Following the Kernels of Successive Powers
For a linear operator, applying the operator repeatedly can reveal how its action changes the vectors it sends to zero. The kernels of its powers form a sequence of subspaces. When the operator is nilpotent, this sequence must eventually reach the entire space; the finite dimension puts a precise bound on how long that can take. A second fact is just as useful: once two consecutive kernels agree, no later power can enlarge them.
Let \(V\) be a finite-dimensional vector space over \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(N:V\to V\) be linear. The power \(N^k\) means that \(N\) is composed with itself \(k\) times, and \(N^0=\operatorname{id}_V\), the identity map. The kernel of a linear map \(S:V\to V\) is \(\ker S=\{\mathbf{v}\in V:S(\mathbf{v})=\mathbf{0}\}\). For each integer \(k\geq 0\), write \(K_k=\ker N^k\). Each \(K_k\) is a subspace of \(V\).
The zeroth kernel is \(K_0=\{\mathbf{0}\}\), since the identity map sends only the zero vector to zero. For every \(k\geq 0\), if \(N^k(\mathbf{v})=\mathbf{0}\), then \(N^{k+1}(\mathbf{v})=N(\mathbf{0})=\mathbf{0}\). Therefore
This inclusion holds for every linear operator, not just a nilpotent one. Nilpotence adds the fact that some kernel is all of \(V\): if \(N^q\) is the zero map, then every vector is in \(\ker N^q\), so \(K_q=V\).
Equality at One Step Persists
Proof. Put \(K_i=\ker N^i\). The ascending-chain inclusion gives \(K_{k+1}\subseteq K_{k+2}\). To prove the reverse inclusion, take \(\mathbf{x}\in K_{k+2}\). Then \(N^{k+2}(\mathbf{x})=\mathbf{0}\), so \(N(\mathbf{x})\in K_{k+1}\). By the assumed equality, \(K_{k+1}=K_k\), and hence \(N(\mathbf{x})\in K_k\). This says \(N^{k+1}(\mathbf{x})=\mathbf{0}\), or \(\mathbf{x}\in K_{k+1}\). Thus \(K_{k+2}=K_{k+1}=K_k\). Applying the same argument with \(k+1\) in place of \(k\), and then repeating it for each later step, proves \(K_j=K_k\) for every \(j\geq k\). \(\square\)
The argument does not require nilpotence: equality persists for any linear operator. Nilpotence is needed for the further conclusion that the stabilized kernel is the whole space. This distinction matters. A sequence can stabilize at a proper subspace if the operator is not nilpotent.
Worked Example: A Nilpotent Shift on \(\mathbb{R}^3\)
Consider the linear operator \(N:\mathbb{R}^3\to\mathbb{R}^3\) with matrix, in the standard basis, given by
For \(\mathbf{x}=(x,y,z)^{\top}\), multiplication gives \(N(\mathbf{x})=(y,z,0)^{\top}\). Applying \(N\) again gives \(N^2(\mathbf{x})=(z,0,0)^{\top}\), and one more application gives \(N^3(\mathbf{x})=\mathbf{0}\). In matrix form, these calculations are
Now solve for each kernel. The equation \(N(\mathbf{x})=\mathbf{0}\) requires \(y=0\) and \(z=0\), with \(x\) unrestricted. The equation \(N^2(\mathbf{x})=\mathbf{0}\) requires \(z=0\), with \(x,y\) unrestricted. Since \(N^3\) is the zero map, every vector is in its kernel. Thus
Check: \(N(\mathbf{e}_1)=\mathbf{0}\), \(N(\mathbf{e}_2)=\mathbf{e}_1\), and \(N(\mathbf{e}_3)=\mathbf{e}_2\). Consequently \(N^2(\mathbf{e}_3)=\mathbf{e}_1\) and \(N^3(\mathbf{e}_3)=\mathbf{0}\), matching the computed kernels. Each inclusion through \(K_3\) is strict, and \(K_3\) is the whole space.
