Title: ETH S22 Mathematics of Information - Hilbert Spaces Prerequisites

Front: [latex]
What are the properties of addition in a vector space?
[/latex]
Back: [latex]
The addition operation denoted by $+$ satisfies
\begin{enumerate}[(1)]
\item associativity, so $\left( x + y \right) + z = x + \left( y + z \right)$ for all $x, y, z \in \X$,
\item commutativity, so $x + y = y + x$ for all $x, y \in \X$.
\end{enumerate}
[/latex]

Front: [latex]
What are the properties of multiplication in a vector space?
[/latex]
Back: [latex]
The multiplication operator denoted by $\cdot$ satisfies
\begin{enumerate}[(1)]
\item $1 \cdot x = x$ for all $x \in \X$,
\item $\alpha \cdot \left( \beta \cdot x \right) = \left( \alpha \beta \right)\cdot x$ for all $\alpha,\beta \in \F$ and $x \in \X$,
\item $\left( \alpha +\beta \right)\cdot x = \alpha \cdot x + \beta \cdot x$ for all $\alpha,\beta \in \F$ and $x \in \X$,
\item $\alpha \cdot \left( x + y \right) = \alpha \cdot x + \alpha \cdot y$ for all $\alpha \in \F$ and $x, y \in \X$.
\end{enumerate}
[/latex]

Front: [latex]
State definition: Vector space.
[/latex]
Back: [latex]
A set $\X$ together with two operations $(+, \cdot)$ is a vector space over $\F$ if the following properties are satisfied:
\begin{enumerate}[(i)]
\item The addition operation denoted by $+$ satisfies
\begin{itemize}
\item associativity, so $\left( x + y \right) + z = x + \left( y + z \right)$ for all $x, y, z \in \X$,
\item commutativity, so $x + y = y + x$ for all $x, y \in \X$.
\end{itemize}
\item There exists an element $0 \in \X$, called the zero vector, such that $x + 0 = x$ for all $x \in \X$.
\item For every $x \in \X$, there exists $-x \in \X$ such that $x + \left( -x \right) = 0$.
\item The multiplication operator denoted by $\cdot$ satisfies
\begin{itemize}
\item $1 \cdot x = x$ for all $x \in \X$,
\item $\alpha \cdot \left( \beta \cdot x \right) = \left( \alpha \beta \right)\cdot x$ for all $\alpha,\beta \in \F$ and $x \in \X$,
\item $\left( \alpha +\beta \right)\cdot x = \alpha \cdot x + \beta \cdot x$ for all $\alpha,\beta \in \F$ and $x \in \X$,
\item $\alpha \cdot \left( x + y \right) = \alpha \cdot x + \alpha \cdot y$ for all $\alpha \in \F$ and $x, y \in \X$.
\end{itemize}
\end{enumerate}
[/latex]

Front: [latex]
What is a subspace of a vector space $\X$ over $\F$?
[/latex]
Back: [latex]
$\Y$ is a subspace of $\X$ if $\Y$ is itself a vector space and a subset of $\X$.
[/latex]

Front: [latex]
What's an easy way to show that $\Y$ is a subspace of $\X$ over $\F$?
[/latex]
Back: [latex]
$\Y$ is a subspace of $\X$ if it contains 0 and if it is stable under linear combinations, i.e.
\[
\alpha \cdot y + \beta \cdot z \in \Y
\] for all $\alpha , \beta \in \F$ and $y, z \in \Y$.
[/latex]

Front: [latex]
How are spaces of $L^{p}(\S)$ functions mapping from $\S$ to $\Co$ related to the space of all functions $\Fc\left( \S, \Co \right)$ from $\S$ to $\Co$?
[/latex]
Back: [latex]
$L^{p}(\S)$ are subspaces of $\Fc\left( \S, \Co \right)$, which itself is a vector space under pointwise addition and multiplication.
[/latex]

Front: [latex]
Suppose $\Y$ and $\Z$ are subspaces of $\X$. Are $\Y \cap \Z$ and $\Y \cup \Z$ also subspaces of $\X$?
[/latex]
Back: [latex]
$\Y \cap \Z$ is a subspace of $\X$, but $\Y \cup \Z$ might not be.
[/latex]

