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\begin{document}

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\section{Eigenvalues and eigenvectors of a linear operator}

In this section we consider a vector space $\cV$ over a scalar field $\nF$. By $\cL(\cV)$ we denote the vector space $\cL(\cV,\cV)$ of all linear operators on $\cV$. The vector space  $\cL(\cV)$ with the composition of operators as an additional binary operation is an algebra in the sense of the following definition.

\begin{definition}
A vector space $\cA$ over a field $\nF$ is an {\em algebra}  over $\nF$ if the following conditions are satisfied:
\begin{enumerate}
\vspace*{-7pt}
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\item
there exist a binary operation $\cdot:\cA \times \cA \to \cA$.
\item
({\em associativity}) for all $x, y, z \in \cA$ we have $(x\cdot y) \cdot z = x \cdot (y \cdot z)$.
\item
({\em right-distributivity}) for all $x, y, z \in \cA$ we have $(x+y) \cdot z = x \cdot z + y \cdot z$.
\item
({\em left-distributivity}) for all $x, y, z \in \cA$ we have $z \cdot (x+y) = z \cdot x + z \cdot y$.
\item
({\em respect for scaling}) for all $x, y \in \cA$ and all $\alpha \in \nF$ we have $\alpha (x \cdot y) = (\alpha x) \cdot y = x \cdot (\alpha y)$.
\end{enumerate} \vspace*{-5pt}
The binary operation in an algebra is often referred to as {\em multiplication}.
\end{definition}

The multiplicative identity in the algebra $\cL(\cV)$ is the identity operator $I_{\cV}$.


For $T \in \cL(\cV)$ we recursively define nonnegative integer powers of $T$ by $T^0 = I_{\cV}$ and for all $n \in \nN$ $T^{n} = T \circ T^{n-1}$.


For $T \in  \cL(\cV)$, set
\[
\cA_T = \lspan \bigl\{T^k: k \in \nN\cup\{0\}\bigr\}.
\]
Clearly $\cA_T$ is a subspace of $\cL(\cV)$. Moreover, we will see below that $\cA_T$ is a commutative subalgebra of $\cL(\cV)$.

Recall that by definition of a span a nonzero $S \in \cL(\cV)$ belongs to $\cA_T$ if and only if $\exists m \in \nN\cup\{0\}$ and $\alpha_0,\alpha_1,\dots, \alpha_m \in \nF$ such that $a_m \neq 0$ and
\begin{equation} \label{eq-lcTs}
S = \sum_{k=0}^m \alpha_k T^k.
\end{equation}

The last expression reminds us of a polynomial over $\nF$. Recall that by $\nF[z]$ we denote the algebra of all polynomials over $\nF$. That is
\[
\nF[z]= \biggl\{\sum_{j=0}^n \alpha_j z^j  \,: \, n\in \nN\cup\{0\}, \  (\alpha_0, \ldots,\alpha_n)\in\nF^{n+1}\biggr\}.
\]
Next we recall the definition of the multiplication in the algebra $\nF[z]$.  Let $m,n \in \nN\cup\{0\}$ and
\begin{equation} \label{eq-pq}
p(z)= \sum_{i=0}^m\alpha_iz^i \in \nF[z] \qquad \text{and} \qquad  q(z)=\sum_{j=0}^n \beta_jz^j \in \nF[z].
\end{equation}
Then by definition
\[
(pq)(z) = \sum_{k=0}^{m+n} \Biggl(
\sum_{\substack{i+j=k\\ i\in\{0,\ldots,m\}\\ j\in\{0,\ldots,n\}}}\alpha_i\beta_j
\Biggr)z^k.
\]
Since the multiplication in $\nF$ is commutative, it follows that $pq = qp$. That is $\nF[z]$ is a commutative algebra.


The obvious alikeness of the expression \eqref{eq-lcTs} and the expression for the polynomial $p$ in \eqref{eq-pq} is the motivation for the following definition. For a fixed $T \in \cL(\cV)$ we define
\[
\Xi_T: \nF[z] \to \cL(\cV)
\]
by setting
\begin{equation} \label{eq-Xi}
\Xi_T(p) = \sum_{i=0}^m \alpha_i T^i \qquad \text{where} \qquad p(z)= \sum_{i=0}^m\alpha_iz^i \in \nF[z].
\end{equation}
It is common to write $p(T)$ for $\Xi_T(p)$.

\begin{theorem}
Let $T \in \cL(\cV)$. The function $\Xi_T: \nF[z] \to \cL(\cV)$ defined in \eqref{eq-Xi} is an algebra homomorphism. The range of $\Xi_T$ is $\cA_T$.
\end{theorem}
\begin{proof}
It is not difficult to prove that $\Xi_T: \nF[z] \to \cL(\cV)$ is linear. We will prove that $\Xi_T: \nF[z] \to \cL(\cV)$ is multiplicative, that is, for all $p, q \in \nF[z]$ we have $\Xi_T(pq) = \Xi_T(p) \Xi_T(q)$. To prove this let $p, q \in \nF[z]$ be arbitrary and given in \eqref{eq-pq}. Then
\begin{alignat*}{2}
\Xi_T(p)\Xi_T(q)& = \Biggl(\sum_{i=0}^m\alpha_iT^i\Biggr)\Biggl(\sum_{j=0}^n \beta_jT^j \Biggr)
 & \qquad & \text{(by definition in \eqref{eq-Xi})} \\
&= \sum_{i=0}^m\sum_{j=0}^n  \alpha_i\beta_jT^{i+j} & \qquad & \text{(since $\cL(\cV)$ is an algebra)} \\
&=  \sum_{k=0}^{m+n}\left(\sum_{i+j=k}\alpha_i\beta_j\right)T^k  & \qquad & \text{(since $\cL(\cV)$ is a vector space)} \\
&=   \Xi_T(pq) & \qquad & \text{(by definition in \eqref{eq-Xi})}.
\end{alignat*}
This proves the multiplicative property of $\Xi_T$.

The fact that $\cA_T$ is the range of $\Xi_T$ is obvious.
\end{proof}


\begin{corollary}
Let $T \in \cL(\cV)$.  The subspace $\cA_T$ of $\cL(\cV)$ is a commutative subalgebra of $\cL(\cV)$.
\end{corollary}
\begin{proof}
Let $Q, S \in \cA_T$. Since $\cA_T$ is the range of $\Xi_T$ there exist $p, q \in \nF[z]$ such that $Q = \Xi_T(p)$ and $S = \Xi_T(q)$. Then, since $\Xi_T$ is an algebra homomorphism we have
\[
QS = \Xi_T(p)\Xi_T(p) = \Xi_T(pq) = \Xi_T(qp) = \Xi_T(q)\Xi_T(p) = SQ.
\]
This sequence of equalities shows that $QS \in \ran(\Xi_T) = \cA_T$ and  $QS = SQ$. That is $\cA_T$ is closed with respect to the operator composition and the operator composition on $\cA_T$ is commutative.
\end{proof}

\begin{corollary} \label{co-linfact}
Let $\cV$ be a complex vector space and let $T \in \cL(\cV)$ be a nonzero operator. Then for every $p \in \nC[z]$ such that $\deg p \geq 1$ there exist a nonzero $\alpha \in \nC$ and $z_1, \ldots, z_m \in \nC$  such that
\[
\Xi_T(p) = p(T) = \alpha(T-z_1I) \, \cdots \,(T-z_m I).
\]
\end{corollary}
\begin{proof}
Let $p \in \nC[z]$ such that $m = \deg p \geq 1$.    Then there exist $\alpha_0, \ldots, \alpha_m \in \nC$ such that $\alpha_m \neq 0$ such that
\[
 p(z) = \sum_{k=0}^m \alpha_j z^j.
\]
By the Fundamental Theorem of Algebra there exist nonzero $\alpha \in \nC$ and $z_1,\dots,z_m \in \nC$ such that
\[
p(z) = \alpha(z-z_1)\cdots(z-z_m).
\]
Here $\alpha = \alpha_m$ and $z_1,\dots,z_m$ are the roots of $p$.
Since $\Xi_T$ is an algebra homomorphism we have
\[
p(T) = \Xi_T(p) = \alpha \, \Xi_T(z-z_1) \, \cdots  \, \Xi_T(z-z_m) =  \alpha(T-z_1 I)\, \cdots \, (T-z_m I).
\]
This completes the proof.
\end{proof}

\begin{lemma} \label{le-inj-n}
Let $n \in \nN$ and $S_1,\ldots,S_n \in \cL(\cV)$. If $S_1,\ldots,S_n$ are all injective, then $S_1 \, \cdots \, S_n$ is injective.
\end{lemma}
\begin{proof}
We proceed by Mathematical Induction. The base step is trivial. It is useful to prove the implication for $n = 2$. Assume that $S, T \in \cL(\cV)$ are injective and let $u,v \in \cV$ be such that $u \neq v$. Then, since $T$ is injective, $T u \neq T v$. Since $S$ is injective, $S(T u) \neq S(T v)$. Thus, $S T$ is injective.

