NoteNextra-origin/content/Math429/Math429_L35.md

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\\
    &=\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