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.

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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.