Front: [latex]
Suppose $\S$ and $\Tc$ are linear subspaces of a vector space $\X$ with $\S \cap \Tc = \left\{ 0 \right\}$. What is the direct sum $\S \oplus \Tc$?
[/latex]
Back: [latex]
$\S \oplus \Tc$ is defined by
\[
\S \oplus \Tc = \left\{ x + y | x \in \S, y \in \Tc \right\}.
\]
[/latex]

Front: [latex]
Suppose $\S$ and $\Tc$ are linear subspaces of a vector space. When is a direct sum defined? What does this imply about elements of $\S \oplus \Tc$?
[/latex]
Back: [latex]
The direct sum is defined when $\S \cap \Tc = \left\{ 0 \right\}$.
This implies that all elements $x \in \S \oplus \Tc$ have a unique decomposition $x = s + t$ where $s \in \S$, $t \in \Tc$.
[/latex]

Front: [latex]
Suppose $\Y$ and $\Z$ are subspaces of $\X$. What is $\Y \cup \Z$?
[/latex]
Back: [latex]
There is no rule for this. $\Y \cup \Z$ is not necessarily a subspace.
[/latex]

Front: [latex]
Suppose $\S$ is a subset of a vector space $\X$ over $\F$. What is the subspace spanned by $\S$?
[/latex]
Back: [latex]
The subspace spanned by $\S$ is the intersection of all subspaces containing $\S$. It can be written as the set of all finite combinations of elements of $\S$, so
\[
\text{span} \left( \S \right) \triangleq \left\{ \sum_{l = 1}^{k} \lambda _l x_l : k \in \N, x_l \in \S, \lambda _l \in \F \right\}.
\]
[/latex]

Front: [latex]
Let $\X$ be a vector space over $\F$. What does it mean for a finite set $\S$ of vectors $\left\{ x_1, x_2, \dots, x_N \right\}$ in $\X$ to be linearly independent? What if $\S$ is infinite?
[/latex]
Back: [latex]
The set of vectors $\left\{ x_1. x_2, \dots, x_N \right\}$ is linearly independent if for every $\lambda _1, \lambda _2, \dots, \lambda _N \in \F$, we have the implication
\[
\lambda _1x_1 + \lambda _2 + x_2 + \cdots + \lambda _N + x_N = 0 \implies \lambda _1 = \lambda _2 = \cdots = \lambda _N = 0.
\]
If $\S$ is infinite, then it is linearly independent if every finite subset of $\S$ is linearly independent.
[/latex]

Front: [latex]
What does it mean for a vector space $\X$ to be $N$-dimensional?
[/latex]
Back: [latex]
$\X$ is $N$-dimensional if there exist $N$ linearly independent vectors in $\X$ and any $N + 1$ vectors in $\X$ are linearly dependent.
[/latex]

Front: [latex]
What is a basis of a vector space $\X$?
[/latex]
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A basis is a set of linearly independent vectors spanning $\X$.
[/latex]

Front: [latex]
Suppose $\X$ is a vector space over $\F$. State the definition of a norm on $\X$.
[/latex]
Back: [latex]
A norm on $\X$ is a function which maps $\X$ to $\R$ and satisfies
\begin{enumerate}[(i)]
\item for all $x \in \X$, $\left\Vert x \right\Vert \ge 0$ and $\left\Vert x \right\Vert = 0$ implies $x = 0$ (positivity),
\item for all $x \in \X$ and for all $\alpha \in \F$, $\left\Vert \alpha x \right\Vert = \left| \alpha \right| \left\Vert x \right\Vert $ (homogeneity),
\item for all $x, y \in \X$, $\left\Vert x + y \right\Vert \le \left\Vert x \right\Vert + \left\Vert y \right\Vert$ (triangle inequality).
\end{enumerate}
[/latex]

