In linear algebra, a nilpotent matrix is a square matrix N such that

${\displaystyle N^{k}=0\,}$

for some positive integer ${\displaystyle k}$. The smallest such ${\displaystyle k}$ is called the index of ${\displaystyle N}$,^[1] sometimes the degree of ${\displaystyle N}$.

More generally, a nilpotent transformation is a linear transformation ${\displaystyle L}$ of a vector space such that ${\displaystyle L^{k}=0}$ for some positive integer ${\displaystyle k}$ (and thus, ${\displaystyle L^{j}=0}$ for all ${\displaystyle j\geq k}$).^[2][3][4] Both of these concepts are special cases of a more general concept of nilpotence that applies to elements of rings.

The matrix

${\displaystyle A={\begin{bmatrix}0&1\\0&0\end{bmatrix}}}$

is nilpotent with index 2, since ${\displaystyle A^{2}=0}$.

More generally, any ${\displaystyle n}$-dimensional triangular matrix with zeros along the main diagonal is nilpotent, with index ${\displaystyle \leq n}$ ^{[citation needed]}. For example, the matrix

${\displaystyle B={\begin{bmatrix}0&2&1&6\\0&0&1&2\\0&0&0&3\\0&0&0&0\end{bmatrix}}}$

is nilpotent, with

${\displaystyle B^{2}={\begin{bmatrix}0&0&2&7\\0&0&0&3\\0&0&0&0\\0&0&0&0\end{bmatrix}};\ B^{3}={\begin{bmatrix}0&0&0&6\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}};\ B^{4}={\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}}}$

The index of ${\displaystyle B}$ is therefore 4.

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

${\displaystyle C={\begin{bmatrix}5&-3&2\\15&-9&6\\10&-6&4\end{bmatrix}}\qquad C^{2}={\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}}}$

although the matrix has no zero entries.

Additionally, any matrices of the form

${\displaystyle {\begin{bmatrix}a_{1}&a_{1}&\cdots &a_{1}\\a_{2}&a_{2}&\cdots &a_{2}\\\vdots &\vdots &\ddots &\vdots \\-a_{1}-a_{2}-\ldots -a_{n-1}&-a_{1}-a_{2}-\ldots -a_{n-1}&\ldots &-a_{1}-a_{2}-\ldots -a_{n-1}\end{bmatrix}}}$

such as

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

or

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

square to zero.

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

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

The first few of which are:

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

These matrices are nilpotent but there are no zero entries in any powers of them less than the index.^[5]

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.

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For an ${\displaystyle n\times n}$ square matrix ${\displaystyle N}$ with real (or complex) entries, the following are equivalent:

• ${\displaystyle N}$ is nilpotent.
• The characteristic polynomial for ${\displaystyle N}$ is ${\displaystyle \det \left(xI-N\right)=x^{n}}$.
• The minimal polynomial for ${\displaystyle N}$ is ${\displaystyle x^{k}}$ for some positive integer ${\displaystyle k\leq n}$.
• The only complex eigenvalue for ${\displaystyle N}$ 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 ${\displaystyle n\times n}$ nilpotent matrix is always less than or equal to ${\displaystyle n}$. For example, every ${\displaystyle 2\times 2}$ 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.

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

${\displaystyle S={\begin{bmatrix}0&1&0&\ldots &0\\0&0&1&\ldots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\ldots &1\\0&0&0&\ldots &0\end{bmatrix}}.}$

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:

${\displaystyle S(x_{1},x_{2},\ldots ,x_{n})=(x_{2},\ldots ,x_{n},0).}$^[6]

This matrix is nilpotent with degree ${\displaystyle n}$, and is the canonical nilpotent matrix.

Specifically, if ${\displaystyle N}$ is any nilpotent matrix, then ${\displaystyle N}$ is similar to a block diagonal matrix of the form

${\displaystyle {\begin{bmatrix}S_{1}&0&\ldots &0\\0&S_{2}&\ldots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\ldots &S_{r}\end{bmatrix}}}$

where each of the blocks ${\displaystyle S_{1},S_{2},\ldots ,S_{r}}$ is a shift matrix (possibly of different sizes). This form is a special case of the Jordan canonical form for matrices.^[7]

For example, any nonzero 2 × 2 nilpotent matrix is similar to the matrix

${\displaystyle {\begin{bmatrix}0&1\\0&0\end{bmatrix}}.}$

That is, if ${\displaystyle N}$ 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.)

A nilpotent transformation ${\displaystyle L}$ on ${\displaystyle \mathbb {R} ^{n}}$ naturally determines a flag of subspaces

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

and a signature

${\displaystyle 0=n_{0}<n_{1}<n_{2}<\ldots <n_{q-1}<n_{q}=n,\qquad n_{i}=\dim \ker L^{i}.}$

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

${\displaystyle n_{j+1}-n_{j}\leq n_{j}-n_{j-1},\qquad {\mbox{for all }}j=1,\ldots ,q-1.}$

Conversely, any sequence of natural numbers satisfying these inequalities is the signature of a nilpotent transformation.

A linear operator ${\displaystyle T}$ is locally nilpotent if for every vector ${\displaystyle v}$, there exists a ${\displaystyle k\in \mathbb {N} }$ such that

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

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

• Beauregard, Raymond A.; Fraleigh, John B. (1973), A First Course In Linear Algebra: with Optional Introduction to Groups, Rings, and Fields, Boston: Houghton Mifflin Co., ISBN 0-395-14017-X
• Herstein, I. N. (1975), Topics In Algebra (2nd ed.), John Wiley & Sons
• Nering, Evar D. (1970), Linear Algebra and Matrix Theory (2nd ed.), New York: Wiley, LCCN 76091646

• Nilpotent matrix and nilpotent transformation on PlanetMath.