chap8.tex

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\section{Operators on Complex Vector Spaces}
\setcounter{subsection}{2}
\subsection{Characteristic and Minimal Polynomial}
\begin{definition}
	The number of times an eignevalue $\lambda$ appears in the matrix is called the algebragic multiplicity of $\lambda$.
\end{definition}
\begin{example}
	Suppose $V$ is a complex vector space and let $T \in \c L(V)$. Supppose $T$ has the following matrix presentation
\[ \bml 
	2 & 1 & 0 \\
	0 & 2 & 1 \\
	0 & 0 & 2 \\
	&&& 3 & 1 \\
	&&& 0 & 3 \\
	&&&&& 2	\bmr\]
	We can see that $\lambda = 2$ has a multiplicity of $4$ and $\lambda = 3$ has a algebraic multiplicaity of $2$.
\end{example}
\begin{definition}
	Suppose $V$ is a complex vector space and $T \in \c L(V)$. Suppose $T$ has eigenvalues $\li \lambda n$ with algebraic multiplicity of $\li dn$. Then the polynomial
	\[ p_{\text{char}}(z) = \prod_{j} (z - \lambda_j)^{d_j} \] is the characteristic polynomial of $T$.
\end{definition}
\begin{theorem}[The Cayley-Hamilton Theorem]
	Suppose V is a complex vector space and $T \in \c L(V)$. Then $p(T) = 0$, where $p$ is the characteristic polynomial. 
\end{theorem}
\begin{proof}
	Trivial by Jordan Normal Form in section 8.d.
\end{proof}	
\begin{definition}
	A minimal polynomial for $T \in \c L(V)$ os a monoic polynomial of the smallest degree that annihilates $T$. i.e. $q(T) = 0$ and $q$ is of smallest degree with this property of leading coefficient $1$.
\end{definition}
\begin{example}
	Consider $T$ in example 8.2. Take the largest block of eahc eigenvalue and raise each term to the size of the block will yield the minimal polynomial
	\[ p_{text{min}} (z) = (z - 2)^3(z - 3)^2\]
\end{example}
\begin{corollary}
	Suppose $h(T) = 0$ for some polynomial $h \not\equiv 0$. Then $h(z) = p_{\text{min}} (z)q(x)$ for some $q$.
\end{corollary}
\begin{proof}
	By the remainder theorem we have 
	\[ h(z) = p_{\text{min}}(z)q(z) + r(z)\]
	where $\deg r < \deg p_{\text{min}}$. By the minimality of $p_{\text{min}}$, $r \equiv 0$.
\end{proof}
\subsection{Jordan Form}
\subsubsection*{Goal} To find te the sparest matrix representation for an arbitary linear operator on a finite dimensional vector space over $\b C$.
\subsubsection{Observation}
``Rough'' decomposition 1
\[ \left[\begin{array}{ccc|ccc} 
\ast & \cdots & \ast & 0 & \cdots & 0 \\ 
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ 
\ast & \cdots & \ast & 0 & \cdots & 0 \\
\hline
0 & \cdots & 0  & \ast & \cdots & \ast  \\ 
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ 
0 & \cdots & 0 & \ast & \cdots & \ast \\

          \end{array}\right]  = \c M(T)\]
          Notice that $T$ has two invariant subspaces that are direct sums of each other.
\begin{definition}
	An operator is called nilpotent if some power of it equals to $0$.
\end{definition}
\begin{proposition}
	For any $T \in \c L(V)$, there exists two subsapces, $V_s$ and $V_r$ such that $V = V_s \oplus V_r$ and $V_s,V_r$ are both $T$-invariant, and $T\vert_{V_s}$ is nilpotent and $T\vert_{V_r}$ is invertible. 
\end{proposition}

\begin{proof}
Consider 
\[ \lb v \rb \subseteq \nul T \subseteq \nul T^2 \subseteq \cdots\]
Because $\dim V \leq \infty$, we must be able to find $q \in \b N$ such that $T^q$ and $T^{q+h}$ for any $h \in \b N$ have the same null space. In other words 
\[ \exists q\in \b N : \nul T^q = \nul T^{q + h} \qquad \forall k \in \b N\]
Take $V_s : = \nul T^q$ and $V_r = \range T^q$. Obsrve that $V_s$ and $V_r$ ae $T$-invariant. 