Front: [latex]
What's a normed space?
[/latex]
Back: [latex]
It's a vector space equipped with a norm $\left\Vert \cdot \right\Vert $.
A norm on $\X$ is a function which maps $\X$ to $\R$ and satisfies
\begin{enumerate}[(i)]
\item for all $x \in \X$, $\left\Vert x \right\Vert \ge 0$ and $\left\Vert x \right\Vert = 0$ implies $x = 0$ (positivity),
\item for all $x \in \X$ and for all $\alpha \in \F$, $\left\Vert \alpha x \right\Vert = \left| \alpha \right| \left\Vert x \right\Vert $ (homogeneity),
\item for all $x, y \in \X$, $\left\Vert x + y \right\Vert \le \left\Vert x \right\Vert + \left\Vert y \right\Vert$ (triangle inequality).
\end{enumerate}
[/latex]

Front: [latex]
What does it mean for an operator $\X \times \X \to \F$ to be bilinear?
[/latex]
Back: [latex]
An operator denoted by $\left\langle \cdot, \cdot \right\rangle $ is bilinear if for all $x, y \in \X$ and $\alpha \in \F$ we have
\begin{enumerate}[(i)]
\item $\left\langle \alpha x, y \right\rangle = \alpha \left\langle x, y \right\rangle $,
\item $\left\langle x + y, z \right\rangle = \left\langle x, z \right\rangle + \left\langle y, z \right\rangle $.
\end{enumerate}
[/latex]

Front: [latex]
What does it mean for an operator $\X \times \X \to \Co$ to be symmetric?
[/latex]
Back: [latex]
\[
\left\langle y, x \right\rangle = \overline{\left\langle x, y \right\rangle }
\]
[/latex]

Front: [latex]
Suppose $\X$ is a vector space over $\F$ ($\R$ or $\Co$). What is an inner product $\X \times \X \to \F$ over $\F$?
[/latex]
Back: [latex]
An inner product over $\F$ is a function $\X \times \X \to \F$ such that for all $x, y \in \X$ and $\alpha
\in \F$, we have
\begin{enumerate}[(i)]
\item symmetry, so $\left\langle y, x \right\rangle = \overline{\left\langle x, y \right\rangle }$,
\item bilinearity, so $\left\langle \alpha x, y \right\rangle = \alpha \left\langle x, y \right\rangle $ and $\left\langle x + y, z \right\rangle = \left\langle x, z \right\rangle + \left\langle y, z \right\rangle $,
\item positivity, so $\left\langle x, x \right\rangle \in \R$, $\left\langle x, x \right\rangle \ge 0$, and $\left\langle x, x \right\rangle = 0 \implies x = 0$.
\end{enumerate}
[/latex]

Front: [latex]
Consider a vector space $\X$ with a norm $\left\Vert \cdot \right\Vert $. What does it mean for the norm to fulfil the parallelogram law?
[/latex]
Back: [latex]
It means that for all $x, y \in \X$
\[
2\left\Vert x \right\Vert ^{2} + 2\left\Vert y \right\Vert ^{2} = \left\Vert x + y \right\Vert ^{2} + \left\Vert x - y \right\Vert ^{2}.
\]
[/latex]

Front: [latex]
Suppose $\X$ is a vector space. If $\left\langle \cdot, \cdot \right\rangle $ is an inner product on $\X$, then how can you use it to induce a norm which fulfils the parallelogram law?
[/latex]
Back: [latex]
The norm $\left\Vert x \right\Vert := \sqrt{\left\langle x, x \right\rangle } $ fulfils the parallelogram law
\[
2\left\Vert x \right\Vert ^{2} + 2\left\Vert y \right\Vert ^{2} = \left\Vert x + y \right\Vert ^{2} + \left\Vert x - y \right\Vert ^{2}.
\]
[/latex]

Front: [latex]
Suppose $\X$ is a vector field over $\R$, and $\left\Vert \cdot \right\Vert $ is a norm which fulfils the parallelogram law. What is the inner product which induces $\left\Vert \cdot \right\Vert $?
[/latex]
Back: [latex]
The inner product $\left\langle \cdot, \cdot \right\rangle $ which induces $\left\Vert \cdot \right\Vert $ is
\[
\left\langle x, y \right\rangle = \frac{1}{4} \left( \left\Vert x + y \right\Vert ^{2} - \left\Vert x - y \right\Vert ^{2} \right).
\]
[/latex]

