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Nilpotent matrix

by , PhD

A square matrix is said to be nilpotent if, by rasing it to a sufficiently high integer power, we get the zero matrix as a result.

Table of Contents

Definition

We start with a definition.

Definition Let A be a $K	imes K$ matrix. We say that A is nilpotent of index k if and only if[eq1]and[eq2]for [eq3].

The index k is often called the index of nilpotency (or nilpotency index) of the matrix.

Null spaces

Let $S$ be the space of all Kx1 vectors.

Remember that the null space of $A^{j}$ is[eq4]and, as we demonstrated in the lecture on matrix powers, the larger the power of a matrix is, the larger its null space is.

Proposition Let $S$ be the space of all Kx1 vectors. Let A be a $K	imes K$ matrix. Then, A is nilpotent of index k if and only if[eq5]and[eq6]where $subset $ denotes strict inclusion.

Proof

Let us prove the "only if" part, starting from the assumption that A is nilpotent of index k. Then, for any $sin S$,[eq7]which implies that [eq8]. Moreover, [eq9]which implies that at least one column of $A^{k-1}$ is different from zero. Suppose it is the $l$-th column. Let $e_{l}in S$ be the vector of the canonical basis such that its $l$-th entry is equal to 1 and all its other entries are equal to 0. Then, $A^{k-1}e_{l}
eq 0$, being equal to the $l$-th column of $A^{k-1}$. As a consequence, [eq10] and [eq11]. Let us now prove the "if" part, starting from the assumption that [eq12] and [eq11]. For any $sin S$, we have[eq14]and, in particular,[eq15]for every vector of the canonical basis $e_{l}$, which implies that all the columns of $A^{k}$ are equal to zero. Hence, $A^{k}=0$. Since [eq16], there exists $sin S$ such that[eq17]which is possible only if at least one column of $A^{k-1}$ is different from zero. Hence, $A^{k-1}
eq 0$.

Range

An important fact is that the range (or column space) of a nilpotent matrix cannot coincide with $S$.

Proposition Let $S$ be the space of all Kx1 vectors. Let A be a $K	imes K$ matrix and denote its range by [eq18]. Then, A is nilpotent only if[eq19]where $subset $ denotes strict inclusion.

Proof

If [eq20], then A is full-rank and $A^{k}$ is full-rank (because the product of two full-rank matrices is full-rank). Hence, $A^{k}
eq 0$ for any k.

Eigenvalues

The following proposition characterizes nilpotent matrices in terms of their eigenvalues.

Proposition A $K	imes K$ matrix A is nilpotent if and only if all its eigenvalues are equal to zero.

Proof

Let us prove the "only if" part, starting from the hypothesis that A is nilpotent of index k. Let $lambda $ be an eigenvalue of A with associated eigenvector $x
eq 0$, that is,[eq21]By pre-multiplying both sides by $A^{k-1}$, we get[eq22]Since $A^{k}=0$, the previous equation becomes[eq23]But $x
eq 0$, which implies $lambda =0$. Since $lambda $ was an arbitrary eigenvalue, all the eigenvalues of A must be equal to zero. Let us prove the "if" part, starting from the hypothesis that all the eigenvalues of A are zero. The proof is by contradiction. Suppose that A has an eigenvalue $lambda 
eq 0$. Then, we can construct a Schur decomposition[eq24]in which $U$ is a unitary matrix and $T$ is an upper triangular matrix such that [eq25]. In other words, the first column of $T$ is[eq26]The matrix $Q^{st }$ is unitary, hence invertible. As a consequence, we can find a vector x such that[eq27]where $e_{1}$ is the first vector of the canonical basis (its first entry is equal to 1 and all its other entries are zero). Then,[eq28]which implies $A^{k}
eq 0$, a contradiction. Hence, A cannot have any non-zero eigenvalue.

In other words, $K	imes K$ nilpotent matrices are those that have a single eigenvalue equal to zero, with algebraic multiplicity equal to K.

Geometric multiplicity of eigenvalues

We now prove a simple fact about geometric multiplicity.

Proposition Let A be a $K	imes K$ nilpotent matrix. Then, A is non-defective (i.e., the geometric multiplicity of the eigenvalue $lambda =0$ is K) if and only if $A=0$.