\noindent Next we want to check that $V_s \cap V_r = \lb \vec 0 \rb$. \\ 
Suppose $\vec v \in V_s \cap V_r$. Then $T^q \vec v = \vec 0$, and $T^q \vec w = \vec v$ for some $\vec w \in V$. So $T^{2q}\vec w = \vec 0$. Hnec eby the choice of $q$ we have $\vec w \in \nul T^q$, so \[T^q \vec w = \vec 0 = \vec v\]
So $\vec v = 0$, and $V_s \cap V_r = \lb \vec 0 \rb$. By Rank-Nullity, $V = V_s \oplus V_r$. \\
$T\vert_{V_s}$ is nilpotent sicne $\left(T\vert_{V_s}\right)^2$ is zero. \\
$T\vert_{V_r}$ is invertible since for any $\vec w \in V_r$ such that $T \vec w = \vec 0$ will also satify $T^q \vec w = \vec 0$, hence $\vec w = \vec 0$, and $T\vert_{V_r}$ being injective implies invertiablity.
\end{proof}
\subsubsection*{Zoom in to the nilpotent part}
	Say, the whole space $V$ satifies the condition $T^q = 0$ and without the loss of generality we can take $q$ minimal with this property. This means there exists $\vec v_0 \in V$ such that $T^{q - 1} \vec v_0 \neq \vec 0$. Take 
	\[ \vec v_0 = \spa \lb \vec v_0 , T\vec v_0, \ldots, T^{q-1} \vec v_0 \rb\]
	Since there exists a vector such that $T^{q - 1} \vec v_0 \neq \vec 0$ we can also there exists $\vec w_0 \in V$ such that $\la T^{q-1}\vec v_0, \vec w_0\ra \neq 0$. Take the following matrix
	\[ \left( \la T^{j-1}\vec v_0 , T^{*^{q-i}} \vec w_0 \ra\right)_{i,j = 1}^q = \left( \la T^{q + j - i - 1} \vec v_0, \vec w_0 \ra\right)_{i,j = 1}^q\]
	Notice that this is a lower triangular matrix with nonzero diagonal matrix.
\begin{corollary}
	The list $\vec v_0 , T\vec v_0, \ldots, T^{q-1} \vec v_0$ is linearly indepedent and so is the list $\vec w_0 , T^*\vec w_0, \ldots, T^{*^{q-1}} \vec w_0 $
\end{corollary}
\begin{proof}
	Take $V_1 := \left( \spa \left( \vec w_0 , T^*\vec w_0, \ldots, T^{*^{q-1}} \vec w_0 \right) \right)^{\perp}$. Notice that if $W$ is $T*$-invariant, $W^\perp$ is $T$-invariant. Indeed, for any $\vec v \in W^\perp$ and any $\vec w \in W$, we have \[\la T \vec v, \vec w \ra =\la  \vec v, T^* \vec w \ra = 0\]
	Hence we have $V = V_0 \oplus V_1$, where $V_0, V_1$ are both $T$-invariant. To see that the sum is direct 
	Suppose \[\alpha_0 \vec v_0 + \alpha_1 T\vec v_0 + \cdots + \alpha^{q-1}T^{q-1}\vec v\]
	is orthogonal to $\vec w_0 , T^*\vec w_0, \ldots, T^{*^{q-1}} \vec w_0 $. Thn the matrix 
	\[ \left( \la T^{j-1} \vec v_0, T^{*^{q-1}} \vec w_0 \ra\right)\] 
	being invertible gurantees that
	\[ \alpha_0 = \alpha_1 = \cdots = \alpha_{q-1} = 0\]
\end{proof}
\subsubsection*{fine decomposition}
We look at $\c M\left( T\vert_{V_1}\right)$ with respect to the basis $\vec v_0 , T\vec v_0, \ldots, T\vec v_0$. Hence we have 
\[ \bml 
	0 & 0 & 0 & \cdots & 0 & 0\\ 
	1 & 0 & 0 & \cdots & 0 & 0\\
	0 & 1 & 0 & \cdots & 0 & 0\\
	\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\

	0 & 0 & 0 & \cdots & 1 & 0\bmr\]
\subsubsection*{Warp-up}
Repeat the process many gives 
\[V = V_1 \oplus V_2 \oplus \cdots \oplus V_n \]
This guarantees a bock-diagonal form where each block looks like 
\[ \bml 
\lambda_j & 1 \\	
& \lambda_j & 1 \\
&& \lambda_j & 1 \\
&&& \lambda_j & 1 \\
&&&& \ddots & \ddots \\
&&&&& \lambda_j & 1 \\
&&&&&& \lambda_j
\bmr\]