Next we prove the inductive step. Let $m \in \nN$ and assume that $S_1 \, \cdots \, S_m$ is injective whenever $S_1,\ldots,S_m \in \cL(\cV)$ are all injective. (This is the inductive hypothesis.) Now assume that $S_1,\ldots,S_m,S_{m+1} \in \cL(\cV)$ are all injective. By the inductive hypothesis the operator $S = S_1 \, \cdots \, S_m$ is injective. Since by assumption $T = S_{m+1}$ is injective, the already proved claim for $n=2$ yields that
\[
ST = S_1 \, \cdots \, S_m \, S_{m+1}
\]
is injective. This completes the proof.
\end{proof}


\begin{theorem} \label{th-ev-ex}
Let $\cV$ be a nontrivial finite dimensional vector space over $\nC$.  Let $T\in \cL(\cV)$. Then there exists a $\lambda \in \nC$ and $v\in \cV$ such that $v\neq 0_v$ and $Tv = \lambda v$.

\begin{proof}
The claim of the theorem is trivial if $T$ is a scalar multiple of the identity operator. So, assume that $T\in \cL(\cV)$ is not a scalar multiple of the identity operator.

Since $\cL(\cV)$ is finite dimensional and $\nC[z]$ is infinite dimensional, by the Rank-nullity theorem the operator $\Xi_T$ is not injective. Thus $\nul(\Xi_T) \neq \{0_v\}$. Hence, there exists a $p\in \nC[z]$ such that $p\neq 0_{\nC[z]}$ and $\Xi_T(p) = p(T) = 0_{\cL(\cV)}$. Since $p\neq 0_{\nC[z]}$ then $\deg p \ge 0$. Note that if $\deg p = 0$ then $p(z) = c$ for some $c\in \nC$ for all $z\in \nC$.  Thus $\Xi_T(p) = p(T) = cI_{\cV}$. This is not possible since we assume that $T$ is not a scalar multiple of the identity. Hence $\deg p > 0$. By Corollary~\ref{co-linfact} there exists $\alpha \neq 0$ and $z_1,\ldots,z_m \in \nC$ such that
\[
0_{\cL(\cV)} = \Xi_T(p) = p(T) = \alpha(T-z_1 I)\, \cdots \, (T-z_m I).
\]
Since $0_{\cL(\cV)}$ is not injective, Lemma~\ref{le-inj-n} implies that there exists $j \in \{1,\ldots,m\}$ such that $T-z_j I$ is not injective. That is, there exists $v \in \cV$, $v \neq 0_\cV$ such that
\[
(T- z_j I) v = 0.
\]
Setting $\lambda = z_j$ completes the proof.
\end{proof}
\end{theorem}

\begin{remark}
Note that the proof in the textbook is different. The proof in the textbook is somewhat more elementary since it does not use the Rank-nullity theorem.
\end{remark}

\begin{definition} Let $\cV$ be a vector space over $\mathbb{F}$, $T\in\cL(\cV)$. A scalar $\lambda \in \nF$ is an {\em eigenvalue} of $T$ if there exists $v \in \cV$ such that $v \neq 0$ and $Tv = \lambda v$.  The subspace $\nul(T-\lambda I)$ of $\cV$ is called the {\em eigenspace} of $T$ corresponding to $\lambda$
\end{definition}

\begin{definition}
Let $\cV$ be a finite dimensional vector space over $\nF$. Let $T \in \cL(\cV)$. The set of all eigenvalues of $T$ is denoted by $\sigma(T)$. It is called the {\em spectrum} of $T$.
\end{definition}




The next theorem can be stated in English simply as: Eigenvectors corresponding to distinct eigenvalues are linearly independent.

\begin{theorem}
Let $\cV$ be a vector space over $\nF$, $T \in \cL(\cV)$ and $n \in \nN$.  If the following two conditions are satisfied:
\begin{enumerate}
\vspace*{-7pt}
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\item
$\lambda_1, \ldots , \lambda_n \in \nF$ are such that $\lambda_i \neq \lambda_j$ for all $i,j \in \{1,\ldots,n\}$ such that $i \neq j$,
\item
$v_1, \ldots , v_n \in \cV$ are such that $Tv_k = \lambda_k v_k$ and $v_k \neq 0$ for all $k \in \{1, \ldots ,n\}$,
\end{enumerate} \vspace*{-5pt}
then $\{v_1, \ldots, v_n\}$ is linearly independent.
\end{theorem}

\begin{proof}
We will prove this by using the mathematical induction on $n$. For the base case, we will prove the claim  for $n = 1$.  Let $\lambda_1 \in \nF$ and let $v_1 \in \cV$ be such that $v_1 \neq 0$ and $Tv_1 = \lambda_1 v_1$. Since $v_1 \neq 0$, we conclude that $\{v_1\}$ is linearly independent.

Next we prove the inductive step.  Let $m \in \nN$ be arbitrary.  The inductive hypothesis is the assumption that the following implication holds.

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 \fcolorbox{green}{white}{
\begin{minipage}{0.79\textwidth}
If the following two conditions are satisfied:
\begin{enumerate}
\vspace*{-7pt}
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\item \label{i-ih-1}
$\mu_1, \ldots , \mu_m \in \nF$ are such that $\mu_i \neq \mu_j$ for all $i,j \in \{1,\ldots,m\}$ such that $i \neq j$,
\item \label{i-ih-2}
$w_1, \ldots , w_m \in \cV$ are such that $Tw_k = \mu_k w_k$ and $w_k \neq 0$ for all $k \in \{1, \ldots , m\}$,
\end{enumerate} \vspace*{-5pt}
then $\{w_1, \ldots, w_m\}$ is linearly independent.
\end{minipage}
}
\end{center}

We need to prove the following implication
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 \fcolorbox{red}{white}{
\begin{minipage}{0.83\textwidth}
If the following two conditions are satisfied:
\begin{enumerate}
\vspace*{-7pt}
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\item \label{i-as-1}
$\lambda_1, \ldots , \lambda_{m+1} \in \nF$ are such that $\lambda_i \neq \lambda_j$ for all $i,j \in \{1,\ldots,{m+1}\}$ such that $i \neq j$,
\item \label{i-as-2}
$v_1, \ldots , v_{m+1} \in \cV$ are such that $Tv_k = \lambda_k v_k$ and $v_k \neq 0$ for all $k \in \{1, \ldots ,{m+1}\}$,
\end{enumerate} \vspace*{-5pt}
then $\{v_1, \ldots, v_{m+1}\}$ is linearly independent.
\end{minipage}
}
\end{center}

Assume (\ref{i-as-1}) and (\ref{i-as-2}) in the red box. We need to prove that $\{v_1, \ldots, v_{m+1}\}$ is linearly independent.

