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 $dim\ V=m$

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

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.