107 lines
3.2 KiB
Markdown
107 lines
3.2 KiB
Markdown
# Lecture 23
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## Chapter V Eigenvalue and Eigenvectors
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### 5D Diagonalizable Operators
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#### Theorem 5.55
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Suppose $V$ is a finite dimensional vector space and $T\in \mathscr{L}(V)$. let $\lambda_1,...,\lambda_m$ be the distinct eigenvalues of $T$, then the followings are equal:
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a) $T$ is diagonalizable
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b) $V$ has a basis of eigenvectors of $T$
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c) $V=E(\lambda, T)\oplus....\oplus E(\lambda_m,T)$
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d) $dim\ V= dim\ E(\lambda_1,T)+...+dim\ E(\lambda_m,T)$
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ideas of Proof:
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$(a)\iff (b)$ look at $M(T)$'
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$(b)\iff (c)$ recall $E(\lambda_1,T)+...+E(\lambda_m,T)$ is always a distinct sum
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$(c)\iff (d)$ again $E(\lambda_1,T)+...+E(\lambda_m,T)$ is always a distinct sum
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Example:
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$T:\mathbb{R}^2\to\mathbb{R}^3$, $M(T)=\begin{pmatrix}
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0&1&0\\
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0&0&1\\
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0&0&0
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\end{pmatrix}$
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Eigenvalues:[0], Eigenvectors $E(0,T)=null\ (T-0I)=Span\{(1,0,0)\}$
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There are no basis of eigenvectors, $\mathbb{R}^3\neq E(0,T)$, $3\neq dim\ (E(0,T))=1$
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#### Theorem 5.58
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Suppose $V$ is a finite dimensional $T\in \mathscr{L}(v)$ and $T$ has $n=\dim $. distinct eigenvalues then T is diagonalizable.
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Proof:
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Let $\lambda_1,...,\lambda_n$ be the distinct elements of$T$.. =
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Then let $v_1,...,v_n$ be eigenvectors of $\lambda_1,...,\lambda_n$ in the same order. Note $v_1,...,v_n$ are eigenvectors for distinct eigenvectors by **Theorem 5.11** they are linearly independent thus they form a basis. So by **Theorem 5.55**, $T$ is diagonalizable.
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Example:
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$$
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M(T)=\begin{pmatrix}
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1& 4& 5 \\
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0&2&6\\
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0&0&3
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\end{pmatrix}
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$$
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is diagonalizable
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#### Theorem 5.62
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Suppose $V$ finite dimensional $T\in \mathscr{L}(V)$. Then $T$ is diagonalizable if and only if the **minimal polynomial** is of the form $(z-\lambda_1)...(z-\lambda_m)$ for distinct $\lambda_1,...,\lambda_m\in\mathbb{F}$
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Proof:
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$\Rightarrow$
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Suppose $T$ is diagonalizable, let $\lambda_1,...,\lambda_m$ be the distinct eigenvalues of $T$. And let $v_1,...,v_n$ for $n=dim\ V$ be a basis of eigenvectors of $T$. We need to show
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$$
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(T-\lambda_1I)...(T-\lambda_mI)=0
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$$
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Consider $(T-\lambda_1I)...(T-\lambda_mI)v_k=(T-\lambda_1I)...(T-\lambda_mI)$, suppose $Tv_k=\lambda_j v_k$. Then $(T-\lambda_1I)...(T-\lambda_mI)=0$
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So $(T-\lambda_1I)...(T-\lambda_mI)=0\implies$ minimal polynomial divides $(z-\lambda_1)...(z-\lambda_m)$ so the minimal polynomial has distinct linear factors.
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$\Leftarrow$
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Suppose $T$ has minimal polynomial $(z-\lambda_1)...(z-\lambda_m)$ with distinct $\lambda_1,...,\lambda_m$
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Induction on $m$,
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Base case: $(m=1)$:
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Then $T-\lambda I=0$, so $T=\lambda I$ is diagonalizable.
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Induction step: $(m>1)$:
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Suppose the statement hold for $<m$, consider $U=range\ (T-\lambda_mI)$, $T\vert_U$ has minimal polynomial $(z-\lambda_1)...(z-\lambda_m)$ so $T\vert_U$ is diagonalizable.
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$null (T-\lambda_m I)=E(\lambda_m,T)$ has $dim\ (E(\lambda_m,T))$ distinct eigenvector.
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Need to show $null\ (T-\lambda_m I)\cap range\ (T-\lambda_m I)=\{0\}$
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#### Corollary
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If $U$ is an invariant subspace of $T$ and $T$ is diagonalizable, then $T\vert_U$ is diagonalizable.
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Proof:
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minimal polynomial $T\vert_U$ divides minimal polynomial of $T$.
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#### Theorem (Gershigorem Disk Theorem)
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The eigenvalue of $T$ satisfies the following:
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$$
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|\lambda-A_{j,j}|\leq \sum_{k=1,k\neq j}^n |A_{j_k}|
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$$
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for some $j$ where $A=M(T)$
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