Nilpotent matrix

Template:Short description 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 $j\ge k$). Both of these concepts are special cases of a more general concept of nilpotence that applies to elements of rings.

Examples

Example 1

The matrix

$A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

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 $\le n$ Template:Citation needed. For example, the matrix

$B=\left[\begin{array}{cccc}0& 2& 1& 6\\ 0& 0& 1& 2\\ 0& 0& 0& 3\\ 0& 0& 0& 0\end{array}\right]$

is nilpotent, with

$B^2=\left[\begin{array}{cccc}0& 0& 2& 7\\ 0& 0& 0& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right];B^3=\left[\begin{array}{cccc}0& 0& 0& 6\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right];B^4=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

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,

$C=\left[\begin{array}{ccc}5& -3& 2\\ 15& -9& 6\\ 10& -6& 4\end{array}\right]C^2=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

although the matrix has no zero entries.

Example 4

Additionally, any matrices of the form

$\left[\begin{array}{cccc}{a}_1& a_1& \cdots & a_1\\ a_2& a_2& \cdots & a_2\\ ⋮& ⋮& \ddots & ⋮\\ -a_1-a_2-\dots -a_{n-1}& -a_1-a_2-\dots -a_{n-1}& \dots & -a_1-a_2-\dots -a_{n-1}\end{array}\right]$

such as

$\left[\begin{array}{ccc}5& 5& 5\\ 6& 6& 6\\ -11& -11& -11\end{array}\right]$

or

$\left[\begin{array}{cccc}1& 1& 1& 1\\ 2& 2& 2& 2\\ 4& 4& 4& 4\\ -7& -7& -7& -7\end{array}\right]$

square to zero.

Example 5

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

$\left[\begin{array}{ccccc}2& 2& 2& \cdots & 1-n\\ n+2& 1& 1& \cdots & -n\\ 1& n+2& 1& \cdots & -n\\ 1& 1& n+2& \cdots & -n\\ ⋮& ⋮& ⋮& \ddots & ⋮\end{array}\right]$

The first few of which are:

$\left[\begin{array}{cc}2& -1\\ 4& -2\end{array}\right]\left[\begin{array}{ccc}2& 2& -2\\ 5& 1& -3\\ 1& 5& -3\end{array}\right]\left[\begin{array}{cccc}2& 2& 2& -3\\ 6& 1& 1& -4\\ 1& 6& 1& -4\\ 1& 1& 6& -4\end{array}\right]\left[\begin{array}{ccccc}2& 2& 2& 2& -4\\ 7& 1& 1& 1& -5\\ 1& 7& 1& 1& -5\\ 1& 1& 7& 1& -5\\ 1& 1& 1& 7& -5\end{array}\right]\dots$

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

Template:Unreferenced section For an $n\times n$ square matrix $N$ with real (or complex) entries, the following are equivalent:

• $N$ is nilpotent.
• The characteristic polynomial for $N$ is $\det (xI-N)=x^n$.
• The minimal polynomial for $N$ is $x^k$ for some positive integer $k\le n$.
• The only complex eigenvalue for $N$ is 0.

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 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#Nilpotency criterion.

Classification

Consider the $n\times n$ (upper) shift matrix:

$S=\left[\begin{array}{ccccc}0& 1& 0& \dots & 0\\ 0& 0& 1& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & 1\\ 0& 0& 0& \dots & 0\end{array}\right].$

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:

$S(x_1,x_2,\dots ,x_n)=(x_2,\dots ,x_n,0).$

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

$\left[\begin{array}{cccc}{S}_1& 0& \dots & 0\\ 0& S_2& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & S_r\end{array}\right]$

where each of the blocks $S_1,S_2,\dots ,S_r$ 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 × 2 nilpotent matrix is similar to the matrix

$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right].$

That is, if $N$ is any nonzero 2 × 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 ${ℝ}^n$ naturally determines a flag of subspaces

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

and a signature

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

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

$n_{j+1}-n_j\le n_j-n_{j-1},\text{for all }j=1,\dots ,q-1.$

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.

References

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External links

• Nilpotent matrix and nilpotent transformation on PlanetMath.