Generalized Eigenspace
Definition1 2
For a matrix $A \in M_{n \times n}(\mathbb{C})$ and its eigenvalue $\lambda$, denote by $W_{\lambda}$ the set of all generalized eigenvectors of $\lambda$, and call it the generalized eigenspace.
$$ W_{\lambda} = \left\{ \mathbf{v} \in \mathbb{C}^{n} : (A - \lambda I)^{k} \mathbf{v} = \mathbf{0} \text{ for some } k \in \mathbb{N} \right\} $$
Explanation
Just as the eigenspace is the space of all eigenvectors corresponding to an eigenvalue, the generalized eigenspace is the space of all generalized eigenvectors corresponding to the eigenvalue. If some $\mathbf{v}$ satisfies $(A - \lambda I)^{k} \mathbf{v} = \mathbf{0}$, then it also satisfies $(A - \lambda I)^{k+1} \mathbf{v} = \mathbf{0}$, so for some $K$ the $W_{\lambda}$ can be expressed as follows.
$$ W_{\lambda} = \ker (A - \lambda I)^{K} $$
Hence the restriction map of $(A - \lambda I)$ to $W_{\lambda}$ is a nilpotent transformation. From now on we will treat the matrix $A$ as both a matrix and a linear transformation.
$$ N_{\lambda} := (A - \lambda I) |_{W_{\lambda}} \text{ is nilpotent.} $$
In the following (d), since the kernel is monotone, the statement also holds for numbers larger than $m$. In general it may also hold for numbers smaller than $m$, but we do not specify that. The point is that taking a sufficiently large number makes (d) true, and $m$ is sufficient as such a sufficiently large number.
Properties
(a) $W_{\lambda}$ is a $A$–invariant subspace.
(a’) For any scalar $\mu$, $W_{\lambda}$ is $(A - \mu I)$–invariant.
(b) For the scalar $\mu \ne \lambda$, the restriction $(A -\mu I)|_{W_{\lambda}}$ of $(A - \mu I)$ to $W_{\lambda}$ is one-to-one.
(b’) Note that above we did not assume $\mu$ is an eigenvalue. In particular, for the eigenvalue $\lambda$ the map $A|_{W_{\lambda}}$ is one-to-one.
(b’’) The only eigenvalue of $A|_{W_{\lambda}}$ is $\lambda$.
Let $\lambda$ be an eigenvalue of $A$ with algebraic multiplicity $m$. Then the following hold.
(c) $\dim (W_{\lambda}) \le m$ (in fact, equality holds).
(d) $W_{\lambda} = \ker (A - \lambda I)^{m}$
Proof
(a)
For an arbitrary $\mathbf{v} \in W_{\lambda}$, suppose $(A - \lambda I)^{k} \mathbf{v} = \mathbf{0}$. Since $A$ and $(A - \lambda I)$ commute, the following holds and hence $A \mathbf{v} \in W_{\lambda}$.
$$ (A - \lambda I)^{k} (A \mathbf{v}) = A (A - \lambda I)^{k} \mathbf{v} = A \mathbf{0} = \mathbf{0} $$
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(a')
For $\mathbf{v} \in W_{\lambda}$, we have $A \mathbf{v} \in W_{\lambda}$ and $\mu I \mathbf{v} \in W_{\lambda}$, so the claim follows.
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(b)
Showing that a linear map is one-to-one is equivalent to showing its kernel is $\left\{ \mathbf{0} \right\}$. Let $\mathbf{v} \in W_{\lambda}$. Let also $(A - \mu I) \mathbf{v} = \mathbf{0}$. Now assume $\mathbf{v} \ne \mathbf{0}$ (and we will derive a contradiction). Let $k$ be the smallest integer for which $(A - \lambda I)^{k} \mathbf{v} = \mathbf{0}$ holds. If we set $\mathbf{y} = (A - \lambda I)^{k-1} \mathbf{v}$ then the following holds, so $\mathbf{y} \in E_{\lambda}$. Here $E_{\lambda} = \left\{ \mathbf{v} : (A - \lambda I)\mathbf{v} = \mathbf{0} \right\}$ is the eigenspace of $\lambda$.
$$ \mathbf{y} \ne \mathbf{0}, \qquad (A - \lambda I)\mathbf{y} = \mathbf{0} $$
Also, since $(A - \lambda I)$ and $(A - \lambda I)$ commute, the following holds.
$$ (A - \mu I) \mathbf{y} = (A - \mu I) (A - \lambda I)^{k-1} \mathbf{v} = (A - \lambda I)^{k-1} (A - \mu I) \mathbf{v} = (A - \lambda I)^{k-1} \mathbf{0} = \mathbf{0} $$
Hence $\mathbf{y} \in E_{\mu}$. But for distinct $\mu$ and $\lambda$ we have $E_{\mu} \cap E_{\lambda} = \left\{ \mathbf{0} \right\}$, so $\mathbf{y} = \mathbf{0}$. This contradicts the earlier fact $\mathbf{y} \ne \mathbf{0}$. Therefore the assumption was false, and any $\mathbf{v}$ satisfying $(A - \mu I)\mathbf{v} = \mathbf{0}$ is the zero vector. Thus the kernel of $(A - \mu I)|_{W_{\lambda}}$ is $\left\{ \mathbf{0} \right\}$, and the map is one-to-one.
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(b')
For $\mathbf{v} \ne \mathbf{0}$, let $A|_{W_{\lambda}} \mathbf{v} = \mu \mathbf{v}$. Since $\mathbf{v} \in W_{\lambda}$, for some $k$ we have $(A - \lambda I)^{k} \mathbf{v} = \mathbf{0}$. Then we obtain
$$ (A - \lambda I)^{k}\mathbf{v} = (\mu - \lambda)^{k} \mathbf{v} = \mathbf{0} $$
and therefore $\mu = \lambda$.
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(c)
Let the characteristic polynomial of $A|_{W_{\lambda}}$ be $h(x)$. Since $W_{\lambda}$ is $A$–invariant, $h$ divides $f$, the characteristic polynomial of $A$. By (b’’), the eigenvalues of $A|_{W_{\lambda}}$ are only $\lambda$, hence
$$ h(x) = (x - \lambda)^{d} $$
Therefore $d = \dim (W_{\lambda}) \le m$.
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(d)
First, $\ker (A - \lambda I)^{m} \subset W_{\lambda}$ is obvious.
$W_{\lambda}$ is $A$–invariant, and the characteristic polynomial $f$ of $A_{W_{\lambda}}$ divides the characteristic polynomial of $A$. Also, since the only eigenvalue of $A|_{W_{\lambda}}$ is $\lambda$, for $d$ with $d \le m$ the following holds.
$$ f(x) = (x - \lambda)^{d} $$
Then by the Cayley–Hamilton theorem the following holds.
$$ f(A_{W_{\lambda}}) = (A_{W_{\lambda}} - \lambda I)^{d} = O $$
$$ \implies (A - \lambda I)^{d} \mathbf{v} = \mathbf{0}, \quad \forall \mathbf{v} \in W_{\lambda} $$
Since $d \le m$, $W_{\lambda} \subset \ker (A - \lambda I)^{m}$
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