110 lines
3.6 KiB
Markdown
110 lines
3.6 KiB
Markdown
# Lecture 36
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## Chapter VIII Operators on complex vector spaces
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### Generalized Eigenvectors and Nilpotent Operators 8A
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If $T\in \mathscr{L}$, is an linear operator on $V$ and $n=dim\ V$.
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$\{0\}\subset null\ T\subset null\ T^2\subset \dots\subset null\ T^n=null\ T^{n+1}$
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#### Definition 8.14
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$T$ is called a nilpotent operator if $null\ T^n=V$. Equivalently, there exists $k>0$ such that $T^k=0$
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#### Lemma 8.16
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$T$ is nilpotent $\iff 0$ is the only eigenvalue of $T$.
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If $\mathbb{F}=\mathbb{C}$, then $0$ is the only eigenvalue $\implies T$ is nilpotent.
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Proof:
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If $T$ is nilpotent, then $T^k=0$ for some $k$. The minimal polynomial of $T$ is $z^m=0$ for some $m$. So $0$ is the only eigenvalue.
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over $\mathbb{C}$, the eigenvalues are all the roots of **minimal polynomial**.
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#### Proposition 8.17
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The following statements are equivalent:
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1. $T$ is nilpotent.
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2. The minimal polynomial of $T$ is $z^m$ for some $m\geq 1$.
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3. There is a basis of $V$ such that the matrix of $T$ is upper triangular with $0$ on the diagonal ($\begin{pmatrix}0&\dots&*\\ &\ddots& \\0 &\dots&0\end{pmatrix}$).
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### Generalized Eigenspace Decomposition 8B
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Let $T\in \mathscr{L}(V)$ be an operator on $V$, and $\lambda$ be an eigenvalue of $T$. We want to study $T-\lambda I$.
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#### Definition 8.19
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The generalized eigenspace $G(\lambda, T)=\{(T-\lambda I)^k v=0\textup{ for some }k\geq 1\}$
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#### Lemma 8.20
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$G(\lambda, T)=null\ (T-\lambda I)^{dim\ V}$
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#### Proposition 8.22
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If $\mathbb{F}=\mathbb{C}$, $\lambda_1,...,\lambda_m$ all the eigenvalues of $T\in \mathscr{L}$, then
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(a) $G(\lambda_i, T)$ is invariant under $T$.
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(b) $(T-\lambda_1)\vert_{G(\lambda_1,T)}$ is nilpotent.
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(c) $V=G(\lambda_1,T)\oplus...\oplus G(\lambda_m,T)$
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Proof:
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(a) follows from $T$ commutes with $T-\lambda_1 I$. If $(T-\lambda_1 I)^k=0$, then $(T-\lambda_i T)^k T(v)=T((T-\lambda_i T)^kv)=0$
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(b) follow from lemma
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(c) $V=G(\lambda_1,T)\oplus...\oplus G(\lambda_m,T)$
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1. $V$ has a basis of generalized eigenvectors $\implies V=G(\lambda_1,T)+...+G(\lambda_m,T)$
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2. If there exists $v_i\in G(\lambda_i,T)$, and $v_1+...+v_m=0$, then $v_i=0$ for each $i$. Because the generalized eigenvectors from distinct eigenvalues are linearly independent, $V=G(\lambda_1,T)\oplus...\oplus G(\lambda_m,T)$.
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#### Definition 8.23
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Let $\lambda$ be an eigenvalue of $T$, the multiplicity of $\lambda$ is defined as $mul(x):= dim\ G(\lambda, T)=dim\ null\ (T-\lambda I)^{dim\ V}$
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#### Lemma 8.25
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If $\mathbb{F}=\mathbb{C}$,
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$$
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\sum^n_{i=1} mul\ (\lambda_i)=dim\ V
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$$
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Proof from proposition part (c).
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#### Definition 8.26
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If $\mathbb{F}=\mathbb{C}$, we defined the characteristic polynomial of $T$ to be
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$$
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q(z):=(z-\lambda_1)^{mul\ (\lambda_1)}\dots (z-\lambda_m)^{mul\ (\lambda_m)}
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$$
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$deg\ q=dim\ V$, and roots of $q$ are eigenvalue of $V$.
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#### Theorem 8.29 Cayley-Hamilton Theorem
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Suppose $\mathbb{F}=\mathbb{C}$, $T\in \mathscr{L}(V)$, and $q$ is the characteristic polynomial of $T$. Then $q(T)=0$.
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Proof:
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$q(T)\in \mathscr{L}(V)$ is a linear operator. To show $q(T)=0$ it is enough to show $q(T)v_1=0$ for a basis $v_1,...,v_n$ of $V$.
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Since $V$ is a sum of vectors in $G(\lambda_1, T),...,G(\lambda_m,T)$.
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$$
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q(T)=(T-\lambda_1 I)^{d_1}\dots (T-\lambda_m I)^{d_m}
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$$
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The operators on the right side of the equation above all commute, so we can
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move the factor $(T-\lambda_k I)^{d_k}$ to be the last term in the expression on the right.
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Because $(T-\lambda_k I)^{d_k}\vert_{G(\lambda_k,T)}= 0$, we have $q(T)\vert_{G(\lambda_k,T)} = 0$, as desired.
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#### Theorem 8.30
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Suppose $\mathbb{F}=\mathbb{C}$, $T\in \mathscr{L}(V)$. Then the characteristic polynomial of $T$ is a polynomial multiple of the minimal polynomial of $T$. |