\begin{example}
	Consider
	\[ \c M(T) =  \bml 
		3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
		0 & 3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
		0 & 0 & 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
		0 & 0 & 0 & 3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
		0 & 0 & 0 & 0 & 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
		0 & 0 & 0 & 0 & 0 & 3 & 0 & 0 & 0 & 0 & 0 & 0  \\
		0 & 0 & 0 & 0 & 0 & 0 & -2 & 1 & 0 & 0 & 0 & 0 \\
		0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 & 0 & 0 & 0 & 0 & \\
		0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\
		0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\
		0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
		0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \bmr\]
Where the empty entries are zero. We can see that the $T$ has eigenvalue $3,-2,1,0$. \\
We can see that \begin{align*} \dim \nul (T - 3 \b I)^j &= 3,5,6,6,6, \ldots \\
\dim \nul (T + 2\b I)^j &= 1,2, \ldots \\
\dim \nul (T - 1 \b I)^j &= 1,2,2,\ldots \\
\dim \nul (T - 0 \b I)^j &= 1,2,2,2, \ldots \\
j = 1,2,3, \ldots & \qquad \forall\ j \in \b N
\end{align*}
\end{example}
\begin{example}
	Suppose $V:= \c P_4(\b C)$. Let $D$ be the differetiation operator. Construct the Jordan Normal Form of $D$ and the Jordan Basis of $V$. \\
	We can compute for $\c M(T)$ with the standard basis
	\[ \bml 
		0 & 1 & 0 & 0 & 0 \\
		0 & 0 & 2 & 0 & 0 \\
		0 & 0 & 0 & 3 & 0 \\
		0 & 0 & 0 & 0 & 4 \\
		0 & 0 & 0 & 0 & 0 		
	\bmr\]
	with some algebraic manipulation we then can see that $D$ has jodran normal form of
	\[ \bml 
		0 & 1 & 0 & 0 & 0 \\
		0 & 0 & 1 & 0 & 0 \\
		0 & 0 & 0 & 1 & 0 \\
		0 & 0 & 0 & 0 & 1 \\
		0 & 0 & 0 & 0 & 0 		
	\bmr\]
	with basis $\displaystyle 1, x, \frac 12 x^2, \frac 13 x^3, \frac 14 x^4$.
\end{example}
\begin{example}
	Suppose $V = \spa \left( e^{ikt} : |k| \leq 3 \right)$. Let $D$ be the differetiation operator. Construct the Jordan Normal Form of $D$ and the Jordan Basis of $V$. \\
	We can see that $V = \spa (e^{-i3t},e^{-i2t},e^{-it},e^{0},e^{it},e^{i2t},e^{i3t})$. We can compute for the matrxi representation with repsect to the standard basis
	\[ \c M(T) = \bml 
	-3 \\
	& -2 \\
	&& -1 \\
	&&& 0 \\
	&&&& 1 \\
	&&&&& 2 \\
	&&&&&& 3 
	\bmr\]
	Notice that if we use basis $\displaystyle V = \spa \left(\frac{1}{-3i} e^{-i3t}, \frac{1}{-2i}e^{-i2t},\frac{1}{-i}e^{-it}, e^{0}, \frac{1}{i}e^{it}, \frac{1}{2i}e^{i2t}, \frac{1}{3i}e^{i3t}\right)$ we can obtain the Jordan Normal Form
	\[ \bml 1 \\
	& 1 \\
	&& 1 \\
	&&& 0 \\
	&&&& 1 \\
	&&&&& 1 \\
	&&&&&& 1 \bmr\]
\end{example}
\newpage
\subsubsection*{Reverse Engineering of the Jordan Normal Form}
\[ \begin{array}{|c|c|c|c|c|c|}
\hline 
& \dim (T - \lambda \b I) & \dim (T - \lambda \b I)^2 & \dim (T - \lambda \b I)^3 & \dim (T - \lambda \b I)^4 & \dim (T - \lambda \b I)^j, j \geq 5 \\
\hline
\hline 
\lambda = i & 3 & 6 & 7 & 7 & 7 \\
\lambda = -i & 2 & 4 & 6 & 8 & 8 \\
\lambda = 1 & 1 & 2 & 3 & \not5 \ \ 4 & 5 \\
\hline
\end{array}\]
What is the Jordan Normal form og $T$ based on this info? Assume all eigenvalues of $T$ are given above.
\begin{proof}[Solution] Notice that the difference in the sequence denotes the number of $1$'s that get send to $0$, hence we have the following matrix

\[ \bml 
	i & 1\\
	& i & 1\\
	&& i \\
	&&& i & 1\\
	&&&& i \\
	&&&&& i & 1\\
	&&&&&& i \\
	&&&&&&& -i & 1 \\
	&&&&&&&& -i & 1 \\
	&&&&&&&&& -i & 1 \\
	&&&&&&&&&& -i  \\
	&&&&&&&&&&& -i & 1 \\
	&&&&&&&&&&&& -i & 1 \\
	&&&&&&&&&&&&& -i & 1\\
	&&&&&&&&&&&&&& -i \\
	&&&&&&&&&&&&&&& 1 & 1 \\
	&&&&&&&&&&&&&&&& 1 & 1 \\
	&&&&&&&&&&&&&&&&& 1 & 1 \\
	&&&&&&&&&&&&&&&&&& 1 & 1 \\
	&&&&&&&&&&&&&&&&&&& 1 \bmr\]
\end{proof}
 \vfill
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