Generalized Eigenvectors

Definition A generalized eigenvector of $T$ for $\lambda$ is a nonzero vector $v \in V$ with $N^k v = 0$ for some positive integer $k$. Its rank is the smallest such $k$, so $v$ has rank $k$ when

The vectors of rank $1$ are exactly the eigenvectors for $\lambda$. A vector of rank $k \ge 2$ is not an eigenvector for $\lambda$, because an eigenvector for $\lambda$ is already killed by $N$. If $v$ has rank $k$, then $N^{k-1} v$ is nonzero and killed by $N$, so it is an eigenvector for $\lambda$. In particular, generalized eigenvectors for $\lambda$ exist only when $\lambda$ is an eigenvalue of $T$.

Definition Let $v$ be a generalized eigenvector of rank $k$ for $\lambda$. The Jordan chain generated by $v$ is the list $v_1, \dots, v_k$ with

It satisfies the chain equations $v_1 \ne 0$, $Nv_1 = 0$ and $Nv_j = v_{j-1}$ for $2 \le j \le k$, and conversely any list with these properties is the chain generated by its top vector $v_k$. In terms of $T$ itself, the equations read

So the span of the chain is invariant under $T$. It is the cyclic subspace generated by $v_k$, under $N$ and under $T$ alike, because polynomials in $T$ and polynomials in $N$ are the same operators. The chain is independent, as the next term shows, and in the ordered basis $(v_1, \dots, v_k)$ of its span the matrix of the restriction of $T$ is the $k \times k$ Jordan block

with $\lambda$ on the diagonal, ones directly above it and zeros elsewhere. This uses the bottom-first order. Listing the chain from the top down moves the ones below the diagonal.

Definition Every Jordan chain $v_1, \dots, v_k$ for $\lambda$ is linearly independent. To prove it, suppose

The vector $v_j$ becomes zero after $j$ applications of $N$, so $N^{k-1}$ kills $v_1, \dots, v_{k-1}$ and sends $v_k$ to $v_1$. The relation becomes $c_k v_1 = 0$, and $v_1 \ne 0$ gives $c_k = 0$. Applying $N^{k-2}$ to the remaining relation leaves $c_{k-1} v_1 = 0$, so $c_{k-1} = 0$, and continuing down to $N^0 = I$ gives $c_j = 0$ for every $j$. Consequently a generalized eigenvector has rank at most $n = \dim V$, since a vector of rank $k$ generates a chain of $k$ independent vectors in $V$.

Definition The kernels of the powers of $N$ form an increasing sequence of subspaces

because $N^k v = 0$ implies $N^{k+1} v = 0$. Each $\ker N^k$ is invariant under $T$, because $T$ commutes with $N^k$, so $N^k v = 0$ gives $N^k(Tv) = T N^k v = 0$. The kernel of $N$ itself is the eigenspace $E_\lambda = \ker(T - \lambda I)$. The generalized eigenvectors of rank $k$ are exactly the vectors in $\ker N^k \setminus \ker N^{k-1}$, and the difference

measures how many independent directions are first killed at step $k$. It concerns only the scalar $\lambda$ fixed in $N$ and says nothing about how many distinct eigenvalues $T$ has.

Definition If $\ker N^m = \ker N^{m+1}$ for some $m \ge 0$, then $\ker N^{m+j} = \ker N^m$ for every $j \ge 0$. To prove it, let $v \in \ker N^{m+2}$. Then $Nv \in \ker N^{m+1} = \ker N^m$, so $N^{m+1} v = 0$ and $v \in \ker N^{m+1}$. Hence $\ker N^{m+2} = \ker N^{m+1}$, and induction gives the rest. In dimension $n$ the kernels stabilize by step $n$. Until the first equality each step raises the dimension by at least $1$, and no dimension exceeds $n$, so $\ker N^m = \ker N^{m+1}$ for some $m \le n$. Therefore

and $\ker N^n$ contains every vector that some power of $N$ kills.

Definition The generalized eigenspace of $T$ for $\lambda$ is

By stabilization it consists of the zero vector together with all generalized eigenvectors for $\lambda$, that is, of all $v$ with $(T - \lambda I)^k v = 0$ for some $k \ge 1$. It is a $T$-invariant subspace that contains the eigenspace $E_\lambda$, and it is nonzero exactly when $\lambda$ is an eigenvalue of $T$. On $G_\lambda(T)$ the power $N^n$ is zero, but $N$ need not be nilpotent on all of $V$, because vectors belonging to other eigenvalues can survive every power of $N$.