126 lines
3.6 KiB
Markdown
126 lines
3.6 KiB
Markdown
# Lecture 22
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## Chapter V Eigenvalue and Eigenvectors
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### Upper Triangular Matrices 5C
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#### Theorem 5.44
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Let $T\in \mathscr{L}(V)$ be a linear operator, then $T$ has an upper triangular matrix (with respect to some basis), if the minimal polynomial is $(z-\lambda_1)...(z-\lambda_m)$ for $\lambda_1,..,\lambda_m\in \mathbb{F}$
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Proof:
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$\implies$ easy
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$\impliedby$ Suppose the minimal polynomial of $T$ is $(z-\lambda_1)...(z-\lambda_m)$
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Then we do induction on $m$.
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Base case: $m=1$, then $T-\lambda_1 I=0$, $T=\lambda I$ but $\lambda I$ has an upper triangular matrix,
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Induction step: $m>1$, Suppose the results holds for smaller $m$. Let $u=range(T-\lambda_m I)$, $U$ is invariant under $T$, consider $T\vert_u$.
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Note that if $u\in U$, $(T-\lambda_1 I)...(T-\lambda_{m-1} I)u=(T-\lambda_1 I)...(T-\lambda_m I)v=0$. Thus the minimal polynomial of $T\vert_U$ divides $(z-\lambda_1)...(z-\lambda_{m-1})$
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#### Corollary 5.47 (staring point for Jordan Canonical Form)
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Suppose $V$ is a finite dimensional complex vector space, and $T\in \mathscr{L}(V)$, then $T$ has an upper triangular matrix with respect to some basis.
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Recall: $T$ is upper triangular $\iff$ $Tv_k\in Span\ (v_1,...,v_k)$. where $v_1,...,v_n$ is a basis.
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Let $u_1,...,u_r$ be a basis for $U$ such that $Tu_k\in Span\ (v_1,...,v_k)$ (such thing exists because $T$ is upper triangular.
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Extend to a basis of $V$, $u_1,..,u_r,v_1,...,v_s$, then
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$$
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Tv_k=((T-\lambda_m I)+\lambda_m I)v_k=(T-\lambda_m I)v_k+\lambda_m v_k
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$$
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and $(T-\lambda_m I)v_k\in U, \lambda_m v_k\in Span\ (u_1,..,u_r,v_k)$
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Thus with respect to the same basis $u_1,..,u_r,v_1,...,v_s$ $T$ is upper triangular.
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$$
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M(T)=\begin{pmatrix}
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M(T\vert_U) &\vert & *\\
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\rule{2cm}{1pt}&&\rule{4cm}{0.4pt}\\
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0 & \vert&\lambda \textup{ on the diagonal line}
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\end{pmatrix}
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$$
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Example:
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$M(T)=\begin{pmatrix}
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2&0&1\\
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0&2&1\\
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1&1&3
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\end{pmatrix}$ and the minimal polynomial is $(z-2)(z-2)(z-3)$
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$v_1=(1,-1,0), v_2=(1,0,-1), v_3=(-1,1,0)$
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$M(T,(v_1,v_2,v_3))=\begin{pmatrix}
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2&1&0\\
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0&2&0\\
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0&0&3
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\end{pmatrix}$ which is upper triangular.
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### 5D Diagonalizable Operations
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#### Definition 5.48
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A **Diagonal matrix** is a matrix where all entries except the diagonal is zero
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Example: $I,0,\begin{pmatrix}
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2&0&0\\
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0&2&0\\
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0&0&3
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\end{pmatrix}$
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#### Definition 5.50
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An operator $T\in\mathscr{L}(V)$ is diagonalizable if $M(T)$ is diagonalizable with respect to some basis.
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Example:
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$T:\mathbb{F}->\mathbb{F^2}$
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$M(T)=\begin{pmatrix}
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3&-1\\
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-1&3&
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\end{pmatrix} v_1=(1,-1), v_2=(1,1)$, $T(v_1)=(4,-4)=4v_1, T(v_2)=(2,2)=2v_2$, so the eigenvalues are $2$ with eigenvector $v_2$, and $4$ with eigenvector $v_1$. The eigenvectors for $z$ are $Span (v_2)\ \{0\}$
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$M(T,(v_1,v_2))=\begin{pmatrix}
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4&0\\
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0&2
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\end{pmatrix}$ and $T$ is diagonalizable.
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#### Definition 5.52
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Let $T\in \mathscr{L}(V),\lambda \in \mathbb{F}$. the **eigenspace** of $T$ corresponding to $\lambda$ is the subspace $E(\lambda, T)\in V$ defined by
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$$
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E(\lambda, T)=null\ (T-\lambda I)=\{ v\in V\vert Tv=\lambda v\}
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$$
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Example:
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$E(2,T)=Span\ (v_2)$ $E(4,T)=Span\ (v_1)$, $E(3,T)=\{0 \}$
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#### Theorem 5.54
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Suppose $T\in \mathscr{L}(V)$ $\lambda_1,...,\lambda_m$ are distinct eigenvalues of $T$, Then
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$$
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E(\lambda_1, T)+...+E(\lambda_m,T)
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$$
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is a direct sum. In particular if $V$ is finite dimensional.
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$$
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dim\ (E(\lambda_1, T))+...+dim\ (E(\lambda_m,T))\leq dim\ V
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$$
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Proof:
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Need to show that if $v_k\in E(\lambda_k,T)$ for $k=1,...,m$ then $v_1+...+v_m=0\iff v_k=0$ for $k=1,...,m$. i.e eigenvectors for distinct eigenvalues are linearly independent. (Prop 5.11)
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