Nilpotent linear map

A linear map $T:V\to V$ is nilpotent if $T^k=0$ for some positive integer $k$.

For a nilpotent linear map with minimal polynomial $x^m$, set $K_i=\ker T^i$ and $d_i=\dim K_i$, with $d_0=0$. ThenFor strictness, choose $w$ with $T^{m-1}w\ne 0$. The vector $T^{m-1-i}w$ belongs to $K_{i+1}\setminus K_i$ for $0\le i<m$. For the second inequality, the rank-nullity theorem givesThe first image is contained in the second because $T(K_{i+1})\subseteq K_i$.

If the Nilpotent Jordan block sizes are $s_1,\dots ,s_r$, thenThus successive increments count the Jordan blocks reaching a given size, and $(d_i-d_{i-1})-(d_{i+1}-d_i)$ counts blocks of size exactly $i$, with $d_{m+1}=d_m$. The nullities determine the entire Jordan normal form. Conversely any positive nonincreasing integer increments summing to $\dim V$ are realized by taking the corresponding numbers of each Nilpotent Jordan block.

A square matrix is nilpotent when one of its positive powers is zero. It represents a nilpotent linear map in a chosen basis; changing basis preserves nilpotence. Over an algebraically closed field, its Jordan normal form has only zero-eigenvalue blocks.

Over the complex numbers, if $N^m=0$, thenHere the generalized binomial coefficient is $\left(\genfrac{}{}{0ex}{}{\alpha }{j}\right)=\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$, with $\left(\genfrac{}{}{0ex}{}{\alpha }{0}\right)=1$. This is an exact finite identity, even when an infinite binomial series would not converge.

One can construct the required polynomial $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$ without any infinite series. At polynomial degree $j$, the coefficient of $q(z{)}^3$ is $3c_j$ plus terms involving only earlier coefficients. Recursively choose it to agree with $1+z$ through polynomial degree $m-1$; the first coefficients are $1/3,-1/9,5/81$. The stated generalized binomial coefficients solve this recursion: near $z=0$, the binomial series for $(1+z{)}^{1/3}$ cubes to $1+z$, and truncating it changes no coefficient below polynomial degree $m$. Thus $q(z{)}^3-(1+z)$ is divisible by $z^m$ as a polynomial identity. Substituting the nilpotent matrix $N$ kills the remaining terms, all divisible by $N^m$. For a Jordan block $\lambda I+N$ with $\lambda \ne 0$, replace $N$ by $N/\lambda$ and multiply $Q$ by a scalar cube root of $\lambda$.

A complex nilpotent matrix of size six is determined up to matrix similarity by its matrix rank and nilpotency index. Its Nilpotent Jordan block sizes form an integer partition of six. If there are $k$ blocks and the largest size is $m$, then its matrix rank is $6-k$ and its minimal polynomial is $t^m$. The eleven possible size lists areNo two have the same pair $(m,k)$, which proves the assertion by Jordan normal form. The dimension restriction matters: in size seven, block lists $(3,3,1)$ and $(3,2,2)$ both have matrix rank four and minimal polynomial $t^3$, but give different Jordan normal forms.

The nilpotency index is the least positive power at which a nilpotent matrix becomes zero.

The binomial theorem for the commuting matrices $I,N$ proves the formula for nonnegative integers: every term containing $N^2$ vanishes. Since $(I+N)(I-N)=I$, the matrix inverse is $I-N$; applying the same argument to its powers gives the negative integers as well. The same identity shows that $I+N$ is invertible even when the nilpotent matrix $N$ is nonzero. In dimension two, the example $N=\left(\begin{array}{cc}0& p\\ 0& 0\end{array}\right)$ describes repeated shear mappings, whose displacement parameters add under composition.

The only diagonalizable nilpotent matrix is the zero matrix. Indeed, if $A^p=0$ and $Av=\lambda v$ with $v\ne 0$, then ${\lambda }^p=0$, so $\lambda =0$. A basis of eigenvectors would make every diagonal entry zero and therefore $A=0$. Nonzero nilpotent matrices require nontrivial Jordan blocks, even if their characteristic polynomial has only one distinct root.

If a diagonalizable matrix satisfies $B^k=0$ for a positive integer $k$, write $B=P\mathrm{diag}({\lambda }_j)P^{-1}$. Then every ${\lambda }_j^k=0$, so every eigenvalue is zero and $B=0$. Diagonalizability is essential: the matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ is nonzero but its square is zero. More generally a nonzero nilpotent operator requires a nontrivial Jordan block.

For a real or complex two-by-two matrix $A$, the Cayley-Hamilton theorem reads $A^2-(\mathrm{tr}A)A+(\det A)I=0$. Hence $A^2=0$ exactly when $\mathrm{tr}A=\det A=0$. Conversely, if $A^m=0$ for any positive integer $m$, every eigenvalue is zero and the same identity gives $A^2=0$. Thus a nonzero nilpotent matrix of this size has nilpotency index two.

There are $q^{n^2-n}$ nilpotent matrices of size n over $F_q$. Every matrix has a unique invariant direct sum on which it is nilpotent and invertible. Counting those ordered decompositions gives $q^{n^2}/|G{L}_n(q)|={\sum }_{k=0}^n{a}_k/|G{L}_k(q)|$. Subtraction at n and n−1 yields $a_n=q^{n^2-n}$. Translation by the identity counts unipotent matrices.

For a nilpotent endomorphism $N$ of a vector space, the commutator action $\mathrm{ad}N:T\mapsto NT-TN$ on its endomorphisms is nilpotent. The commuting left and right multiplication maps give $(\mathrm{ad}N{)}^r(T)={\sum }_{j=0}^r(-1{)}^j\left(\genfrac{}{}{0ex}{}{r}{j}\right)N^{r-j}T{N}^j$. If $N^q=0$ and $r\ge 2q-1$, every summand vanishes. This works in every characteristic of a field and is useful when applying Engel theorem to an Adjoint representation.