Let $\alpha_1, \ldots, \alpha_{m+1} \in \nF$ be such that
\begin{equation} \label{eq-li-st1}
\alpha_1 v_1 + \cdots + \alpha_{m} v_{m} + \alpha_{m+1} v_{m+1} = 0.
\end{equation}
Applying $T \in \cL(\cV)$ to both sides of \eqref{eq-li-st1}, using the linearity of $T$ and  assumption (\ref{i-as-2}) we get
\begin{equation} \label{eq-li-st2}
\alpha_1 \lambda_1 v_1 + \cdots + \alpha_{m} \lambda_{m} v_{m} + \alpha_{m+1} \lambda_{m+1} v_{m+1} = 0.
\end{equation}
Multiplying both sides of  \eqref{eq-li-st1} by $\lambda_{m+1}$ we get
\begin{equation} \label{eq-li-st3}
\alpha_1 \lambda_{m+1} v_1 + \cdots + \alpha_{m} \lambda_{m+1} v_{m} + \alpha_{m+1} \lambda_{m+1} v_{m+1} = 0.
\end{equation}
Subtracting \eqref{eq-li-st3} from \eqref{eq-li-st2} we get
\begin{equation*} % \label{eq-li-st4}
\alpha_1 (\lambda_1 - \lambda_{m+1}) v_1 + \cdots + \alpha_{m} (\lambda_{m}- \lambda_{m+1}) v_{m} = 0.
\end{equation*}
Since by assumption (\ref{i-as-1}) we have $\lambda_j - \lambda_{m+1} \neq 0$ for all $j \in \{1,\ldots,m\}$, setting
\[
w_j = (\lambda_j - \lambda_{m+1}) v_j, \qquad j \in \{1,\ldots,m\},
\]
and taking into account (\ref{i-as-2}) we have
\begin{equation} \label{eq-li-st5}
w_j \neq 0 \qquad \text{and} \qquad Tw_j = \lambda_jw_j \qquad \text{for all} \qquad j \in \{1,\ldots,m\}.
\end{equation}
Thus, by (\ref{i-as-1}) and \eqref{eq-li-st5}, the scalars $\lambda_1, \ldots, \lambda_m$ and vectors $w_1, \ldots, w_m$ satisfy assumptions (\ref{i-ih-1}) and (\ref{i-ih-2}) of the inductive hypothesis (the green box). Consequently, the vectors $w_1, \ldots, w_m$ are linearly independent.
Since by \eqref{eq-li-st5} we have
\begin{equation*}
\alpha_1 w_1 + \cdots + \alpha_{m} w_{m} = 0,
\end{equation*}
it follows that $\alpha_1 = \cdots = \alpha_m = 0$. Substituting these values in \eqref{eq-li-st1} we get $\alpha_{m+1} v_{m+1} = 0$. Since by (\ref{i-as-2}), $v_{m+1} \neq 0$ we conclude that $\alpha_{m+1} = 0$. This completes the proof of the linear independence of $v_1, \ldots, v_{m+1}$. \end{proof}

\textbf{Corollary 3:} Let $\cV$ be a finite dimensional vector space over $\nF$ and let $T\in\cL(\cV).$ Then $T$ has at most $n=\dim\cV$ distinct eigenvalues.

\begin{proof}

Let $\cB$ be a basis of $\cV$ where $\cB=\{u_1,...,u_n\}.$ Then $|\cB|=n$ and $\lspan\cB=\cV.$ Let $\cC=\{v_1,...,v_m\}$ be eigenvectors corresponding to $m$ distinct eigenvalues. Then $\cC$ is a linearly independent set with $|\cC|=m$. By the Steinitz Exchange Lemma, $m\le n$. Consequently, $T$ has at most $n$ distinct eigenvalues.
\end{proof}

\begin{definition} Let $\cV$ be a vector space over $\mathbb{F}$ and $T\in\cL(V)$. A subspace $\cU$ of $\cV$  is called an {\em invariant subspace} under $T$ if $T(\cU)\subseteq \cU$.
\end{definition}


The following proposition is straightforward.
\begin{proposition}
Let $S, T \in \cL(\cV)$ be such that $ST = TS$. Then $\nul T$ is invariant under $S$ and $\nul S$ is invariant under $T$.  In particular, all eigenspaces of $T$ are invariant under $T$.
\end{proposition}


\begin{definition} A matrix $A\in \mathbb{F}^{n\times n}$ with entries  $a_{ij}$, $i,j \in \{1,\dots,n\}$ is called  {\em upper triangular} if $a_{i,j} = 0$ for all $i,j \in \{1,\dots,n\}$ such that $i > j$.
\end{definition}

\begin{definition}
Let $\cV$ be a finite dimensional vector space over $\nF$ with $n = \dim \cV > 0$. Let $T \in \cL(\cV)$. A sequence of nontrivial subspaces $\cU_1, \ldots, \cU_n$ of $\cV$ such that
\begin{equation} \label{eq-fan-sss}
\cU_1 \subsetneq \cU_2 \subsetneq \cdots \subsetneq \cU_n
\end{equation}
 and
\[
T \cU_k \subseteq \cU_k \qquad \text{for all} \qquad k \in \{1,\ldots,n\}
\]
is called a {\em fan} for $T$ in $\cV$.  A basis $\{v_1, \ldots, v_n\}$ of $\cV$ is called a {\em fan basis} corresponding to $T$ if the subspaces
\[
\cV_k = \lspan\{v_1,\ldots,v_k\}, \qquad k \in \{1,\ldots,n\},
\]
form a fan for $T$.
\end{definition}

Notice that \eqref{eq-fan-sss} implies
\[
1 \leq \dim \cU_1 < \dim \cU_2 < \cdots < \dim \cU_n \leq n.
\]
Consequently, if $\cU_1, \ldots, \cU_n$ is a fan for $T$ we have $\dim \cU_k = k$ for all $k \in \{1,\ldots,n\}$. In particular $\cU_n = \cV$.


\begin{theorem}[Theorem 5.12] \label{th-utc}
Let $\cV$ be a finite dimensional vector space over $\nF$ with $\dim\cV=n$ and let $T\in\cL(\cV)$. Let $\cB=\{v_1,\dots,v_n\}$ be a basis of $\cV$ and set
\[
\cV_k = \lspan\{v_1,\ldots,v_k\},  \qquad k\in\{1,\ldots,n\}.
\]
The following statements are equivalent.
\begin{enumerate}
\vspace*{-7pt}
% \setlength{\topsep}{-30pt}\setlength{\partopsep}{-10pt}
\renewcommand*\theenumi{\alph{enumi}}
\renewcommand*\labelenumi{\rm{(\theenumi)}}
\setlength{\itemsep}{-2pt}% \itemsep0em
\item \label{i-utc-1}
$M^{\cB}_{\cB}(T)$ is upper-triangular.
\item \label{i-utc-2}
$Tv_k\in \cV_k$ for all $k\in\{1,\ldots,n\}$.
\item \label{i-utc-3}
$T\cV_k\subseteq\cV_k$ for all $k\in\{1,\ldots,n\}$.
\item \label{i-utc-4}
$\cB$ is a fan basis corresponding to $T$.
\end{enumerate}
\end{theorem}

\begin{proof}
(\ref{i-utc-1}) $\Rightarrow$ (\ref{i-utc-2}). Assume that $M^{\cB}_{\cB}(T)$ is upper triangular. That is
\[
M^{\cB}_{\cB}(T)=\begin{bmatrix}
a_{11}&a_{12}&\cdots&a_{1k}&\cdots&a_{1n}\\
 0&a_{22}&\cdots&a_{2k}&\cdots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots&&\vdots\\
0&0&\cdots&a_{kk}&\cdots&0\\
 \vdots&\vdots&&\vdots&\ddots&\vdots\\
0&0&\cdots&0&\cdots&a_{nn}\end{bmatrix}.
\]
Let $k\in\{1,\ldots,n\}$ be arbitrary. Then, by the definition of  $M^{\cB}_{\cB}(T)$,
\[
C_\cB(Tv_k) = \begin{bmatrix}
  a_{1k}\\
  \vdots\\
  a_{kk}\\
  0\\
 \vdots\\
  0\\
\end{bmatrix}.
\]
Consequently, by the definition of $C_\cB$, we have
\[
Tv_k = a_{1k} v_1 + \cdots + a_{kk} v_k \in \lspan\{v_1, \ldots,v_k\} = \cV_k.
\]
Thus, (\ref{i-utc-2}) is proved.