Why the Dimension Gives a Bound
Proof. If \(n=0\), then \(V=\{\mathbf{0}\}\), so \(K_0=V\) and the claim holds. Now suppose \(n\geq 1\). Nilpotence gives an integer \(q\geq 1\) such that \(K_q=V\). Consider the \(n+1\) inclusions from \(K_0\) through \(K_{n+1}\). If all were strict, each inclusion would increase dimension by at least one: whenever a subspace \(U\) is properly contained in a finite-dimensional subspace \(W\), a vector in \(W\setminus U\) can be added to a basis of \(U\), so \(\dim W\geq\dim U+1\). It would follow that \(\dim K_{n+1}\geq\dim K_0+(n+1)=n+1\), contradicting \(K_{n+1}\subseteq V\) and \(\dim V=n\). Hence \(K_s=K_{s+1}\) for some \(s\) with \(0\leq s\leq n\). By the persistence theorem, all kernels from \(K_s\) onward agree. Since \(K_q=V\), this forces \(K_s=V\): if \(q\geq s\), persistence gives \(K_s=K_q\); if \(q<s\), the ascending chain gives \(V=K_q\subseteq K_s\subseteq V\). Since \(s\leq n\), \(K_n=V\) as well. Finally, \(K_{n+1}=V\) too, so \(K_n=K_{n+1}\). \(\square\)
The dimension argument captures why the bound is \(n\). Starting at the zero subspace, every strict step must add at least one dimension. There can be at most \(n\) such increases before reaching an \(n\)-dimensional space. If an equality occurs sooner, the persistence theorem says the sequence cannot start growing again. For a nilpotent operator, the sequence must eventually be the whole space, so that earlier equality already has to be at the whole space.
Worked Example: Stabilization Before the Dimension Bound
Define \(N:\mathbb{R}^3\to\mathbb{R}^3\) by \(N(x,y,z)=(z,0,0)\). Its matrix is
For any \((x,y,z)^{\top}\), \(N^2(x,y,z)^{\top}=\mathbf{0}\), so \(N^2\) is the zero map. The kernel of \(N\) consists of the vectors with \(z=0\), while \(K_2=\mathbb{R}^3\). Therefore
Check: Matrix multiplication gives \(\mathbf{N}^2=\mathbf{O}\), and \(N(\mathbf{e}_3)=\mathbf{e}_1\ne\mathbf{0}\), so \(\mathbf{e}_3\notin K_1\) but \(\mathbf{e}_3\in K_2\). Thus \(K_1\) is proper in \(K_2\), and equality begins at \(K_2=K_3\), before the dimension bound \(n=3\) is needed.
What Equality Does Not Say Without Nilpotence
Worked Example: Equality at a Proper Kernel
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) have matrix \(\mathbf{T}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\). For \((x,y)^{\top}\), \(T(x,y)^{\top}=(x,0)^{\top}\), and multiplying the matrix by itself gives \(\mathbf{T}^2=\mathbf{T}\). Therefore \(T^k=T\) for every \(k\geq 1\), and
Check: \(T(0,y)^{\top}=(0,0)^{\top}\), but \(T(1,0)^{\top}=(1,0)^{\top}\), so the displayed kernel is correct and is not all of \(\mathbb{R}^2\). Also \(T^k(1,0)^{\top}=(1,0)^{\top}\ne\mathbf{0}\) for every \(k\geq 1\), so \(T\) is not nilpotent. Equality persists, as the theorem predicts, but nilpotence is essential if we want to conclude that the stabilized kernel is the whole space.
Common Pitfall
A tempting but incorrect argument is: “If \(K_k=K_{k+1}\), then \(K_k=V\).” The example with matrix \(\mathbf{T}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\) disproves this: \(K_1=K_2=\operatorname{span}\{\mathbf{e}_2\}\), which is a proper subspace. The correction is to keep the hypotheses separate. Equality always implies that later kernels agree; to conclude that this common kernel is \(V\), also require that the operator is nilpotent. Only then do we know that some later kernel equals \(V\), forcing the stabilized kernel to equal \(V\) as well.
It is also important to count from \(K_0=\ker\operatorname{id}_V=\{\mathbf{0}\}\). If \(V\) has dimension \(n\), there are only \(n\) possible strict increases in dimension from \(K_0\) to \(V\). The theorem does not claim that each of \(K_1,\ldots,K_n\) must be different: the second worked example shows that equality can occur earlier. Rather, it guarantees that a nilpotent operator has reached the whole space by \(K_n\), whether by strict growth at every step or by an earlier equality that persists.
Check Your Understanding
Use the definitions and results above to justify each answer.
- For a linear operator \(N\), explain why \(\ker N^k\subseteq\ker N^{k+1}\) for every \(k\geq 0\).
- If \(\ker N^3=\ker N^4\), what can be concluded about \(\ker N^j\) for \(j\geq 3\)? State the result that justifies your answer.
- A nilpotent operator acts on a space of dimension \(5\). What does the stabilization bound say about \(\ker N^5\)?
- For the operator \(N(x,y,z)=(z,0,0)\), find \(K_1\) and \(K_2\), and identify the first equality between consecutive kernels.
- Why does equality of consecutive kernels alone not imply that the common kernel is the entire space?
