# Nilpotent matrix

In linear algebra, a nilpotent matrix is a square matrix N such that

$N^k = 0\,$

for some positive integer k. The smallest such k is sometimes called the degree or index of N.[1]

More generally, a nilpotent transformation is a linear transformation L of a vector space such that Lk = 0 for some positive integer k (and thus, Lj = 0 for all jk).[2][3][4] Both of these concepts are special cases of a more general concept of nilpotence that applies to elements of rings.

## Examples

The matrix

$M = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

is nilpotent, since M2 = 0. More generally, any triangular matrix with 0s along the main diagonal is nilpotent. For example, the matrix

$N = \begin{bmatrix} 0 & 2 & 1 & 6\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 3\\ 0 & 0 & 0 & 0 \end{bmatrix}$

is nilpotent, with

$N^2 = \begin{bmatrix} 0 & 0 & 2 & 7\\ 0 & 0 & 0 & 3\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} ;\ N^3 = \begin{bmatrix} 0 & 0 & 0 & 6\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} ;\ N^4 = \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}.$

Though the examples above have a large number of zero entries, a typical nilpotent matrix does not. For example, the matrix

$N = \begin{bmatrix} 5 & -3 & 2 \\ 15 & -9 & 6 \\ 10 & -6 & 4 \end{bmatrix}$

squares to zero, though the matrix has no zero entries.

## Characterization

For an n × n square matrix N with real (or complex) entries, the following are equivalent:

The last theorem holds true for matrices over any field of characteristic 0 or sufficiently large characteristic. (cf. Newton's identities)

This theorem has several consequences, including:

• The degree of an n × n nilpotent matrix is always less than or equal to n. For example, every 2 × 2 nilpotent matrix squares to zero.
• The determinant and trace of a nilpotent matrix are always zero. Consequently, a nilpotent matrix cannot be invertible.
• The only nilpotent diagonalizable matrix is the zero matrix.

## Classification

Consider the n × n shift matrix:

$S = \begin{bmatrix} 0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & 1 \\ 0 & 0 & 0 & \ldots & 0 \end{bmatrix}.$

This matrix has 1s along the superdiagonal and 0s everywhere else. As a linear transformation, the shift matrix “shifts” the components of a vector one slot to the left:

$S(x_1,x_2,\ldots,x_n) = (x_2,\ldots,x_n,0).$[5]

This matrix is nilpotent with degree n, and is the “canonical” nilpotent matrix.

Specifically, if N is any nilpotent matrix, then N is similar to a block diagonal matrix of the form

$\begin{bmatrix} S_1 & 0 & \ldots & 0 \\ 0 & S_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & S_r \end{bmatrix}$

where each of the blocks S1S2, ..., Sr is a shift matrix (possibly of different sizes). This theorem is a special case of the Jordan canonical form for matrices.[6]

For example, any nonzero 2 × 2 nilpotent matrix is similar to the matrix

$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$

That is, if N is any nonzero 2 × 2 nilpotent matrix, then there exists a basis b1b2 such that Nb1 = 0 and Nb2 = b1.

This classification theorem holds for matrices over any field. (It is not necessary for the field to be algebraically closed.)

## Flag of subspaces

A nilpotent transformation L on Rn naturally determines a flag of subspaces

$\{0\} \subset \ker L \subset \ker L^2 \subset \ldots \subset \ker L^{q-1} \subset \ker L^q = \mathbb{R}^n$

and a signature

$0 = n_0 < n_1 < n_2 < \ldots < n_{q-1} < n_q = n,\qquad n_i = \dim \ker L^i.$

The signature characterizes L up to an invertible linear transformation. Furthermore, it satisfies the inequalities

$n_{j+1} - n_j \leq n_j - n_{j-1}, \qquad \mbox{for all } j = 1,\ldots,q-1.$

Conversely, any sequence of natural numbers satisfying these inequalities is the signature of a nilpotent transformation.

$(I + N)^{-1} = I - N + N^2 - N^3 + \cdots,$
where only finitely many terms of this sum are nonzero.
• If N is nilpotent, then
$\det (I + N) = 1,\!\,$
where I denotes the n × n identity matrix. Conversely, if A is a matrix and
$\det (I + tA) = 1\!\,$
for all values of t, then A is nilpotent. In fact, since $p(t) = \det (I + tA) - 1$ is a polynomial of degree $n$, it suffices to have this hold for $n+1$ distinct values of $t$.

## Generalizations

A linear operator T is locally nilpotent if for every vector v, there exists a k such that

$T^k(v) = 0.\!\,$

For operators on a finite-dimensional vector space, local nilpotence is equivalent to nilpotence.

## Notes

1. ^ Herstein (1964, p. 250)
2. ^ Beauregard & Fraleigh (1973, p. 312)
3. ^ Herstein (1964, p. 224)
4. ^ Nering (1970, p. 274)
5. ^ Beauregard & Fraleigh (1973, p. 312)
6. ^ Beauregard & Fraleigh (1973, pp. 312,313)
7. ^ R. Sullivan, Products of nilpotent matrices, Linear and Multilinear Algebra, Vol. 56, No. 3