, then $v-v^{\mathrm{\prime }}\in W$, so $T(v)-T(v^{\mathrm{\prime }})=T(v-v^{\mathrm{\prime }})\in T(W)\subseteq W$ and therefore $T(v)+W=T(v^{\mathrm{\prime }})+W$; thus $\bar{T}$ is well defined.

For scalars $a,b$, the operations of [L3] give $a(v+W)+b(u+W)=(av+bu)+W$, so by [L1] and the linearity of $T$, $\bar{T}(a(v+W)+b(u+W))=T(av+bu)+W=(aT(v)+bT(u))+W=a(T(v)+W)+b(T(u)+W)$, and $\bar{T}$ is linear; also $\bar{T}(\pi (v))=\bar{T}(v+W)=T(v)+W=\pi (T(v))$ proves $\bar{T}\pi =\pi T$; the same calculation covers $W=0$, $W=V$, and $V=0$.

Induction on $k$ gives $(T{\mathrm{\mid }}_W{)}^k=T^k{\mathrm{\mid }}_W$ and ${\bar{T}}^k(v+W)=T^{k}v+W$, beginning with the identity at $k=0$ and using [L1] at the successor step; summing with the coefficients of $p$ yields both displayed identities.

Taking $p={\mu }_T$ makes $p(T)=0$, so step 1.1 gives $p(T{\mathrm{\mid }}_W)=0$ and $p(\bar{T})=0$; [L3] then gives ${\mu }_{T{\mathrm{\mid }}_W}\mid {\mu }_T$ and ${\mu }_{\bar{T}}\mid {\mu }_T$, including $W=0$, $W=V$, and $V=0$.

Choose an ordered basis of $W$, a basis of $V\mathrm{/}W$, and representatives of the latter; [L2] gives an adapted basis of $V$, in which invariance makes the matrix of $T$ block upper triangular, with upper-left block representing $T{\mathrm{\mid }}_W$ and lower-right block representing the well-defined operator $\bar{T}$.

Applying [L3] to that matrix and then [L4] to identify its diagonal-block polynomials gives ${\chi }_T={\chi }_{T{\mathrm{\mid }}_W}{\chi }_{\bar{T}}$; if $W=0$, $W=V$, or $V=0$, the missing block has characteristic polynomial $1$, so the same identity remains valid.

If the matrix is upper triangular, its $j$-th column has no nonzero entry below row $j$, so $T(v_j)\in V_j$; linearity then gives $T(V_j)\subseteq V_j$ for every $j$.

Conversely, if every $V_j$ is invariant, then $T(v_j)\in V_j$, so the $j$-th matrix column has zero entries below row $j$ and the matrix is upper triangular; the statements include the empty flag for $V=0$ and the flag $0\subset V$ in dimension one.

If $T$ is triangularisable, then $xI-[T]$ is upper triangular and its determinant is the product of its diagonal entries $x-a_{jj}$, so ${\chi }_T$ splits; by [L3], ${\chi }_T$ splits exactly when ${\mu }_T$ splits.

If $\dim V=0$, the empty basis and the polynomial $1$ give all three conditions.

Assume $\dim V>0$, that ${\chi }_T$ splits, and that the reverse implication holds in smaller dimensions; choose a root $\lambda \in F$ of ${\chi }_T$, then [L4] supplies a nonzero eigenvector $v$, and $W=Fv$ is a one-dimensional invariant subspace.

By [L2], ${\chi }_T=(x-\lambda ){\chi }_{\bar{T}}$, so ${\chi }_{\bar{T}}$ splits; the induction hypothesis triangularises $\bar{T}$ on $V\mathrm{/}W$, and [L5] lifts its triangular basis after $v$ to a basis whose initial spans are $T$-invariant, so [L1] triangularises $T$.

Step 1.1 gives $1\Rightarrow 3\iff 2$, while steps 1.2-2.1 give $3\Rightarrow 1$ in every finite dimension, completing all three equivalences.

A monic polynomial of degree $0$ is $1$, hence is the empty product of linear factors.

Let $f\in F[x]$ be monic of positive degree and assume every monic polynomial of smaller degree splits; [L2] gives a root $a$, and [L3] writes $f=(x-a)q$ with $q$ monic of degree one less, so the induction hypothesis makes $q$, and therefore $f$, split.