Front: [latex]
What's an inner product space?
[/latex]
Back: [latex]
It's a vector space equipped with an inner product $\left\langle \cdot, \cdot \right\rangle $.
An inner product over $\F$ is a function $\X \times \X \to \F$ such that for all $x, y \in \X$ and $\alpha
\in \F$, we have
\begin{enumerate}[(i)]
\item symmetry, so $\left\langle y, x \right\rangle = \overline{\left\langle x, y \right\rangle }$,
\item bilinearity, so $\left\langle \alpha x, y \right\rangle = \alpha \left\langle x, y \right\rangle $ and $\left\langle x + y, z \right\rangle = \left\langle x, z \right\rangle + \left\langle y, z \right\rangle $,
\item positivity, so $\left\langle x, x \right\rangle \in \R$, $\left\langle x, x \right\rangle \ge 0$, and $\left\langle x, x \right\rangle = 0 \implies x = 0$.
\end{enumerate}
[/latex]

Front: [latex]
State the Cauchy-Schwarz inequality.
[/latex]
Back: [latex]
Let $\X$ be an inner product space. Then for all $x, y \in \X$,
\[
\left| \left\langle x, y \right\rangle \right| \le \left\Vert x \right\Vert \left\Vert y \right\Vert
\] with equality if and only if $x$ and $y$ are linearly dependent.
[/latex]

Front: [latex]
Let $\X$ be a normed vector space. Suppose $\left( x_{n} \right)_{n \geq 1} $ is a sequence which converges to $x \in \X$. What can you say about $\lim_{n \to \infty} \left\Vert x_n \right\Vert $?
[/latex]
Back: [latex]
If $\X$ is a normed vector space then $\left\Vert \cdot \right\Vert $ is continuous, so \[\lim_{n \to \infty} \left\Vert x_n \right\Vert = \left\Vert x \right\Vert. \]
[/latex]

Front: [latex]
Let $\X$ be an inner product space. Suppose $\left( x_{n} \right)_{n \geq 1} $ and $\left( y_{n} \right)_{n \geq 1} $ are sequences with limits to $x \in \X$ and $y \in \X$ respectively. What can you say about $\lim_{n \to \infty} \left\langle x_n, y_n \right\rangle?$
[/latex]
Back: [latex]
If $\X$ is an inner product space then $\left\langle \cdot, \cdot \right\rangle $ is continuous, so \[\lim_{n \to \infty} \left\langle x_n, y_n \right\rangle = \left\langle x, y \right\rangle . \]
[/latex]

Front: [latex]
Consider a normed space $\X$. What does it mean for a sequence $\left( x_{n} \right)_{n \geq 1} $ of elements of $\X$ to be Cauchy?
[/latex]
Back: [latex]
The sequence $\left( x_{n} \right)_{n \geq 1} $ is Cauchy if for every $\varepsilon > 0$ there exists an integer $N\left( \varepsilon \right)$ such that $\left\Vert x_n - x_m \right\Vert \le \varepsilon $ whenever $n, m \ge N\left( \varepsilon \right)$.
[/latex]

Front: [latex]
What does it mean for a set $\X$ to be a complete space?
[/latex]
Back: [latex]
$\X$ is a complete space if $\X$ is a normed space and if every Cauchy sequence in $\X$ converges in $\X$.
[/latex]

Front: [latex]
What are
\begin{enumerate}[(1)]
\item a Banach space,
\item a Hilbert space, and
\item a pre-Banach or pre-Hilbert space?
\end{enumerate}
[/latex]
Back: [latex]
\begin{enumerate}[(1)]
\item A Banach space is a normed space which is complete.
\item A Hilbert space is an inner product space which is complete.
\item The pre-spaces are same as the above, but are not complete.
\end{enumerate}
[/latex]

Front: [latex]
What does it mean for a subset $\S$ of a normed space $\X$ to be closed?
[/latex]
Back: [latex]
$\S$ is closed if it contains all of its limit points.
[/latex]

