8. Optional supplement: Jordan form

The zero-dimensional case has the empty chain basis. In the remaining case take $V\ne0$. Choose an ordered basis and write $A=[T]$, so $\chi_T(t)=\det(tI-A)$. We first establish the polynomial relation and the direct-sum decomposition without assuming Jordan form.

Throughout, $V$ is a nonzero finite-dimensional vector space over a field $F$, and $T\in\operatorname{End}_F(V)$. No algebraic closure is assumed unless stated.

For $p(t)=\sum_{j=0}^dc_jt^j$, define $p(T)=\sum c_jT^j$, with $T^0=I$. Because powers of the same operator commute, $(pq)(T)=p(T)q(T)$ and evaluation is a ring homomorphism $F[t]\to\operatorname{End}(V)$. This does not permit replacing variables by arbitrary noncommuting matrices in every matrix-coefficient identity.

Theorem. $\chi_T(T)=0$.

Proof. Let $\chi_A(t)=\sum_{k=0}^nc_kt^k$ with $c_n=1$. The adjugate identity over the commutative polynomial ring $F[t]$ gives

Set $B_{-1}=B_n=0$. Comparing coefficients yields $B_{k-1}-AB_k=c_kI$ for $0\le k\le n$. Multiply the $k$th identity on the left by $A^k$ and sum. The terms telescope:

This is $\chi_A(A)=0$. It uses coefficient identities and associativity only, not an unexplained substitution into a matrix polynomial with potentially noncommuting coefficients. ∎

Since Cayley–Hamilton supplies a nonzero annihilating polynomial, choose the monic one of least degree and call it $m_T$. Polynomial division proves

Indeed, write $p=qm_T+r$ with $\deg r<\deg m_T$. If $p(T)=0$, then $r(T)=0$, and minimality forces $r=0$. Uniqueness follows from the same argument. In particular $m_T\mid\chi_T$. A constant nonzero polynomial cannot annihilate a nonzero space, so $m_T$ has positive degree.

Two-factor decomposition. Suppose $f,g\in F[t]$ are relatively prime and $f(T)g(T)=0$. Then

Proof. Polynomial Euclid gives $af+bg=1$. For any $v$,

The first term lies in $\ker g(T)$ and the second in $\ker f(T)$, because polynomial operators commute and $f(T)g(T)=0$. If $v$ lies in both kernels, the same identity gives $v=0$. ∎

Applying this repeatedly to the factorization

into distinct monic irreducibles gives the primary decomposition

Each summand is invariant. Over a field where $m_T$ splits into linear factors, these are generalized eigenspaces. Over $\mathbb R$, an irreducible quadratic may remain and must not be relabelled as a real eigenspace. The real rotation matrix with $m=t^2+1$ illustrates this obstruction.

Since $m_T\mid\chi_T$ and $\chi_T$ splits over $F$, write $m_T=\prod_\lambda(t-\lambda)^{s_\lambda}$. The displayed primary decomposition gives $V=\bigoplus_\lambda\ker(T-\lambda I)^{s_\lambda}$. Each summand is invariant and the restriction of $T-\lambda I$ to it is nilpotent. We now prove the required chain-basis theorem by induction on the image subspace.

An operator $N$ is nilpotent if $N^s=0$ for some positive integer $s$. Its least such exponent, on a nonzero space, is its nilpotency index.

Theorem. A nilpotent operator on a finite-dimensional space has a basis consisting of chains.

Proof by induction on $\dim V$. The zero space has the empty basis. Let $V\ne0$ and set $W=NV$. Nilpotency implies $W$ is a proper subspace: if $N$ were surjective then every power would be surjective, contradicting a zero power on nonzero $V$. The restriction of $N$ to $W$ is nilpotent, so by induction choose its chain basis

The case $W=0$ has no such chains, so $k=0$. The top vectors $w_{i,r_i}$ form a basis of $W/NW$ modulo $NW$: in a chain basis, $NW$ is spanned by all entries except each chain's top.

Every top vector lies in $W=NV$, so choose $v_i\in V$ with $Nv_i=w_{i,r_i}$. Then the extended chains $w_{i,1},\ldots,w_{i,r_i},v_i$ are jointly independent. Indeed, if $\sum_i a_iv_i+w=0$ with $w\in W$, apply $N$ to get $\sum_i a_iw_{i,r_i}+Nw=0$. Their quotient-basis property makes every $a_i=0$, after which independence of the basis of $W$ makes $w=0$.

Let $E$ be the span of the extended chains. We have $W\subseteq E$, $NE=W$, and $\dim E=\dim W+k$, so rank–nullity on $N|_E$ gives $\dim(E\cap\ker N)=k$. Choose a basis of $\ker N$ extending a basis of this intersection, and append its extra vectors to the extended chains as chains of length one. They are independent of $E$ by the definition of a complement inside $\ker N$. The total number of vectors is

by rank–nullity for $N$ on $V$. Thus they form the required full chain basis. ∎

This proof also shows why arbitrarily choosing bases of successive kernels is insufficient: chain tops must be liftable and compatible. A chain construction is more than knowing the kernel dimensions.

Apply this theorem on every primary summand and add $\lambda I$ back to each nilpotent restriction. Combining the chain bases of the direct summands proves Jordan existence on all of $V$. This is an image-induction existence proof; the later kernel-filtration chapter proves the distinct construction using compatible quotient representatives.

For uniqueness and counts, a $\lambda$-block of length $r$ contributes $\min(k,r)$ to $d_k$. On a block for $\mu\ne\lambda$, put $a=\mu-\lambda\ne0$ and let $R^r=0$ be its nilpotent part. The finite identity

is verified by multiplying: all interior powers cancel and the remaining $R^r$ term is zero. Such a block contributes no kernel. Thus $d_k=\sum_r b_r\min(k,r)$, where $b_r$ is the number of $\lambda$-blocks of size $r$. Taking one difference gives $d_k-d_{k-1}=\sum_{r\ge k}b_r$; taking a second gives $2d_k-d_{k-1}-d_{k+1}=b_k$. These intrinsic nullities determine all block sizes, proving uniqueness up to order.