Proof

Let $S$ be the space of all Kx1 vectors. Let us prove the "only if" part, starting from the hypothesis that A in non-defective. Then, the eigenvectors of A corresponding to the eigenvalue $lambda =0$ span $S$. In other words, $S$ is the eigenspace of $lambda $. Then, for any $sin S$, we have[eq29]which implies $A=0$. Let us now prove the "if" part, starting from the hypothesis that $A=0$. Then, for any $sin S$, we have[eq30]which implies that the eigenspace of $lambda =0$ is $S$. Hence, A is non-defective.

Since a matrix is diagonalizable if and only if it is non-defective, the previous proposition implies that the only nilpotent diagonalizable matrix is the zero matrix.

Minimal polynomial

Remember that the minimal polynomial p of a matrix A is the lowest-degree monic polynomial such that [eq31].

The minimal polynomial of a matrix having distinct eigenvalues [eq32] with algebraic multiplicities [eq33] can be written as[eq34]where [eq35] for $j=1,ldots ,m$.

Thus, when A is nilpotent, its minimal polynomial becomes[eq36]where $1leq 
u leq K$.

Proposition Let A be a $K	imes K$ matrix. Then, A is nilpotent of index k if and only if the minimal polynomial of A is[eq37]

Proof

Let us prove the "only if" part, starting from the assumption that A is nilpotent of index k. Then, $A^{k}=0$ and $A^{
u }
eq 0$ if $
u <k$. This implies that the exponent $
u $ of the minimal polynomial must be equal to k (if it was larger, the minimal polynomial would not be the lowest-degree monic annihilating polynomial). Let us now prove the "if" part, starting from the assumption that the minimal polynomial is [eq38]. Then, since the minimal polynomial is annihilating, [eq39]Moreover,[eq9]otherwise p would not be minimal.

Wow moment

With the previous proposition, although extremely simple and intuitive, we are at a wow moment!

Three different concepts we previously defined turn out to be the same thing when we confine our attention to nilpotent matrices:

  1. the nilpotency index;

  2. the exponent of the linear factor in the minimal polynomial;

  3. the index of the matrix (i.e., the smallest power after which null spaces stop growing).

We highly recommend revising the lecture on the minimal polynomial while having the previous proposition in mind.

Nilpotent operator

The concept of a nilpotent matrix can be generalized to that of a nilpotent operator.

Definition Let $S$ be a vector space. Let $f:S
ightarrow S$ be a linear operator. We say that $f$ is nilpotent of index k if and only if, for any $sin S$,[eq41]and there exists a non-zero $sin S$ such that[eq42]

In other words, a linear operator $f$ is nilpotent of index k if and only if[eq43]and[eq44]where $subset $ denotes strict inclusion and $QTR{rm}{null}$ denotes the null space (or kernel) of the operators.

When $S$ is finite-dimensional, we can as usual represent:

By the proposition above on null spaces, $f$ is a nilpotent mapping if and only if [eq48] is a nilpotent matrix for any basis $B$.

We end this section with a simple but useful fact.

Proposition Let $S$ be a vector space and $f:S
ightarrow S$ a nilpotent operator. If [eq49], then $f$ is the zero mapping.

Proof

Choose a basis $B$, which is made up of a single vector since [eq50]. Therefore, the matrix of the operator [eq51] is a scalar. Suppose that the operator is nilpotent of index k. Then, the condition[eq52]can be true only if [eq53].

Application to generalized eigenspaces

Let us consider again the space $S$ of all Kx1 vectors.

Let A be a $K	imes K$ matrix, $lambda $ one of its eigenvalues, k a strictly positive integer and $N_{lambda ,k}$ a generalized eigenspace associated to $lambda $:[eq54]

We can consider the restricted linear operator [eq55] defined, for any [eq56], by[eq57]

For the definition to make sense, we need to check that the range of $f_{lambda ,k}$ is included in $N_{lambda ,k}$, that is, [eq58] whenever [eq56].

But [eq56] implies that[eq61]

If we multiply both sides of the last equation by [eq62], we get[eq63]which implies that [eq64]. Thus, $f_{lambda ,k}$ is indeed an operator because we can consider its codomain to be equal to its domain.

Moreover, $f_{lambda ,k}$ is a nilpotent operator: by the very definition of the domain $N_{lambda ,k}$, we have[eq65]for every [eq56]. Note that $f_{lambda ,k}$ is not necessarily idempotent of index k, as there might be a smaller integer $j$ such that[eq67]for every $sin S$.

How to cite

Please cite as:

Taboga, Marco (2017). "Nilpotent matrix", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/nilpotent-matrix.

The book

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