(\ref{i-utc-2}) $\Rightarrow$ (\ref{i-utc-1}). Assume that $Tv_k\in \cV_k$ for all $k \in\{1,\ldots,n\}$.  Let $a_{ij}$, $i,j \in \{1,\dots,n\}$, be the entries of $M^{\cB}_{\cB}(T)$. Let $j \in \{1,\ldots,n\}$  be arbitrary. Since $Tv_j \in \cV_j$ there exist $\alpha_1, \ldots, \alpha_j \in \nF$ such that
\[
Tv_j = \alpha_1 v_1 + \cdots + \alpha_j v_j.
\]
By the definition of $C_\cB$ we have
\[
C_\cB(Tv_j) = \begin{bmatrix}
  \alpha_1\\
  \vdots\\
  \alpha_j\\
  0\\
 \vdots\\
  0\\
\end{bmatrix}.
\]
On the other side, by the definition of  $M^{\cB}_{\cB}(T)$, we have
\[
C_\cB(Tv_j) = \begin{bmatrix}
  a_{1k}\\
  \vdots\\
  a_{jj}\\
  a_{j+1,j}\\
 \vdots\\
  a_{nj}\\
\end{bmatrix}.
\]
The last two equalities, and the fact that $C_\cB$ is a function, imply $a_{ij} = 0$ for all $i \in \{j+1,\ldots,n\}$. This proves (\ref{i-utc-1}).

(\ref{i-utc-2}) $\Rightarrow$ (\ref{i-utc-3}). Suppose $Tv_k\in\cV_k= {span}\{v_1,\ldots,v_k\}$ for all $k\in\{1,\ldots,n\}$. Let $v \in \cV_k$.  Then $v=\alpha_1v_1+\cdots+\alpha_kv_k$.  Applying $T$, we get $Tv=\alpha_1 Tv_1+\cdots+\alpha_kTv_k$.  Thus,
\begin{equation} \label{eq-Tv-span1}
Tv \in \lspan\{Tv_1,\ldots,Tv_k\}.
 \end{equation}
Since
\[
Tv_j \in \cV_j \subset \cV_k \qquad \text{for all} \qquad j \in \{1,\ldots,k\},
\]
we have
\[
\lspan\{Tv_1,\ldots,Tv_k\} \subseteq \cV_k.
\]
Together with \eqref{eq-Tv-span1}, this proves (\ref{i-utc-3}).

(\ref{i-utc-3}) $\Rightarrow$ (\ref{i-utc-2}). Suppose $T\cV_k \subseteq \cV_k$ for all  $k\in\{1,\ldots,n\}$. Then since $v_k\in \cV_k$, we have $Tv_k\in \cV_k$ for each $k\in\{1,\ldots,n\}$.

(\ref{i-utc-3}) $\Leftrightarrow$ (\ref{i-utc-4}) follows from the definition of a fan basis corresponding to $T$.
\end{proof}

\begin{theorem} \label{th-invc}
Let $\cV$ be a finite dimensional vector space over $\nF$ with $\dim\cV=n$, and let $T\in\cL(\cV)$. Let $\cB=\{v_1,\dots,v_n\}$ be a basis of $\cV$ such that $M_{\cB}^{\cB}(T)$ is upper triangular with diagonal entries $a_{jj}$, $j\in\{1,\dots,n\}$. Then $T$ is not injective if and only if there exists $j\in\{1,\dots,n\}$ such that $a_{jj} = 0$.
\end{theorem}

\begin{proof}
In this proof we set
\[
\cV_k = \lspan\{v_1,...,v_k\}, \qquad k\in\{1,...,n\}.
\]
Then
\begin{equation} \label{eq-fan-sub}
\cV_1\subsetneq\cV_2\subsetneq \ldots \subsetneq\cV_n
\end{equation}
and by Theorem~\ref{th-utc}, $T\cV_k \subseteq \cV_k$.

We first prove the ``only if'' part.  Assume that $T$ is not injective.
Consider the set
\[
\nK = \bigl\{k \in\{1,...,n\}\, : \, T\cV_k \subsetneq \cV_k \bigr\}
\]
Since $T$ is not injective, $\nul T \neq \{0_\cV\}$. Thus by the Rank-Nullity Theorem, $\ran T \subsetneq \cV = \cV_n$. Since $T\cV_n=\ran T$, it follows that $T\cV_n\subsetneq\cV_n$.  Therefore $n\in\nK$. Hence the set $\nK$ is a nonempty set of positive integers. Hence, by the Well-Ordering principle $\min \nK$ exists. Set $j=\min\nK$.

If $j=1$, then $\dim\cV_1 = 1$, but since $T\cV_1\subsetneq\cV_1$ it must be that $\dim T\cV_1=0$. Thus $T\cV_1 = \{0_\cV\}$, so $Tv_1=0_v$. Hence $C_\cB(T)=[0\, \cdots\, 0]^\top$ and so $a_{jj}=0$.
If $j>1$, then $j-1 \in \{1,\ldots,n\}$ but $j-1 \not\in \nK$. By Theorem~\ref{th-utc}, $T\cV_{j-1} \subseteq \cV_{j-1}$ and, since $j-1 \not\in \nK$, $T\cV_{j-1}\subsetneq \cV_{j-1}$ is not true.   Hence $T\cV_{j-1}=\cV_{j-1}$. Since $j \in \nK$, we have $T\cV_{j}\subsetneq \cV_{j}$. Now we have
\begin{equation*}
\cV_{j-1} = T \cV_{j-1} \subseteq T \cV_{j} \subsetneq \cV_{j}.
\end{equation*}
Consequently,
\begin{equation*}
j-1 = \dim \cV_{j-1}  \leq  \dim \bigl(T \cV_{j}\bigr) < \dim \cV_{j} = j,
\end{equation*}
which implies $\dim \bigl(T \cV_{j}\bigr) = j-1$ and therefore $T \cV_{j} = \cV_{j-1}$. This implies that there exist $\alpha_1,\ldots, \alpha_{j-1} \in \nF$ such that
\begin{equation*}
Tv_j  = \alpha_1 v_1+\cdots+ \alpha_{j-1}v_{j-1}.
\end{equation*}
By the definition of $M^{\cB}_{\cB}$ this implies that $a_{jj} = 0$.


Next we prove the ``if'' part. Assume that there exists $j\in\{1,...,n\}$ such that $a_{jj}=0$. Then
\begin{equation} \label{eq-inv-if1}
Tv_j = a_{1j} v_1 + \cdots + a_{j-1,j} v_{j-1} + 0 v_j \in \cV_{j-1}.
\end{equation}
By Theorem~\ref{th-utc} and \eqref{eq-fan-sub} we have
\begin{equation} \label{eq-inv-if2}
Tv_i  \in \cV_i \subseteq \cV_{j-1} \qquad \text{for all} \qquad i \in \{1,\ldots,j-1\}.
\end{equation}
Now \eqref{eq-inv-if1} and \eqref{eq-inv-if2} imply $Tv_i  \in \cV_{j-1}$ for all $i \in \{1,\ldots,j\}$ and consequently $T\cV_j \subseteq \cV_j$.
To complete the proof, we apply the Rank-Nullity theorem to the restriction $T|_{\cV_j}$ of $T$ to the subspace $\cV_j$:
\[
\dim \nul\bigl(T|_{\cV_j}\bigr) + \dim \ran\bigl(T|_{\cV_j}\bigr) = j.
\]
Since $T\cV_j \subseteq \cV_j$ implies $\dim \ran\bigl(T|_{\cV_j}\bigr) \leq j-1$, we conclude
\[
\dim \nul\bigl(T|_{\cV_j}\bigr) \geq 1.
\]
Thus $\nul\bigl(T|_{\cV_j}\bigr) \neq \{0_\cV\}$, that is, there exists $v \in \cV_j$ such that $v\neq 0$ and $Tv = T|_{\cV_j}v = 0$. This proves that $T$ is not invertible.
\end{proof}

