Nilpotent matrix

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In mathematics, a nilpotent matrix is an n×n square matrix M such that

M^q = 0\,

for some positive integer q. Similarly, a nilpotent transformation is a linear transformation L with Lq = 0 for some integer q.

These are special cases of a more general concept of nilpotence that applies not only to matrices and linear transformations but to members of rings.

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[edit] Examples

Consider the following matrix:

N = \begin{bmatrix}  0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0  \end{bmatrix}.

This is an example of a 4×4 nilpotent matrix (in fact, matrices of this form are called shift matrices). Notice the non-zero superdiagonal. The characteristic feature of this matrix is:

N^2 =   \begin{bmatrix}                      0 & 0 & 1 & 0\\                     0 & 0 & 0 & 1\\                     0 & 0 & 0 & 0\\                     0 & 0 & 0 & 0                   \end{bmatrix}   ;\  N^3 =   \begin{bmatrix}                      0 & 0 & 0 & 1\\                     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}.

The super-diagonal keeps 'shifting' diagonally up, until one gets the null matrix.

The corresponding nilpotent transformation L : R4R4 is defined by:

L(x_1,x_2,x_3,x_4) = (x_2,x_3,x_4,0). \,

There is a classification theorem showing that this is typical: a nilpotent matrix is similar to a block matrix, with diagonal square blocks generalizing this type, and other blocks zero.

[edit] Properties

Let M be an n×n nilpotent matrix.

[edit] Classification theorem

The above example is typical, as the following result shows. Every nilpotent matrix is similar to a block diagonal matrix

\begin{bmatrix}     N_1 & 0 & 0 & \ldots & 0 \\     0 & N_2 & 0 & \ldots & 0 \\    0 & 0 & N_3 & \ldots & 0 \\    \vdots & \vdots & \vdots & \ddots & \vdots \\    0 & 0 & 0 & \ldots & N_k  \end{bmatrix}

where the blocks Ni have ones on the superdiagonal and zeros everywhere else:

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


This fact follows from the Jordan decomposition theorem, together with the result that every matrix similar to a nilpotent matrix is also nilpotent.

[edit] 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 = U

and a signature

0 = n_0 < n_1 < n_2 < \ldots < q_{k-1} < q_k = n,\qquad n_i = \dim \ker N^i.

The signature characterizes L up to a 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.

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