NoteNextra-origin/pages/Math429/Math429_L37.md

Lecture 37

Chapter VIII Operators on complex vector spaces

Generalized Eigenspace Decomposition 8B

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Review

Definition 8.19

The generalized eigenspace of T for \lambda \in \mathbb{F} is G(\lambda,T)=\{v\in V\vert (T-\lambda I)^k v=0\textup{ for some k>0}\}

Theorem 8.20

G(\lambda, T)=null((T-\lambda I)^{dim\ V})

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New materials

Theorem 8.31

Suppose v_1,...,v_n is a basis where M(T,(v_1,...,v_k)) is upper triangular. Then the number of times \lambda appears on the diagonal is the multiplicity of \lambda as an eigenvalue of T.

Proof:

Let \lambda_1,...,\lambda_n be the diagonal entries, S be such that M(S,(v_1,...,v_n)) is upper triangular. Note that if \mu_1,...,\mu_n are the diagonal entires of M(S), then the diagonal entires of M(S^n) are \mu_1^n,...,\mu_n^n

\begin{aligned}
dim(null\ S^n)&=n-dim\  range\ (S^n)\leq n-\textup{ number of non-zero diagonal entries on } S^n\\
&=\textup{ number of zero diagonal entries of }S^n
\end{aligned}

plus in S=T-\lambda I, then

\begin{aligned}
dim G(\lambda, T)&=dim(null\ (T-\lambda I)^n)\\
&\leq \textup{number times where }\lambda \textup{ appears on the diagonal of }M(T)\\
\end{aligned}

Note:

V=G(\lambda_1, T)\oplus \dots \oplus G(\lambda_k, T)

for distinct \lambda_1,...,\lambda_k thus n=dim\ G(\lambda_1,T)+\dots +dim\ (\lambda_k, T)

on the other hand n=\textup{ number of times }\lambda_1 \textup{ appears as a diagonal entry}+\dots +\textup{ number of times }\lambda_k \textup{ appears as a diagonal entry}+\dots

So dim\ G(\lambda_i, T)= number of times where \lambda_i appears oas a diagonal entry.

Definition 8.35

A block diagonal matrix is a matrix of the form $\begin{pmatrix} A_1& & 0\ & \ddots &\ 0& & A_m \end{pmatrix}$ where A_k is a square matrix.

Example:

$ \begin{pmatrix} 1&0&0 & 0&0\ 0 & 2 &1&0&0\ 0 & 0 &2&0&0\ 0& 0&0& 4&1\ 0& 0&0& 0&4\ \end{pmatrix}$

Theorem

Let V be a complex vector space and let \lambda_1,...,\lambda_m be the distinct eigenvalue of T with multiplicity d_1,...,d_m, then there exists a basis where $\begin{pmatrix} A_1& & 0\ & \ddots &\ 0& & A_m \end{pmatrix}$ where A_k is a d_k\times d_k matrix upper triangular with only \lambda_k on the diagonal.

Proof:

Note that (T-\lambda_k I)\vert_{G(\lambda_k,T)} is nilpotent. So there is a basis of G(\lambda_k,T) where (T-\lambda_k I)\vert_{G(\lambda_k,T)} is upper triangular with zeros on the diagonal. Then (T-\lambda_k I)\vert_{G(\lambda_k,T)} is upper triangular with \lambda_k on the diagonal.

Jordan Normal Form 8C

Nilpotent operators

Example: $T(x,y,z)=(0,x,y), M(T)=\begin{pmatrix} 0&1&0\ 0&0&1\ 0&0&0 \end{pmatrix}$

Definition 8.44

Let T\in \mathscr{L}(V) a basis of V is a Jordan basis of T if in that basis $\begin{pmatrix} A_1& & 0\ & \ddots &\ 0& & A_p \end{pmatrix}$ where each $A_k=\begin{pmatrix} \lambda_1& 1& & 0\ & \ddots& \ddots &\ &&\ddots& 1\ 0&&&\lambda_k\ \end{pmatrix}$

Theorem 8.45

Suppose T\in \mathscr{L}(V) is nilpotent, then there exists a basis of V that is a Jordan basis of T.

Sketch of Proof:

Induct on dim\ V, if dim\ V=1, clear.

if dim\ V>1, then let m be such that T^m=0 and T^{m-1}\neq 0. Then \exists u\in V such that T^{m-1}u\neq 0, then Span (u,Tu, ...,T^{m-1}u) is m dimensional.

Theorem 8.46

Suppose V is a complex vector space T\in \mathscr{L}(V) then T has a Jordan basis.

Proof:

take V=G(\lambda_1, T)\oplus \dots \oplus G(\lambda_m, T), then look at (T-\lambda_k I)\vert_{G(\lambda_k,T)}