%Thm 5.16
\begin{corollary}[Theorem 5.16] \label{co-invc}
Let $\cV$ be a finite dimensional vector space over $\nF$ with $\dim\cV=n$, and let $T\in\cL(\cV)$. Let $\cB$ be a basis of $\cV$ such that $M_{\cB}^{\cB}(T)$ is upper triangular with diagonal entries $a_{jj}$, $j\in\{1,\dots,n\}$. The following statements are equivalent.
\begin{enumerate}
\vspace*{-7pt}
% \setlength{\topsep}{-30pt}\setlength{\partopsep}{-10pt}
\renewcommand*\theenumi{\alph{enumi}}
\renewcommand*\labelenumi{\rm{(\theenumi)}}
\setlength{\itemsep}{-2pt}% \itemsep0em
\item \label{i-invc-1}
$T$ is not injective.
\item \label{i-invc-2}
$T$ is not invertible.
\item \label{i-invc-3}
$0$ is an eigenvalue of $T$.
\item \label{i-invc-4}
$\prod_{i=1}^n a_{ii}=0$.
\item \label{i-invc-5}
There exists $j\in\{1,\ldots,n\}$ such that $a_{jj}=0$.
\end{enumerate}
\end{corollary}

\begin{proof}
The equivalence (\ref{i-invc-1}) $\Leftrightarrow$ (\ref{i-invc-2}) follows from the Rank-nullity theorem and it has been proved earlier. The equivalence (\ref{i-invc-1}) $\Leftrightarrow$ (\ref{i-invc-3}) is almost trivial. The equivalence (\ref{i-invc-1}) $\Leftrightarrow$ (\ref{i-invc-5}) was proved in Theorem~\ref{th-invc} and The equivalence (\ref{i-invc-4}) $\Leftrightarrow$ (\ref{i-invc-5}) is should have been proved in high school.
\end{proof}

\begin{theorem} \label{th-sp-di}
Let $\cV$ be a finite dimensional vector space over $\nF$ with $\dim\cV=n$, and let $T\in\cL(\cV)$. Let $\cB$ be a basis of $\cV$ such that $M_{\cB}^{\cB}(T)$ is upper triangular with diagonal entries $a_{jj}$, $j\in\{1,\dots,n\}$.  Then
\[
\sigma(T) = \bigl\{a_{jj} : j \in \{1, ..., n\}\bigr\}.
\]
\end{theorem}
\begin{proof}
Notice that $M_\cB^\cB : \cL(V) \to \nF^{n\times n}$ is a linear operator. Therefore
\[
M_\cB^\cB(T - \lambda I) = M_\cB^\cB(T) - \lambda M_\cB^\cB(I) = M_\cB^\cB(T) - \lambda I_n.
\]
Here $I_n$ denotes the identity matrix in $\nF^{n\times n}$. As $M_\cB^\cB(T)$ and $M_\cB^\cB(I) = I_n$ are upper triangular, $M_\cB^\cB(T - \lambda I)$ is upper triangular as well with diagonal entries $a_{jj} - \lambda$, $j \in \{1, ...,n\}$.

To prove a set equality we need to prove two inclusions.

First we prove $\subseteq$.
Let $\lambda \in \sigma(T)$. Because $\lambda$ is an eigenvalue, $T - \lambda I$ is not injective. Because $T - \lambda I$ is not injective, by Theorem~\ref{th-invc} one of its diagonal entries is zero. So there exists $i \in \{1, ..., n\}$ such that $a_{ii} - \lambda = 0$. Thus $\lambda = a_{ii}$.  So $\sigma(T) \subseteq \bigl\{a_{jj} : j \in \{1, ..., n\}\bigr\}$.

Next we prove $\supseteq$.
Let $a_{ii} \in \bigl\{a_{jj} : j \in \{1, ..., n\}\bigr\}$ be arbitrary. Then $a_{ii} - a_{ii} = 0$. By Theorem~\ref{th-invc} and the note at the beginning of this proof $T - a_{ii} I$ is not injective. This implies that $a_{ii}$ is an eigenvalue of $T$. Thus $a_{ii} \in \sigma(T)$.
This completes the proof.
\end{proof}

\begin{remark}
Theorem~\ref{th-sp-di} is identical to Theorem~5.18 in the textbook.
\end{remark}

%Thm 5.13
\begin{theorem}[Theorem~5.13] \label{th-ex-up}
Let $\cV$ be a nonzero finite dimensional complex vector space.
If $\dim \cV = n$ and  $T\in\cL(\cV)$, then there exists a basis $\cB$ of $\cV$ such that $M^{\cB}_{\cB}(T)$ is upper-triangular.
\end{theorem}

\begin{proof}
We proceed by the complete induction on $n=\dim(\cV)$.

The base case is trivial. Assume $\dim \cV = 1$ and $T\in\cL(\cV)$.  Set $\cB=\{v\}$, where $\,u\in\cV\backslash\{0_u\}$ is arbitrary. Then there exists $\lambda \in \nC$ such that $Tu = \lambda u$. Then, $M^{\cB}_{\cB}(T) = \bigl[\lambda\bigr]$.

Now we prove the inductive step. Let $m \in \nN$ be arbitrary. The inductive hypothesis is

\setlength{\fboxrule}{2pt}%
 \fboxsep 1mm

\vspace*{-7pt}
\begin{center}
 \fcolorbox{green}{white}{
\begin{minipage}{0.85\textwidth}
For every $k \in \{1,\ldots,m\}$ the following implication holds: If $\dim \cU = k$ and  $S\in\cL(\cU)$, then there exists a basis $\cA$ of $\cU$ such that $M^{\cA}_{\cA}(S)$ is upper-triangular.
\end{minipage}
}
\end{center}

We complete the inductive step, we need to prove the implication:

\setlength{\fboxrule}{2pt}%
 \fboxsep 1mm

\vspace*{-7pt}
\begin{center}
 \fcolorbox{red}{white}{
\begin{minipage}{0.85\textwidth}
If $\dim \cV = m+1$ and  $T\in\cL(\cV)$, then there exists a basis $\cB$ of $\cV$ such that $M^{\cB}_{\cB}(T)$ is upper-triangular.
\end{minipage}
}
\end{center}

To prove the red implication assume that $\dim \cV = m+1$ and  $T\in\cL(\cV)$. By Theorem~\ref{th-ev-ex} the operator $T$ has an eigenvalue.  Let $\lambda$ be an eigenvalue of $T$.  Set $\cU = \ran(T-\lambda I)$.  Because $(T-\lambda I)$ is not injective, it is not surjective, and thus $ k = \dim(\cU) < \dim(\cV) = m+1$. That is $k \in \{1,\ldots,m\}$.

Moreover, $T\cU=\cU$.  To show this, let $u\in\cU$.  Then $Tu=(T-\lambda I)u+\lambda u$.  Since $(T-\lambda I)u \in \cU$ and $\lambda u\in\cU$, $\,Tu\in\cU$.  Hence, $S = T|_\cU$ is an operator on $\cU$.

