The Chain-Basis Proof for Nilpotent Operators

From Chains to a Basis

For a nilpotent operator, constructing individual Jordan chains is not enough: a basis must contain enough chains to span the whole space, and their vectors must not overlap in a way that creates linear dependence. The kernel filtration provides a systematic way to choose chain generators while controlling both issues. We will work over \(\mathbb{R}\); the same proof works over \(\mathbb{C}\).

Let \(V\) be a finite-dimensional vector space and \(T:V\to V\) a linear operator. Recall that \(T\) is nilpotent if \(T^m=\mathbf{0}\) for some positive integer \(m\), where \(T^m\) means \(m\) repeated compositions of \(T\). The least such positive integer is the nilpotency index. Define \(K_j=\ker T^j\), and set \(K_0=\{\mathbf{0}\}\). If \(m\) is the nilpotency index, then \(K_m=V\); for convenience, set \(K_{m+1}=V\) as well.

The order in this definition runs from the generator down to the vector killed by \(T\). The central choice is which vectors should serve as generators of chains of each length. A vector in \(K_j\) is killed by \(T^j\), but it may already be in \(K_{j-1}\), in which case it cannot generate a chain of length \(j\). Also, some vectors in \(K_j\) are obtained by applying \(T\) to vectors in \(K_{j+1}\); these belong naturally to chains that continue higher up the filtration.

Selecting the Generators

For each \(j=1,\dots,m\), consider the subspace \(K_{j-1}+T(K_{j+1})\) inside \(K_j\). The inclusion holds because \(T(K_{j+1})\subseteq K_j\). Choose a subspace \(H_j\) such that

Such a complement exists in a finite-dimensional vector space: extend a basis of \(K_{j-1}+T(K_{j+1})\) to a basis of \(K_j\), and let \(H_j\) be the span of the added basis vectors. Thus \(H_j\) represents the part of \(K_j\) not already accounted for by the lower kernel or by images from the next kernel. Choose a basis of each \(H_j\); each basis vector will generate one chain of length \(j\).

Why These Chains Have the Right Number of Vectors

We first calculate the number of generators selected at each length. Write \(d_j=\dim K_j\) for \(0\le j\le m+1\). In particular, \(d_0=0\) and \(d_{m+1}=d_m=\dim V\). Let \(q_j=\dim H_j\). The direct-sum choice gives \(q_j=\dim K_j-\dim(K_{j-1}+T(K_{j+1}))\).

Lemma. For each \(j=1,\dots,m\), \(q_j=2d_j-d_{j-1}-d_{j+1}\).

Proof. We first show \(K_{j-1}\cap T(K_{j+1})=T(K_j)\). If \(\mathbf{x}\) belongs to the left side, then \(\mathbf{x}=T\mathbf{y}\) for some \(\mathbf{y}\in K_{j+1}\), and \(\mathbf{x}\in K_{j-1}\). It follows that \(T^j\mathbf{y}=T^{j-1}\mathbf{x}=\mathbf{0}\), so \(\mathbf{y}\in K_j\), giving \(\mathbf{x}\in T(K_j)\). Conversely, \(T(K_j)\subseteq K_{j-1}\), and \(T(K_j)\subseteq T(K_{j+1})\), proving the equality. By the dimension formula for a sum of subspaces, \(\dim(K_{j-1}+T(K_{j+1}))=d_{j-1}+\dim T(K_{j+1})-\dim T(K_j)\). By the Rank-Nullity Theorem applied to \(T\) restricted to \(K_r\), whose kernel is \(K_1\) for \(r\ge1\), \(\dim T(K_r)=d_r-d_1\). Substitution yields \(\dim(K_{j-1}+T(K_{j+1}))=d_{j-1}+d_{j+1}-d_j\), including \(j=m\) because \(K_{m+1}=K_m\). Subtracting this from \(d_j\) proves the formula. \(\square\)

Consequently, the total number of vectors in the selected chains is \(\sum_{j=1}^m j q_j\). Substituting the lemma gives a telescoping sum:

For the last equality in the telescoping calculation, the coefficients of \(d_1,\dots,d_{m-1}\) cancel, and the coefficient of \(d_m\) is \(1\), using \(d_{m+1}=d_m\). This count alone does not prove that the chains form a basis; we must also establish that they span.

