2.9 KiB
Lecture 34
Chapter VIII Operators on complex vector spaces
Generalized Eigenvectors and Nilpotent Operators 8A
\mathbb{F}=\mathbb{R} or \mathbb{C}
Let V be a finite dimensional vector space over m, and T\in\mathscr{L}(V) be an linear operator
null\ T^2=\{v\in V,T(T(v))=0\}
Since T(0)=0, null\ T\subseteq null\ T^2\subseteq\dots \subseteq null\ T^n
Lemma 8.1
null\ T^m\subseteq null\ T^{m+1} for any m\geq 1.
Lemma 8.2
If null\ T^m=null\ T^{m+1} for some m\geq, then null\ T^m=null\ T^{m+n} for any n\geq 1
Proof:
We proceed by contradiction. If there exists n\geq 1 such that null\ T^{m+n}\cancel{\subseteq}null\ T^{m+n}, then there exists v\neq 0,v\in V such that T^{m+n+1}v=T^{m+1}(T^n v)=0 and T^{m+n}v=T^m(T^n v)\neq 0.
So we gets contradiction that T^n v\neq 0, T^n v\in null\ T^{m+1} but T^n v\cancel{\in}null\ T^m, which contradicts with null T^m=null T^{m+1}
Lemma 8.3
Let m=dim\ V, then null\ T^m =null\ T^{m+1} for any T\in \mathscr{L}(V)
Proof:
Since \{0\}\subsetneq null\ T\subsetneq null\ T^2\subsetneq \dots \subsetneq null\ T^m,m=dim\ V, by Lemma 8.2, if null\ T^m\cancel{\subsetneq} null\ T^{m+1}, then all null\ T^n\cancel{\subsetneq} null\ T^{n+1} for any n\leq m. Since all null\ T^n are sub vector space of V, then null\ T^n\cancel{\subsetneq} null T^{n+1}\implies dimension goes up by at least one, dim\ V=m which contradicts dim\ null\ T^{m+1}\geq m=1
Lemma 8.4
Let dim\ V=m
V=null\ T^m\oplus range\ T^m
Proof:
We need to show that V=null\ T^m+range\ T^m, and null\ T^m\cup range\ T^m=\{0\}
First we show null\ T^m\cup range\ T^m=\{0\}.
If v\in null\ T^m\cup range T^m, T^m v=0,T^m u=v for u\in V.
T^m(u)=v, T^m (T^m(u))=T^m(u)=0
u\in null\ T^{2m},
By Lemma 8.3, null\ T^{2m}=null T^m, T^m u=0=v.
Then form null\ T^m\cup range\ T^m=\{0\} we know that
null\ T^m+range\ T^m=null\ T^m\oplus range\ T^m
and dim(null\ T^m)+dim(range\ T^m)=dim V
Let V be a complex vector spaces, T\in \mathscr{L}(v), \lambda be an eigenvalue of T, S=T-\lambda be an linear operator.
Note: there is v\neq 0 such that Sv=Tv-\lambda v=0, so null\ S\neq \{0\}, and it contains all eigenvectors of T with respect to the eigenvalue \lambda.
Definition 8.8
Suppose T\in \mathscr{L}(V) and \lambda is an eigenvalue of T. A vector v\in V is called a generalized eigenvector of T corresponding to \lambda if v\neq 0 and
(T-\lambda I)^k v=0
for some positive integer k.
Theorem 8.9
If V is a complex vector space and T\in \mathscr{L}(V), then V has a basis of generalized eigenvectors of T.
Lemma 8.11
Any generalized eigenvector v corresponds to an unique eigenvalue \lambda.
Lemma 8.12
Generalized eigenvectors corresponding to different eigenvalues are linearly independent.