119 lines
3.2 KiB
Markdown
119 lines
3.2 KiB
Markdown
# Lecture 19
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## Chapter V Eigenvalue and Eigenvectors
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### Invariant Subspaces 5A
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#### Proposition 5.11
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Suppose $T\in \mathscr{L}(V)$, let $v_1,...,v_n$ be eigenvectors for distinct eigenvalues $\lambda_1,...,\lambda_m$. Then $v_1,...,v_n$ is linearly independent.
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Proof:
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Suppose $v_1,...,v_m$ is linearly dependent, we can assume that $v_1,...,v_{m-1}$ is linearly independent. So let $a_1,...,a_{m}$ not all $=0$. such that $a_1v_1+...+a_nv_m=0$, then we apply $(T-\lambda_m I)$ (map $v_n$ to 0)
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$$
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(T-\lambda_m I)v_k=(\lambda_k-\lambda_m)v_k
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$$
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so
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$$
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(T-\lambda_m I)=a_1(\lambda_1-\lambda_m)v_1+...+a_{m-1}(\lambda_{m-1}-\lambda_{m})v_m
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$$
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but not all of the $a_1,...,a_{m-1}$ are zero and $\lambda_k-\lambda_m\neq 0$ for $1\leq k\leq \lambda$ so they must be linearly independent.
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#### Theorem 5.12
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Suppose $dim\ V=n$ and $T\in \mathscr{L}(V)$ then $T$ has at most $n$ distinct eigenvalues
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Proof:
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Since $dim\ V=n$ no linearly independent list has length than $n$ so by **Proposition 5.11**, there are at most $n$ distinct eigenvalues.
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#### Polynomials on operators
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$p(z)=z+3z+z^3\in \mathscr{P}(\mathbb{R})$
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let $T=\begin{pmatrix}
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1&1\\
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0&1
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\end{pmatrix}\in \mathscr{L}(\mathbb{R}^2)$
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$P(T)=2I+3T+T^3=2I+3T+\begin{pmatrix}
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1&3\\
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0&1
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\end{pmatrix}=\begin{pmatrix}
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6&4\\
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0&6
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\end{pmatrix}$
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#### Notation
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$T^m=TT...TT$ (m times) $T$ must be an operator within the same space
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$T^0=I$
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$T^{-m}=(T^{-1})^m$ (where $T$ is invertible)
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if $p\in \mathscr{P}(\mathbb{F})$ with $p(z)=\sum_{i=0}^na_iz^i$ and $T\in \mathscr{L}(V)$ $V$ is a vector space over $\mathbb{F}$
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$$
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p(T)\sum_{i=0}^na_iT^i
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$$
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#### Lemma 5.17
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Given $p,q\in \mathscr{P}(\mathbb{F})$, $T\in \mathscr{L}(V)$
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then
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a) $(pq)T=p(T)q(T)$
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b) $p(T)q(T)=q(T)p(T)$
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#### Theorem 5.18
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Suppose $T\in \mathscr{L}(V),p\in \mathscr{P}(\mathbb{F})$, then $null\ (P(T))$ and $range\ (P(T))$ are invariant with respect to $T$.
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### 5B The Minimal Polynomial
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#### Theorem 5.15
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Every operator on **finite dimensional complex vector space** has at least on eigenvalues.
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Proof:
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Let $dim\ V=n,T\in \mathscr{L}(V), v\in V$ be a nonzero vector.
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Now consider $v,Tv,T^2 v,...,T^n v$. Since this list is of length $n+1$, there is a linear dependence. Let $m$ be the smallest integer such that $v,Tv,..T^m v$ is linearly dependent, then
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$$
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a_0 v+a_1Tv+...+a_m T^m v=0
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$$
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Let $p(z)=a_0+a_1 z+...+a_m z^m$, then $p(T)(v)=0,p(z)\neq 0$
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$p(z)$ factors as $(z-\lambda) q(z)$ where $degree\ q< degree\ p$
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$$
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p(T)(v)=((T-\lambda I)q(T))(v)=0
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$$
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$$
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(T-\lambda I)(q(T)(v))=0
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$$
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but $m$ was minimal so that $p(z)=a_0+a_1 z+...+a_m z^m$ were linearly independent, so $q(T)(v)\neq 0$, so $\lambda$ is an eigenvalue with eigenvector $q(T)(v)$
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#### Definition 5.24
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Suppose $V$ is finite dimensional $T\in\mathscr{L}(V),p\in \mathscr{P}(\mathbb{F})$, then the **minimal polynomial** is the unique monic (the coefficient of the highest degree is 1) polynomial of minimal degree such that $p(T)=0$
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#### Theorem 5.27
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Let $V$ be finite dimensional, and $T\in\mathscr{L}(V)$, $p(z)$ the minimal polynomial.
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1. The roots of $p(z)$ are exactly the eigenvalues of $T$.
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2. If $\mathbb{F}=\mathbb{C}$, $p(z)=(z-\lambda_1)...(z-\lambda_m)$ where $\lambda_1,...,\lambda_m$ are all the eigenvalues.
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