By the inductive hypothesis (the green box),  there exists a basis $\cA=\{u_1,\ldots,u_k\}$ of $\cU$ such that $M^{\cA}_{\cA}(S)$ is upper-triangular.  That is,
\[
Tu_j = Su_j \in \lspan \{u_1,\ldots,u_j\} \qquad \text{for all} \qquad j\in\{1,\ldots,k\}.
\]
Extend $\cA$ to a basis $\cB =  \{u_1,\ldots,u_k,v_1,\ldots,v_{n-k}\}$ of $\cV$.  Since
\[
Tv_j=(T-\lambda I)v_j + \lambda v_j,  \qquad j \in \{1,\ldots,n-k\}
\]
where $(T-\lambda I)v_j \in \cU$, we have
\[
Tv_j \in \lspan\{u_1,\ldots,u_m,v_j\} \subseteq \lspan\{u_1,\ldots,u_m,v_1,\ldots,v_j\} \qquad \text{for all} \qquad j\in\{1,\ldots,n-k\}.
\]
By Theorem~\ref{th-utc}, $M^{\cB}_{\cB}(T)$ is upper-triangular.
\end{proof}



%%%%CHAPER 6%%%%%%
\section{Inner Product Spaces}

%%%%DEFINE INNER PRODUCT%%%%%

We will first introduce several ``dot-product-like'' objects. We start with the most general.

\begin{definition}
Let $\cV$ be a vector space over a scalar field $\nF$.  A function
\[
[\,\cdot\,,\cdot\,]: \cV\times\cV \to \nF
\]
is a {\em sesquilinear form} on $\cV$ if the following two conditions are satisfied.
\begin{enumerate}
\renewcommand*\theenumi{\alph{enumi}}
\renewcommand*\labelenumi{(\theenumi)}
\setlength{\itemsep}{-2pt}
\item
(linearity in the first variable) \ \ $\forall \, \alpha, \beta \in \nF \quad \forall \, u, v, w  \in \cV \quad [\alpha u + \beta v, w] = \alpha [u,w] + \beta [v,w]$.
\item
(anti-linearity in the second variable) \ \ $\forall \, \alpha, \beta \in \nF \quad \forall \, u, v, w  \in \cV \quad [u, \alpha v + \beta w] = \co{\alpha} [u,v] + \co{\beta} [u,w]$.
\end{enumerate}
\end{definition}

\begin{example} \label{ex-sfnh}
Let $M \in \nC^{n\times n}$ be arbitrary. Then
\[
[\V{x}, \V{y}] = (M\V{x})\cdot \V{y} , \qquad \V{x}, \V{y} \in \nC^n,
\]
is a sesquilinear form on the complex vector space $\nC^n$. Here $\cdot$ denotes the usual dot product in $\nC$.
\end{example}

\begin{theorem}
Let $\cV$ be a vector space over a scalar field $\nF$ and let $[\,\cdot\,,\cdot\,]: \cV\times\cV \to \nF$ be a sesquilinear form  on $\cV$. If $\I \in \nF$, then
\begin{equation} \label{eq-pi}
[u,v] = \frac{1}{4} \sum_{k=0}^3 \I^k \bigl[u + \I^k v, u + \I^k v\bigr]
\end{equation}
for all $u,v \in \cV$.
\end{theorem}

\begin{corollary}
Let $\cV$ be a vector space over a scalar field $\nF$ and let $\kip: \cV\times\cV \to \nF$ be a sesquilinear form  on $\cV$. If $\I \in \nF$ and $[v,v]= 0$ for all $v \in \cV$, then $[u,v] = 0$ for all $u, v \in \cV$.
\end{corollary}

\begin{definition}
Let $\cV$ be a vector space over a scalar field $\nF$.  A sesquilinear form  $\kip: \cV\times\cV \to \nF$ is {\em hermitian} if
\begin{enumerate}
\setcounter{enumi}{2}
\renewcommand*\theenumi{\alph{enumi}}
\renewcommand*\labelenumi{(\theenumi)}
\item
(hermiticity) \ \ $\forall \, u, v \in \cV \quad \co{[u,v]} = [v,u]$.
\end{enumerate}
A hermitian sesquilinear form is also called an {\em inner product}.
\end{definition}

Let $\kip$ be an inner product on $\cV$. The hermiticity of $\kip$ implies that $\co{[v,v]} = [v,v]$ for all $v \in \cV$. Thus $[v,v] \in \nR$ for all $v \in \cV$.  The natural trichotomy that arises is the motivation for the following definition.

\begin{definition}
An inner product $\kip$ on $\cV$ is called {\em nonnegative} if $\quad [v,v] \geq 0$ for all $v \in \cV$, it is called {\em nonpositive} if $\quad [v,v] \leq 0$ for all $v \in \cV$, and it is called {\em indefinite} if there exist $u \in \cV$ and $v \in \cV$ such that $[u,u] < 0$ and $[v,v] > 0$.
\end{definition}


The following implication that you might have learned in high school will be useful below.

\begin{theorem}[High School Theorem]
Let $a, b, c$ be real numbers. Assume $a \geq 0$. Then the following implication holds:
\begin{equation} \label{eq-impl}
\forall \, x \in {\mathbb Q} \quad a x^2 + b x + c \geq 0 \qquad \Rightarrow \qquad b^2-4ac \leq 0.
\end{equation}
\end{theorem}

\begin{theorem}[Cauchy-Bunyakovsky-Schwartz Inequality]
Let ${\mathcal V}$ be a vector space over $\nF$ and let $\ahip$ be a nonnegative inner product on $\cV$. Then
\begin{equation} \label{eq-CSBi}
\forall \, u, v  \in {\mathcal V} \quad  |\langle u, v \rangle |^2  \leq \langle u,u \rangle \langle v,v \rangle.
\end{equation}
The equality occurs in \eqref{eq-CSBi} if and only if there exists $\alpha, \beta \in \nF$ not both $0$ such that $\la \alpha u + \beta v, \alpha u + \beta v \ra = 0$. 
\end{theorem}
\begin{proof}
Let $u, v \in {\mathcal V}$ be arbitrary.  Since $\langle\cdot,\cdot\rangle$ is nonnegative we have
\begin{equation} \label{eq-CSBst1}
\forall \, t \in {\mathbb Q} \qquad  \bigl\langle u + t \langle u,v\rangle v, u + t \langle u,v\rangle v \bigr\rangle \geq 0.
\end{equation}
Since $\langle\cdot,\cdot\rangle$ is a sesquilinear hermitian form on $\mathcal V$, \eqref{eq-CSBst1} is equivalent to
\begin{equation} \label{eq-CSBst2}
\forall \, t \in {\mathbb Q} \qquad  \langle u, u \rangle  +  2 t |\langle u,v\rangle|^2 + t^2 |\langle u,v\rangle|^2 \langle v,v\rangle  \geq 0.
\end{equation}
As $\langle v,v \rangle \geq 0$, the High School Theorem applies and \eqref{eq-CSBst2} implies
\begin{equation} \label{eq-CSBst3}
   4 |\langle u,v\rangle|^4 - 4 |\langle u,v\rangle|^2 \langle u, u \rangle \langle v,v\rangle  \leq 0.
\end{equation}
Again, since $\langle u, u \rangle \geq 0$ and  $\langle v,v\rangle \geq 0$, \eqref{eq-CSBst3} is equivalent to
\begin{equation*}
    |\langle u,v\rangle|^2 \leq \langle u, u \rangle \langle v,v\rangle.
\end{equation*}
Since $u, v \in {\mathcal V}$ were arbitrary, \eqref{eq-CSBi} is proved.
\end{proof}

\begin{corollary} \label{co-nonn-deg}
Let ${\mathcal V}$ be a vector space over $\nF$ and let $\ahip$ be a nonnegative inner product on $\cV$. Then the following two implications  are equivalent.
\begin{enumerate}
\renewcommand*\theenumi{\roman{enumi}}
\renewcommand*\labelenumi{\rm (\theenumi)}
\setlength{\itemsep}{-2pt}
\item \label{i-nondeg}
If $v \in \cV$ and $\la u , v\ra = 0$ for all $u \in \cV$, then $v = 0$.
\item \label{i-pd}
If $v\in \cV$ and  $\la v, v \ra =0$, then $v =0$.
\end{enumerate}
\end{corollary}
\begin{proof}
Assume that the implication (\ref{i-nondeg}) holds and let $v\in\cV$ be such that $\la v,v\ra = 0$. Let $u \in \cV$ be arbitrary. By the the CBS inequality
\begin{equation*}
    |\langle u,v\rangle|^2 \leq \langle u, u \rangle \langle v,v\rangle = 0.
\end{equation*}
Thus, $\langle u,v\rangle = 0$ for all $u \in \cV$. By (\ref{i-nondeg}) we conclude $v =0$. This proves (\ref{i-pd}).