Apply steps 1.1-1.2 to the monic polynomial ${\chi }_T$; it splits over $F$, so [L1] triangularises $T$, with $V=0$ covered by the degree-zero base case.

If $\dim V=0$, the empty basis simultaneously triangularises every family.

Assume $\dim V>0$ and the theorem in smaller dimensions; if $\mathcal{A}$ is empty or all its members are scalar, any basis works, while otherwise choose a nonscalar $A\in \mathcal{A}$, use [L1] to obtain a nonzero proper eigenspace $E$, observe that every $S\in \mathcal{A}$ preserves $E$ because $SA=AS$, and use [L2] plus induction on $E$ to obtain a common eigenvector $v\in E$.

Put $W=Fv$; it is invariant under every $S\in \mathcal{A}$, the induced quotient operators commute by direct evaluation on cosets, and [L2] shows each quotient characteristic polynomial splits, so induction gives a common triangular basis of $V\mathrm{/}W$.

Lift that quotient basis after $v$ by [L5]; its initial spans are invariant for every $S$ by the quotient construction, so [L3] makes every representing matrix upper triangular in the same basis, and this also covers the empty and all-scalar branches.

Suppose $n>0$. By [L1], $N^k=0$ for some positive $k$ exactly when ${\mu }_N\mid x^k$, which is exactly when ${\mu }_N=x^r$ for some positive $r$; [L5] then bounds the least such $r$ by $n$ once ${\chi }_N=x^n$.

If ${\mu }_N=x^r$, [L2] says that $x$ is the only irreducible factor of ${\chi }_N$, and [L4] forces ${\chi }_N=x^n$; conversely ${\chi }_N=x^n$ and [L5] give $N^n=0$.

If $N$ is nilpotent, step 1.2 makes ${\chi }_N$ split, so [L3] gives an upper-triangular matrix; its diagonal entries are roots of ${\chi }_N=x^n$, hence are all zero, making it strictly upper triangular. Conversely, the $n$th power of a strictly upper-triangular $n\times n$ matrix is zero.

For $V=0$, [L1] and [L4] give ${\mu }_N={\chi }_N=1=x^0$, the unique empty matrix is strictly upper triangular, and the convention in Nilpotent endomorphisms and their nilpotency index gives index $1$, so conditions 1, 3 and 4 hold; condition 2 asks for $r$ with $1\le r\le 0$ and no such integer exists, and the exponent $0$ of ${\mu }_N$ differs from the index $1$, which is why the equivalence is stated for $n>0$. Together with steps 1.1-2.1 this proves every asserted case.

If $T^{k}v=0$, then $T^{k+1}v=0$, and every $T^{k+1}v$ equals $T^k(Tv)$; hence the displayed kernel and image inclusions hold, and [L2] turns them into the asserted dimension inequalities.

Assume $\ker T^m=\ker T^{m+1}$. The equality $\ker T^{m+0}=\ker T^m$ is the base case.

By [L3], equality of the two consecutive kernel dimensions is equivalent to equality of the two consecutive ranks; together with the inclusions in step 1.1, either dimension equality is equivalent to equality of the corresponding subspaces.

If $\ker T^{m+r}=\ker T^m$ and $v\in \ker T^{m+r+1}$, then $T^{r}v\in \ker T^{m+1}=\ker T^m$, so $v\in \ker T^{m+r}$; the reverse inclusion is in step 1.1, completing the induction on $r$.

Applying [L3] to every $T^{m+r}$ shows that the later images all have the same dimension as $\mathrm{im}{T}^m$; the nested image inclusions from step 1.1 therefore make them equal, completing the claim.

If $x\in \ker T^m\cap \mathrm{im}{T}^m$, write $x=T^{m}y$; then $T^{2m}y=0$, so [L1] gives $y\in \ker T^m$ and hence $x=0$.

By [L2] applied to $T^m$, the dimensions of $\ker T^m$ and $\mathrm{im}{T}^m$ sum to $\dim V$; with the zero intersection from step 1.1, their sum is therefore all of $V$.

Fact [L3] now gives the asserted internal direct sum, including $m=0$ and $V=0$.