Nilpotent matrix

Inlinear algebra,anilpotent matrixis asquare matrixNsuch that

$N^{k}=0\,$

for some positiveinteger$k$.The smallest such$k$is called theindexof$N$,sometimes thedegreeof$N$.

More generally, anilpotent transformationis alinear transformation$L$of avector spacesuch 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 ofnilpotencethat applies to elements ofrings.

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$n$-dimensionaltriangular matrixwith zeros along themain diagonalis 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.

Example 6

Consider the linear space ofpolynomialsof a bounded degree. Thederivativeoperator 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

For an$n\times n$square matrix$N$withreal(orcomplex) entries, the following are equivalent:

• $N$is nilpotent.
• Thecharacteristic polynomialfor$N$is$\det \left(xI-N\right)=x^{n}$.
• Theminimal polynomialfor$N$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 anyfieldof 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.
• Thedeterminantandtraceof a nilpotent matrix are always zero. Consequently, a nilpotent matrix cannot beinvertible.
• The only nilpotentdiagonalizable matrixis the zero matrix.

See also:Jordan–Chevalley decomposition#Nilpotency criterion.

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 thesuperdiagonaland 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 thecanonicalnilpotent matrix.

Specifically, if$N$is any nilpotent matrix, then$N$issimilarto ablock diagonal matrixof 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 theJordan canonical formfor 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 basisb₁,b₂such thatNb₁= 0 andNb₂=b₁.

Thisclassification theoremholds for matrices over anyfield.(It is not necessary for the field to be algebraically closed.)

Flag of subspaces

A nilpotent transformation$L$on$\mathbb {R} ^{n}$naturally determines aflagof 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

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

The signature characterizes$L$up toan invertiblelinear 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.

Additional properties

• If$N$is nilpotent of index$k$,then$I+N$and$I-N$areinvertible,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$t$.
• Everysingular matrixcan be written as a product of nilpotent matrices.
• A nilpotent matrix is a special case of aconvergent matrix.

Generalizations

Alinear operator$T$islocally nilpotentif 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

• 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.,ISBN0-395-14017-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,LCCN76091646

External links

• Nilpotent matrixandnilpotent transformationonPlanetMath.