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 $\leq n$ [citation needed]. 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

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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.

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 $n\times n$ (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 × 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 of index $k$ , 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}^{k}\left(-N\right)^{m}=I-N+N^{2}-N^{3}+N^{4}-N^{5}+N^{6}-N^{7}+\cdots +(-N)^{k}\\(I-N)^{-1}&=\displaystyle \sum _{m=0}^{k}N^{m}=I+N+N^{2}+N^{3}+N^{4}+N^{5}+N^{6}+N^{7}+\cdots +N^{k}\\\end{aligned)) • If $N$ is nilpotent, then
$\det(I+N)=1.$ 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 $t$ .
• Every singular matrix can be written as a product of nilpotent matrices.
• A nilpotent matrix is a special case of a convergent matrix.

## 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.

1. ^ Herstein (1975, p. 294)
2. ^ Beauregard & Fraleigh (1973, p. 312)
3. ^ Herstein (1975, p. 268)
4. ^ Nering (1970, p. 274)
5. ^ Mercer, Idris D. (31 October 2005). "Finding "nonobvious" nilpotent matrices" (PDF). idmercer.com. self-published; personal credentials: PhD Mathematics, Simon Fraser University. Retrieved 5 April 2023.
6. ^ Beauregard & Fraleigh (1973, p. 312)
7. ^ Beauregard & Fraleigh (1973, pp. 312, 313)
8. ^ R. Sullivan, Products of nilpotent matrices, Linear and Multilinear Algebra, Vol. 56, No. 3