Why the Chains Span

Fix \(j\) with \(1\le j\le m\). The decomposition defining \(H_j\) says that, modulo \(K_{j-1}\), every vector in \(K_j\) is a sum of a vector in \(H_j\) and a vector in \(T(K_{j+1})\). To continue this statement upward, take any \(\mathbf{x}\in K_j\). Write \(\mathbf{x}=\mathbf{u}_j+T\mathbf{y}_{j+1}+\mathbf{z}_{j-1}\), with \(\mathbf{u}_j\in H_j\), \(\mathbf{y}_{j+1}\in K_{j+1}\), and \(\mathbf{z}_{j-1}\in K_{j-1}\). Decompose \(\mathbf{y}_{j+1}\) using the definition of \(H_{j+1}\). Its \(K_j\) part becomes, after applying \(T\), a vector in \(T(K_j)\subseteq K_{j-1}\); its remaining parts give a vector in \(T(H_{j+1})\) and a vector in \(T^2(K_{j+2})\). Repeating this step gives, modulo \(K_{j-1}\), a sum of vectors from \(H_j,T(H_{j+1}),\dots,T^{m-j}(H_m)\). The final remainder lies in \(T^{m-j+1}(K_{m+1})=T^{m-j+1}(K_m)\), which is contained in \(K_{j-1}\) because applying \(T^{j-1}\) to it gives a vector in \(T^m(V)=\{\mathbf{0}\}\).

Thus each layer \(K_j/K_{j-1}\) is spanned by the chain vectors \(T^{s-j}\mathbf{h}\), where \(s\ge j\) and \(\mathbf{h}\) ranges over the chosen bases of \(H_s\). Beginning with \(K_0=\{\mathbf{0}\}\) and applying this conclusion successively for \(j=1,\dots,m\), we obtain that the full collection of chain vectors spans \(K_m=V\).

Why the Spanning List Is Independent

It remains to check that these are genuine chains and that the spanning list has no redundancy. If \(\mathbf{h}\ne\mathbf{0}\) belongs to \(H_j\), then \(\mathbf{h}\notin K_{j-1}\): otherwise it would belong to \(H_j\cap(K_{j-1}+T(K_{j+1}))=\{\mathbf{0}\}\). Therefore \(T^{j-1}\mathbf{h}\ne\mathbf{0}\), while \(T^j\mathbf{h}=\mathbf{0}\), so this is a chain of exactly length \(j\). Since the total number of its vectors, over all chains, is \(\dim V\), the spanning list has exactly \(\dim V\) elements. In a finite-dimensional vector space, a spanning list with \(\dim V\) elements is a basis: if it were dependent, one vector could be removed without changing its span, leaving fewer than \(\dim V\) vectors spanning \(V\), contrary to the definition of dimension. This proves the theorem. \(\square\)

Examples of the Selection

Common Pitfall: Choosing Vectors from a Kernel Arbitrarily

A vector in \(K_j\) is not automatically a suitable generator for a chain of length \(j\). In the first example, choosing \(\mathbf{e}_1\) as a length-three generator would give the proposed list \((\mathbf{e}_1,T\mathbf{e}_1,T^2\mathbf{e}_1)=(\mathbf{e}_1,\mathbf{0},\mathbf{0})\). This is not a length-three Jordan chain: its last vector is zero, and the list cannot span \(\mathbb{R}^3\). The correction is to choose \(\mathbf{e}_3\), which lies in \(K_3\) but not in \(K_2\). The complement condition also prevents counting vectors already supplied by images from longer chains as new generators at shorter lengths.

The proof has two distinct jobs. The quotient choices control how the chain vectors reach every level of the kernel filtration, giving spanning. The dimension count then turns that spanning list into a basis, which proves linear independence without guessing it from the appearance of the chains.