# Lecture 35

## Chapter VIII Operators on complex vector spaces

### Generalized Eigenvectors and Nilpotent Operators 8A

Recall: 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 $k$.

Example:

For $T\in\mathscr{L}(\mathbb{F})$

The matrix for $T$ is $\begin{pmatrix} 0&1\\0&0 \end{pmatrix}$

When $\lambda=0$, $\begin{pmatrix} 1 & 0 \end{pmatrix}$ is an eigenvector  $\begin{pmatrix} 0&1 \end{pmatrix}$ is not and eigenvector but it is a generalized eigenvector.

In fact $\begin{pmatrix} 0&1\\0&0 \end{pmatrix}^2=\begin{pmatrix} 0&0\\0&0 \end{pmatrix}$, so any nonzero vector is a generalized eigenvector. is a generalized eigenvector of $T$ corresponding to eigenvalue $0$.

Fact: $v\in V$ is a generalized eigenvector of $T$ corresponding to $\lambda\iff (T-\lambda I)^{dim\ V}v=0$

#### Theorem 8.9

Suppose $\mathbb{F}=\mathbb{C}$ and $T\in \mathscr{L}(V)$ Then $\exists$ basis of $V$ consisting of generalized eigenvector of $T$.

Proof: Let $n=dim\ V$ we will induct on $n$.

Base case $n=1$, Every nonzero vector in $V$ is an eigenvector of $T$.

Inductive step: Let $n=dim\ V$, assume the theorem is tru for all vector spaces with $dim<n$.

Using **Theorem 8.4** $V=null(T-\lambda I)^n\oplus range(T-\lambda I)^n$. If $null(T-\lambda I)^n=V$, then every nonzero vector is a generalized eigenvector of $T$

So we may assume $null(T-\lambda I)^n\neq V$, so $range(T-\lambda I)^n\neq \{0\}$.

Since $\lambda$ is an eigenvalue of $T$, $null(T-\lambda I)^n\neq \{0\}$, $range(T-\lambda I)^n\neq V$.

Furthermore, $range(T-\lambda I)$n$ is invariant under $T$ by **Theorem 5.18**. (i.e $v\in range\ (T-\lambda I)^n\implies Tv\in range\ (T-\lambda I)^n$.)

Let $S\in \mathscr{L}(range\ (T-\lambda I)^n)$, be the restriction of $T$ to $range\ (T-\lambda I)^n$. By induction, $\exists$ basis of $range\ (T-\lambda I)^n$ consisting of generalized eigenvectors of $S$. These are also generalized eigenvectors of $T$. So we have

$$
V=null\ (T-\lambda I)^n\oplus range\ (T-\lambda I)^n
$$

which gives our desired basis for $V$.

Example:

$T\in \mathscr{L}(\mathbb{C}^3)$ matrix is $\begin{pmatrix}0&0&0\\4&0&0\\0&0&5\end{pmatrix}$ by lower triangular matrix, eigenvalues are $0,5$.

The generalized eigenvector can be obtained $\begin{pmatrix}0&0&0\\4&0&0\\0&0&5\end{pmatrix}^3=\begin{pmatrix}0&0&0\\0&0&0\\0&0&125\end{pmatrix}$

So the generalized eigenvectors for eigenvalue $0$ are $(z_1,z_2,0)$,

So the standard basis for $\mathbb{C}^3$ consists of generalized eigenvectors of $T$.

Recall: If $v$ is an eigenvector of $T$ of eigenvalue $\lambda$ and $v$ is an eigenvector of $T$ of eigenvalue $\alpha$, then $\lambda=\alpha$. 

Proof:

$Tv=\lambda v,Tv=\alpha v$, then $\lambda v=\alpha v,\lambda-\alpha=0$

More generalized we have

#### Theorem 8.11

Each generalized eigenvectors of $T$ corresponds to only one eigenvalue of $T$.

Proof:

Suppose $v\in V$ is a generalized eigenvector of $T$ corresponds to eigenvalues $\lambda$ and $\alpha$.

Let $n=dim\ V$, we know $(T-\lambda I)^n v=0,(T-\alpha I)^n v=0$. Let $m$ be the smallest positive integer such that $(T-\alpha I)^m v=0$ (so $(T-\alpha I)^{m-1}v\neq 0$).

Then, let $A=\alpha I-\lambda I$, $B=T-\alpha I$, and $AB=BA$

<!-- $$
\begin{aligned}
    0&=(T-\lambda I)^n v\\
    &=(B+A)^n v\\
    &=\left(A^n+nA^{n-1}B+\begin{pmatrix}
        n\\2
    \end{pmatrix} A^{n-2}B^2+\dots+B^n
    \right)v
\end{aligned}
$$ this proof is confusing, use the lower one for better-->

$$
\begin{aligned}
    0&=(T-\lambda I)^n v\\
    &=(B+A)^n v\\
    &=\sum^n_{k=0} \begin{pmatrix}
        n\\k
    \end{pmatrix} A^{n-k}B^kv
\end{aligned}
$$

Then we apply $(T-\alpha I)^{m-1}$, which is $B^{m-1}$ to both sides

$$
\begin{aligned}
    0&=A^nB^{m-1}v
\end{aligned}
$$

Since $(T-\alpha I)^{m-1}\neq 0$, $A=0$, then $\alpha I-\lambda I=0$, $\alpha=\lambda$