Lecture 37

Chapter VIII Operators on complex vector spaces

Generalized Eigenspace Decomposition 8B

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

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

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 $\mu_1^n,...,\mu_n^n$

plus in $S=T-\lambda I$, then

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 $(T-\lambda_k I)\vert_{G(\lambda_k,T)}$