The converse is trivial. However, here is a proof. Assume that the implication (\ref{i-pd}) holds. To prove (\ref{i-nondeg}), let $v \in \cV$ and assume  $\la u , v\ra = 0$ for all $u \in \cV$. Setting $u = v$ we get $\la v , v\ra = 0$. Now (\ref{i-pd}) yields $v=0$.
\end{proof}


\begin{definition}
Let $\cV$ be a vector space over a scalar field $\nF$.  An inner product $\kip$ on $\cV$ is {\em nondegenerate} if the following implication holds
\begin{enumerate}
\setcounter{enumi}{3}
\renewcommand*\theenumi{\alph{enumi}}
\renewcommand*\labelenumi{(\theenumi)}
\item
(nondegenerecy) \ \ $u \in \cV$ and $[u,v] = 0$ for all $v \in \cV$ \ implies \ $u = 0$.
\end{enumerate}
\end{definition}

It follows from Corollary~\ref{co-nonn-deg} that a nonnegative inner product $\ahip$ on $\cV$ is nondegenerate if and only if $\la v, v \ra = 0$ implies $v = 0$. A nonnegative nondegenerate inner product is also called {\em positive definite inner product}. Since this is the most often encountered inner product we give its definition as it commonly given in textbooks.

\begin{definition}
Let $\cV$ be a vector space over a scalar field $\nF$. A function
$\ahip: \cV \times \cV \rightarrow \nF$ is called a {\em positive definite inner product} on $\cV$ if the following conditions are satisfied;
\begin{enumerate}
\renewcommand*\theenumi{\alph{enumi}}
\renewcommand*\labelenumi{(\theenumi)}
\setlength{\itemsep}{-2pt}
\item
$\ \forall \, u, v, w \in \cV \ \ \forall \, \alpha, \beta \in \nF \quad \la \alpha u+ \beta v, v \ra = \alpha \la u, w \ra + \beta \la v, w \ra$,
\item
$\ \forall \, u, v\in \cV \quad\la u, v \ra = \co{\la v, u \ra}$,
\item
$\ \forall \, v \in \cV \quad \la v, v \ra \geq 0$,
\item
If $v\in\cV$ and $\la v, v \ra = 0$, then $v = 0$.
\end{enumerate}

\end{definition}

\bigskip

\noindent\rule{7pt}{7pt} \dotfill \rule{7pt}{7pt} \noindent\rule{7pt}{0pt} Branko \'{C}urgus revised up to here. \rule{7pt}{0pt} \rule{7pt}{7pt} \dotfill \rule{7pt}{7pt}

\bigskip


%%%%PYTHAGOREAN THEOREM%%%%%
\begin{theorem} Pythagorean Theorem

Let $u,v \in \cV$. Then $\la u,v \ra = 0 \implies \la u +v, u+v \ra = \la u,u \ra + \la v, v \ra$

Furthermore, if $v_1, \cdots ,v_n \in \cV$ and $\la v_j, j_k \ra = 0$ whenever $j \neq k$ then $\la \sum_{j=1}^n v_j, \sum_{k=1}^n v_k \ra = \sum_{j=1}^n \la v_j, v_j \ra$
\end{theorem}

\begin{proof} For two vectors.
\begin{align*}
 \la u+v, u+v \ra & = \la u, u+v \ra + \la v, u+v \ra \\
&= \la u, u \ra + \la u, v \ra + \la v, u \ra + \la v, v \ra \\
&=  \la u, u \ra + 2 Re \la u, v \ra + \la v, v \ra \\
&= \la u, u \ra + \la v, v \ra
\end{align*}
\end{proof}




{\large November 8} (The Gram-Schmidt orthogonalization was proven the previous day)

\begin{theorem}[Gram-Schmidt]
If $\cV$ is a finite dimensional vector space with positive definite inner product $\ahip$, then $\cV$ has an orthonormal basis.
\end{theorem}

\begin{corollary}
If $\cV$ is a complex vector space with positive definite inner product and $T \in \cL(\cV)$ then there exists an orthonormal basis $B$ such that $\mathsf{M}^B_B(T)$ is upper-triangular.
\end{corollary}

\begin{definition}
Let $(\cV, \ahip)$ be a finite dimensional positive definite inner product space and $A \subset \cV$. We define $A^\perp = \{v \in \cV : \la v, a \ra = 0 \; \forall \; a \in A \}$.
\end{definition}

Claim (Not proven in class): $A^\perp$ is a subspace of $\cV$.

\begin{theorem}
If $\cU$ is a subspace of $\cV$, then $\cV = \cU \oplus \cU^\perp$.

\begin{proof}
Let $v \in \cU$ and $v \in \cU^\perp$. Then $\la v, v \ra = 0$. Since the $\ahip$ is positive definite, this implies $v = 0_\cV$. Note that since $\cU$ is a subspace of $\cV$, $\cU$ inherits the positive definite inner product space. Thus $\cU$ is a finite dimensional positive definite inner product space. Thus there exists an orthonormal basis of $\cU$, $\cB = \{u_1, u_2, \dots u_k \}$.

Let $v \in \cV$ be arbitrary. By the Gram-Schmidt process,
\[
v = \left( \sum_{j=1}^k \la v, u_j \ra u_j \right) + \left(v - \sum_{j=1}^k \la v, u_j \ra u_j \right),
\]
where the first summand is in $\cU$ and the second summand is in $\cU^\perp$. More succinctly, we write this as $v = w + (v - w)$ where $w = \sum_{j=1}^k \la v, u_j \ra u_j$. We prove $w$ is unique: $u \in \cU^\perp$ if and only if $\la w, u_j \ra = 0$ for all $j \in \{1, \dots k\}$. The forward direction is trivial (from the definition of $\cU^\perp$). To prove the reverse direction, let $u \in \cU$ be arbitrary. Then there exist $\alpha_j \in \nF$ such that $u = \sum_{j=1}^k \alpha_j u_j$. Now calculate
\[
\la w, u \ra = \left\langle w, \sum_{j=1}^k \alpha_j u_j \right\rangle = \sum_{j=1}^k \bar{\alpha}_j \la w, u_j \ra = 0.
\]
The last equality follows from the assumption. Thus $u \in \cU^\perp$.

Now for every $i \in \{1, \dots k\}$,
\[
\la v-w, u_i \ra = \left\la v - \sum_{j=1}^k \la v, u_j \ra u_j, u_i \right\ra = \la v, u_i \ra - \sum_{j=1}^k \la v, u_j \ra \la u_j, u_i \ra = \la v, u_i \ra - \la v, u_i \ra = 0.
\]
\end{proof}
\end{theorem}

\begin{definition}
By the previous theorem, if $\cU$ is a subspace of $\cV$, then $\cV = \cU \oplus \cU^\perp$ implies for all $v \in \cV$, there exists a unique $u \in \cU$ such that $(v - u) \in \cU^\perp$ and $v = u + (v - u)$. This defines a function which we call the {\bf orthogonal projection} of $v$ onto $\cU$ as $P_\cU : \cV \to \cU$ such that $P_\cU(v) = u$.
\end{definition}

Since $\cU$ is a subspace of $\cV$, $P_\cU \in \cL(\cV)$. Furthermore, $\ran P_\cU = \cU$, $\nul P_\cU = \cU^\perp$, and $(P_\cU)^2 = P_\cU$ (idempotent).

\begin{proposition}
Let $\cU$ be a subspace of $\cV$, $v \in \cV$ be arbitrary. Let $u_0 \in \cU$. Then $\|v - u_0\| \le \|v - u\|$ for every $u \in \cU$ if and only if $P_\cU(v) = u_0$ and $v - u_0 \in \cU^\perp$.

\begin{proof}
$(\Longleftarrow)$: Assume $v \in \cV$, $u, u_0 \in \cU$, $v - u_0 \in \cU^\perp$. Then $\|v - u\|^2 = \|v - u_0 + u_0 + u\|^2$, where $v - u_0 \in \cU^\perp$ and $u_0 + u \in \cU$. By the pythagorean theorem,
\[
\|v - u_0 + u_0 + u\|^2 = \|v - u_0\|^2 + \|u_0 - u\|^2 \ge \|v - u_0\|^2.
\]

