Nilpotent Operators and Nilpotency Index

Nilpotency: When Repeated Application Reaches Zero

An operator can fail to be zero while still becoming zero after it is applied several times. For instance, it may send one vector to another, then send that second vector to zero. Nilpotency measures how many repeated applications are needed before every vector has been sent to zero. The key quantity is the least such number of applications—not the dimension of the space.

Let \(V\) be a vector space over \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(T:V\to V\) be a linear operator. For a nonnegative integer \(k\), \(T^k\) denotes the \(k\)-fold composition of \(T\), with \(T^0=\operatorname{id}_V\), the identity map on \(V\). Thus \(T^2=T\circ T\) and \(T^3=T\circ T\circ T\). The zero operator is the map sending every vector to \(\mathbf{0}\).

The word “least” matters: if \(T^m\) is zero, then \(T^{m+1}=T\circ T^m\) is also zero, as are all higher powers. The index records the first power at which zero is reached. If \(T\) is represented by a matrix \(\mathbf{A}\) in a basis, then the corresponding matrix powers are used to calculate this index.

Calculating the Index from Matrix Powers

Suppose \(V\) is finite-dimensional and \(\mathcal{B}\) is an ordered basis. Write \([T]_{\mathcal{B}\leftarrow\mathcal{B}}=\mathbf{A}\) for the matrix representing \(T\) in that basis. The entries of \(\mathbf{A}\) depend on the chosen basis, but whether a power of \(T\) is zero does not.

Proof. For \(k=0\), the matrix of \(T^0=\operatorname{id}_V\) is the identity matrix. For \(k=1\), the claim is the definition of \(\mathbf{A}\). Suppose the matrix of \(T^k\) is \(\mathbf{A}^k\). The matrix of a composition of two linear maps represented in the same bases is the product of their matrices, so the matrix of \(T^{k+1}=T\circ T^k\) is \(\mathbf{A}\mathbf{A}^k=\mathbf{A}^{k+1}\). Induction proves the claim for every \(k\). A map is zero exactly when its matrix in a basis is the zero matrix: its columns are the coordinate vectors of the images of the basis vectors, and all those images are zero exactly when all columns are zero. Hence \(T^k\) is zero exactly when \(\mathbf{A}^k\) is zero. The least positive \(k\) with this property is the same in either description. \(\square\)

To find the index in practice, compute successive powers until the first zero matrix appears. It is not enough to find a zero power without checking the smaller positive powers: the index is the first such exponent. The following examples show how the dimension and index can coincide, differ, or fail to describe the same quantity at all.

A Dimension Bound for the Index

Although the index need not equal the dimension, it cannot be arbitrarily large on a nonzero finite-dimensional space. Each additional nonzero power requires room for a longer list of independent vectors. The next result makes that statement precise without requiring a special matrix form.

Proof. Since \(V\ne\{\mathbf{0}\}\), choose a nonzero vector \(x\) if \(m=1\). If \(m>1\), minimality of \(m\) means \(T^{m-1}\) is not the zero operator, so choose \(x\in V\) with \(T^{m-1}x\ne\mathbf{0}\). In either case, \(T^{m-1}x\ne\mathbf{0}\). We show that the list \(x,Tx,\ldots,T^{m-1}x\) is linearly independent. Suppose \(\sum_{j=0}^{m-1}c_jT^jx=\mathbf{0}\), with scalars \(c_j\in\mathbb{F}\), and suppose some coefficient is nonzero. Let \(r\) be the least index with \(c_r\ne0\). Apply \(T^{m-1-r}\) to the equation. Every term with \(j<r\) has coefficient zero. The term with \(j=r\) becomes \(c_rT^{m-1}x\). Every term with \(j>r\) contains a power \(T^{m-1-r+j}\) with exponent at least \(m\), so it is zero because \(T^m\) is zero. We obtain \(c_rT^{m-1}x=\mathbf{0}\). This is impossible: \(c_r\ne0\), and \(T^{m-1}x\ne\mathbf{0}\). Thus the list is independent. A finite-dimensional space of dimension \(\dim V\) cannot contain more than \(\dim V\) linearly independent vectors, so \(m\leq\dim V\). \(\square\)

The hypothesis that \(V\) is nonzero is needed for the stated bound. On the zero space, the unique operator is zero and has index \(1\), while the dimension is \(0\). The usual applications of the bound concern nonzero spaces.

Index and Dimension Are Different Measurements

The dimension counts how many vectors a basis contains. The nilpotency index counts the first power of the operator that vanishes. These quantities can be equal, as in the first example, or different, as in the next two. The dimension bound only says that on a nonzero finite-dimensional space the index is no larger than the dimension; it does not say they must match.

Nor does every operator have a nilpotency index. For example, the identity operator on a nonzero space has \(T^k=\operatorname{id}_V\) for every positive \(k\), so no power is zero. More generally, when testing a matrix, seeing that a few powers remain nonzero does not by itself prove that every power remains nonzero. A non-nilpotence argument needs a reason that no positive power can vanish.

Common Pitfall: Replacing the Index with the Dimension

A common mistake is to infer the nilpotency index from the size of a square matrix. For the \(4\times4\) matrix \(\mathbf{B}\) in the second example, that reasoning would give index \(4\). It is wrong: the calculation shows \(\mathbf{B}^2=\mathbf{O}\) and \(\mathbf{B}\ne\mathbf{O}\), so its index is \(2\), not \(4\). The correction is to compute successive powers and identify the first one that is zero. The dimension gives an upper bound for the index on a nonzero finite-dimensional space, not its value.

A related error is to call an operator’s index \(0\) because the operator is already zero. Under the definition here, the index is the least positive integer \(m\), so the zero operator has index \(1\). Keep the two questions separate: the dimension asks how large the space is, while the index asks how many iterations are needed to make the operator vanish.