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 called the index of $$N$$, sometimes the degree of $$N$$.

More generally, a nilpotent transformation is a linear transformation $$L$$ of a vector space such that $$L^k = 0$$ for some positive integer $$k$$ (and thus, $$L^j = 0$$ for all $$j \geq k$$). Both of these concepts are special cases of a more general concept of nilpotence that applies to elements of rings.

Example 1
The matrix

A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$ is nilpotent with index 2, since $$A^2 = 0$$.

Example 2
More generally, any $$n$$-dimensional triangular matrix with zeros along the main diagonal is nilpotent, with index $$\le n$$. For example, the matrix

B=\begin{bmatrix} 0 & 2 & 1 & 6\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 3\\ 0 & 0 & 0 & 0 \end{bmatrix} $$ is nilpotent, with



B^2=\begin{bmatrix} 0 & 0 & 2 & 7\\ 0 & 0 & 0 & 3\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}

B^3=\begin{bmatrix} 0 & 0 & 0 & 6\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}

B^4=\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} $$

The index of $$B$$ is therefore 4.

Example 3
Although the examples above have a large number of zero entries, a typical nilpotent matrix does not. For example,

C=\begin{bmatrix} 5 & -3 & 2 \\ 15 & -9 & 6 \\ 10 & -6 & 4 \end{bmatrix} \qquad C^2=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ although the matrix has no zero entries.

Example 4
Additionally, any matrices of the form



\begin{bmatrix} a_1 & a_1 & \cdots & a_1 \\ a_2 & a_2 & \cdots & a_2 \\ \vdots & \vdots & \ddots & \vdots \\ -a_1-a_2-\ldots-a_{n-1} & -a_1-a_2-\ldots-a_{n-1} & \ldots & -a_1-a_2-\ldots-a_{n-1} \end{bmatrix}$$

such as



\begin{bmatrix} 5 & 5 & 5 \\ 6 & 6 & 6 \\ -11 & -11 & -11 \end{bmatrix} $$

or


 * $$\begin{bmatrix}

1 & 1 & 1 & 1 \\ 2 & 2 & 2 & 2 \\ 4 & 4 & 4 & 4 \\ -7 & -7 & -7 & -7 \end{bmatrix} $$

square to zero.

Example 5
Perhaps some of the most striking examples of nilpotent matrices are $$n\times n$$ square matrices of the form:


 * $$\begin{bmatrix}

2 & 2 & 2 & \cdots & 1-n \\ n+2 & 1 & 1 & \cdots & -n \\ 1 & n+2 & 1 & \cdots & -n \\ 1 & 1 & n+2 & \cdots & -n \\ \vdots & \vdots & \vdots & \ddots & \vdots \end{bmatrix}$$

The first few of which are:


 * $$\begin{bmatrix}

2 & -1 \\ 4 & -2 \end{bmatrix} \qquad \begin{bmatrix} 2 & 2 & -2 \\ 5 & 1 & -3 \\ 1 & 5 & -3 \end{bmatrix} \qquad \begin{bmatrix} 2 & 2 & 2 & -3 \\ 6 & 1 & 1 & -4 \\ 1 & 6 & 1 & -4 \\ 1 & 1 & 6 & -4 \end{bmatrix} \qquad \begin{bmatrix} 2 & 2 & 2 & 2 & -4 \\ 7 & 1 & 1 & 1 & -5 \\ 1 & 7 & 1 & 1 & -5 \\ 1 & 1 & 7 & 1 & -5 \\ 1 & 1 & 1 & 7 & -5 \end{bmatrix} \qquad \ldots $$

These matrices are nilpotent but there are no zero entries in any powers of them less than the index.

Example 6
Consider the linear space of polynomials of a bounded degree. The derivative operator is a linear map. We know that applying the derivative to a polynomial decreases its degree by one, so when applying it iteratively, we will eventually obtain zero. Therefore, on such a space, the derivative is representable by a nilpotent matrix.

Characterization
For an $$n \times n$$ 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)
 * $$N$$ is nilpotent.
 * The characteristic polynomial for $$N$$ is $$\det \left(xI - N\right) = x^n$$.
 * The minimal polynomial for $$N$$ is $$x^k$$ for some positive integer $$k \leq n$$.
 * The only complex eigenvalue for $$N$$ is 0.

This theorem has several consequences, including:
 * The index of an $$n \times n$$ nilpotent matrix is always less than or equal to $$n$$. For example, every $$2 \times 2$$ 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.

See also: Jordan–Chevalley decomposition.

Classification
Consider the $$n \times n$$ (upper) 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}.$$ This matrix has 1s along the superdiagonal and 0s everywhere else. As a linear transformation, the shift matrix "shifts" the components of a vector one position to the left, with a zero appearing in the last position:
 * $$S(x_1,x_2,\ldots,x_n) = (x_2,\ldots,x_n,0).$$

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} $$ where each of the blocks $$S_1,S_2,\ldots,S_r$$ is a shift matrix (possibly of different sizes). This form is a special case of the Jordan canonical form for matrices.

For example, any nonzero 2 &times; 2 nilpotent matrix is similar to the matrix
 * $$ \begin{bmatrix}

0 & 1 \\  0 & 0 \end{bmatrix}. $$ That is, if $$N$$ is any nonzero 2 &times; 2 nilpotent matrix, then there exists a basis b1, b2 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 $$\mathbb{R}^n$$ 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.

Generalizations
A linear operator $$T$$ is locally nilpotent if for every vector $$v$$, there exists a $$k\in\mathbb{N}$$ such that
 * $$T^k(v) = 0.\!\,$$

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