# 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.

## Examples

### 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 $\leq n$ . 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.

## Characterization

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 $\det \left(xI-N\right)=x^{n}$ .
• The minimal polynomial for $N$  is $x^{k}$  for some positive integer $k\leq n$ .
• The only complex eigenvalue for $N$  is 0.
• tr(Nk) = 0 for all $k>0$ .

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.

## Classification

Consider the $n\times n$  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 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 × 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 $\mathbb {R} ^{n}$  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}

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.

• If $N$  is nilpotent, then $I+N$  and $I-N$  are invertible, where $I$  is the $n\times n$  identity matrix. The inverses are given by
{\begin{aligned}(I+N)^{-1}&=\displaystyle \sum _{m=0}^{\infty }\left(-N\right)^{m}=I-N+N^{2}-N^{3}+N^{4}-N^{5}+N^{6}-N^{7}+\cdots ,\\(I-N)^{-1}&=\displaystyle \sum _{m=0}^{\infty }N^{m}=I+N+N^{2}+N^{3}+N^{4}+N^{5}+N^{6}+N^{7}+\cdots \\\end{aligned}}

As long as $N$  is nilpotent, both sums converge, as only finitely many terms are nonzero.

• If $N$  is nilpotent, then
$\det(I+N)=1,\!\,$
where $I$  denotes the $n\times n$  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 $k\in \mathbb {N}$  such that

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

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