$(\Longrightarrow)$ Assume $\|v - u_0 \| \le \|v - u\|$ for all $u \in \cU$. We show $v - u_0 \in \cU^\perp$. This direction of the proof was given on November 9.
\end{proof}
\end{proposition}


\bigskip

\noindent\rule{7pt}{7pt} \dotfill \rule{7pt}{7pt} \noindent\rule{7pt}{0pt} Stuff from November 19, 2013 \rule{7pt}{0pt} \rule{7pt}{7pt} \dotfill \rule{7pt}{7pt}

\bigskip

\begin{lemma} \label{le-Uinv}
Let $\cV$ be a vector space over $\nF$ and let $\ahip$ be a positive definite inner product on $\cV$. Let $\cU$ be a subspace of $\cV$ and let $T \in \cL(\cV)$. The subspace $\cU$ is invariant under $T$ if and only if the subspace $\cU^\perp$ is invariant under $T^*$.
\end{lemma}
\begin{proof}
By the definition of adjoint we have
\begin{equation} \label{eq-ad-1}
\la T u , v \ra = \la u , T^* v \ra
\end{equation}
for all $u, v \in \cV$. Assume $T\cU \subset \cU$. From \eqref{eq-ad-1} we get
\begin{equation*} % \label{eq-ad-2}
0 = \la T u , v \ra = \la u , T^* v \ra \qquad \forall\, u \in \cU \quad \text{and} \quad \forall v \in \cU^\perp.
\end{equation*}
Therefore, $T^* v \in \cU^\perp$ for all $v \in \cU^\perp$. This proves ``only if'' part.

The proof of the ``if'' part is similar.
\end{proof}

In the proof of the next theorem we use $\delta_{ij}$ to represent the Kronecker delta function, that is $\delta_{ij} = 1$ if $i=j$ and $\delta_{ij} = 0$ otherwise.

\begin{theorem}[Spectral theorem for normal operators]
Let $\cV$ be a finite dimensional complex vector space with a positive definite inner product $\ahip$. Let $T \in \cL(\cV)$.  Then $T$ is normal if and only if there exists an orthonormal basis of $\cV$ which consists of eigenvectors of $T$.
\end{theorem}
\begin{proof}
Set $n = \dim \cV$. We first prove ``only if'' part. Assume that $T$ is normal. Set
\[
\nK = \Biggl\{ k \in \{1,\ldots,n \} \, : \, \begin{array}{l}
 \exists \, w_1,\ldots,w_k \in \cV  \quad \text{and} \quad \exists \, \lambda_1,\ldots,\lambda_k \in \nC   \\
 \text{such that} \ \la w_i,w_j \ra = \delta_{ij} \ \text{and} \ Tw_j = \lambda_j w_j \\
 \text{for all} \quad i,j \in \{1,\ldots,k\}
 \end{array} \Biggr\}
\]
Clearly $1 \in \nK$. Since $\nK$ is finite, $m = \max \nK$ exists. Clearly, $m \leq n$.

Next we will prove that $k \in \nK$ and $k < n$ implies that $k+1 \in \nK$. Assume $k \in \nK$ and $k < n$. Let $w_1,\ldots,w_k \in \cV$ and $\lambda_1,\ldots,\lambda_k \in \nC$ be such that $\la w_i,w_j \ra = \delta_{ij}$ and $Tw_j = \lambda_j w_j$ for all $i,j \in \{1,\ldots,k\}$. Set
\[
\cW = \lspan\{w_1,\ldots,w_k\}.
\]
Since $w_1,\ldots,w_k$ are eigenvectors of $T$ we have $T\cW \subseteq \cW$.  By Lemma~\ref{le-Uinv}, $T^* \bigl(\cW^\perp\bigr) \subseteq \cW^\perp$. Thus, $T^*|_{\cW^\perp} \in \cL\bigl(\cW^\perp\bigr)$. Since $\dim \cW = k < n$ we have $\dim \bigl(\cW^\perp\bigr) = n-k \geq 1$. Since $\cW^\perp$ is a complex vector space the operator $T^*|_{\cW^\perp}$ has an eigenvalue $\mu$ with the corresponding unit eigenvector $u$. Clearly, $u \in {\cW^\perp}$ and $T^*u = \mu u$. Since $T^*$ is normal, we have $Tu = \co{\mu} u$. Since $u \in {\cW^\perp}$ and $Tu = \co{\mu} u$, setting $w_{k+1} = u$ and $\lambda_{k+1} = \co{\mu}$ we have
\[
\la w_i,w_j \ra = \delta_{ij} \quad \text{and} \quad Tw_j = \lambda_j w_j  \quad \text{for all} \quad i,j \in \{1,\ldots,k,k+1\}.
\]
Thus $k+1 \in \nK$. Consequently, $k < m$. Thus, for $k \in \nK$, we have proved the implication
\[
k < n \qquad \Rightarrow \qquad k < m.
\]
The contrapositive of this implication is: For $k \in \nK$, we have
\[
k \geq m \qquad \Rightarrow \qquad k \geq n.
\]
In particular, for $m \in \nK$ we have $m = m$ implies $m \geq n$. Since $m \leq n$ is also true, this proves that $m = n$. That is, $n \in \nK$. This implies that there exist $u_1,\ldots,u_n \in \cV$ and $\lambda_1,\ldots,\lambda_n \in \nC$ such that $\la u_i,u_j \ra = \delta_{ij}$ and $Tu_j = \lambda_j u_j$ for all $i,j \in \{1,\ldots,n\}$.

Since $u_1,\ldots,u_n$ are orthonormal, they are linearly independent. Since $n = \dim \cV$, it turns out that $u_1,\ldots,u_n$ form a basis of $\cV$. This completes the proof.

To prove the converse assume that there exist an orthonormal basis of $\cV$ which consist of eigenvectors of $T$. That is, assume that there exists $u_1,\ldots,u_n \in \cV$ and $\lambda_1,\ldots,\lambda_n \in \nC$ such that $\la u_i,u_j \ra = \delta_{ij}$ and $Tu_j = \lambda_j u_j$ for all $i,j \in \{1,\ldots,n\}$.

Let $j \in \{1,\ldots,n\}$ be arbitrary.
Since $u_1,\ldots,u_n$ form an orthonormal basis we have
\begin{align*}
T^*u_j & = \la T^*u_j,u_1\ra \, u_1 + \la T^*u_j,u_2\ra\, u_2 + \cdots + \la T^*u_j,u_n\ra \, u_n \\
& = \la u_j,T u_1\ra \, u_1 + \la  u_j,T u_2\ra\, u_2 + \cdots + \la u_j, T u_n\ra \, u_n \\
& = \la u_j,\lambda_1 u_1\ra \, u_1 + \la  u_j,\lambda_2 u_2\ra\, u_2 + \cdots + \la u_j, \lambda_n u_n\ra \, u_n \\
& = \co{\lambda_1} \la u_j,  u_1\ra \, u_1 + \co{\lambda_2} \la  u_j, u_2\ra\, u_2 + \cdots + \co{\lambda_n} \la u_j, u_n\ra \, u_n \\
& = \co{\lambda_j} u_j.
\end{align*}
Thus, $T^*u_j = \co{\lambda_j} u_j$ for all $j \in \{1,\ldots,n\}$. Consequently,
\[
TT^*u_j = T\bigl(\co{\lambda_j} u_j\bigr) = \co{\lambda_j} T u_j = \co{\lambda_j} \lambda_j u_j = |\lambda_j|^2 u_j,
\]
and also
\[
T^*Tu_j = T^*\bigl(\lambda_j u_j\bigr) = \lambda_j T^* u_j =  \lambda_j \co{\lambda_j} u_j = |\lambda_j|^2 u_j.
\]
Thus, $TT^*u_j = T^*Tu_j$ for all $j \in \{1,\ldots,n\}$. Since $u_1,\ldots,u_n$ form a basis of $\cV$ this implies $TT^* v = T^*T v$ for all $v\in\cV$, that is, $T$ is normal.
\end{proof}









\end{document}