Front: [latex]
What is the closure of a subset $\S$ of a normed space $\X$?
[/latex]
Back: [latex]
The closure $\overline{\S}$ is the intersection of all closed subsets of $\X$ which are supersets of $\S$.
[/latex]

Front: [latex]
Let $\A \subset \B$ be subsets of a normed space $\X$. What does it mean for $\A$ to be dense in $\B$?
[/latex]
Back: [latex]
$\A$ is dense in $\B$ if for the closure of $\A$ we have $\overline{\A} = \B$.
Equivalently, $\A$ is dense in $\B$ if for every $x \in \B$ and every $\varepsilon > 0$ there exists a $y \in \A$ such that $\left\Vert x - y \right\Vert < \varepsilon $.
[/latex]

Front: [latex]
Suppose $\X$ is a normed space with norm $\left\Vert \cdot \right\Vert $. What is the completion of $\X$? Does it always exist? Is it unique?
[/latex]
Back: [latex]
The completion of $\X$ is a Banach space $\widehat{\X}$ with norm $\widehat{\left\Vert \cdot \right\Vert }$ such that
\begin{enumerate}[(i)]
\item $\X$ is a subspace of $\widehat{\X}$,
\item on $\X$, $\widehat{\left\Vert \cdot \right\Vert } = \left\Vert \cdot \right\Vert $,
\item $\widehat{\X}$ is the closure of $\X$.
\end{enumerate}
It always exists, and is unique up to linear isometries.
[/latex]

Front: [latex]
What does it mean for for vectors $x$ and $y$ in an inner product space $\X$ to be orthogonal? What does orthogonality mean for subsets $\U, \V \subseteq \X$?
[/latex]
Back: [latex]
We say that the vectors $x$ and $y$ are orthogonal, written as $x \perp y$, if
\[
\left\langle x, y \right\rangle = 0.
\]
We say that the sets $\U$ and $\V$ are orthogonal if for all $x \in \U$ and $y \in \V$, $x \perp y$.
[/latex]

Front: [latex]
Suppose $\S$ is a nonempty subset of an inner product space $\X$. What is its orthogonal complement?
[/latex]
Back: [latex]
The orthogonal complement $S^{\perp}$ is defined as
\[
S^{\perp} := \left\{ x \in \X: x \perp \S \right\}.
\]
[/latex]

Front: [latex]
Suppose $\S$ is a nonempty subset of the inner product space $\X$. What are the properties of $\S^{\perp}$?
[/latex]
Back: [latex]
$\S^{\perp}$ is a closed vector subspace of $\X$.
[/latex]

Front: [latex]
Suppose $\S$ is a closed subspace of a Hilbert space $\X$, and take some $x \in \X$. Does the point $y$ such that
\[
\left\Vert x - y \right\Vert = \min_{z \in \S}\left\Vert x - z \right\Vert
\]
exist? Is it unique?
[/latex]
Back: [latex]
Yes, it exists, and yes it is unique. $y$ is the closest point in $\S$ to $x$.
[/latex]

Front: [latex]
Suppose $\S$ is a closed subspace of a Hilbert space $\X$, and take some $x \in \X$. Let $y \in \S$ be the closest point in $\S$ to $x$. What can you say about the vector $\left( x - y \right)$?
[/latex]
Back: [latex]
$\left( x - y \right)$ is orthogonal to all points in $\S$, so\[ \left( x - y \right) \in S^{\perp}.\]
\begin{center}\includegraphics[width=0.8\textwidth]{hilbert-closest-orthogonal}\end{center}
[/latex]

Front: [latex]
Let $\S$ be a closed subspace of a Hilbert space $\X$. What are $\S \oplus \S^{\perp}$ and $\S^{\perp\perp}$?
[/latex]
Back: [latex]
\begin{enumerate}[(i)]
\item $\S \oplus \S^{\perp} = \X$,
\item $\S^{\perp\perp} = \S$.
\end{enumerate}
[/latex]

Front: [latex]
Let $\S$ be a subset of a Hilbert space $\X$. How is $S^{\perp}$ related to $\text{span} \left( \S \right)$?
[/latex]
Back: [latex]
\begin{enumerate}[(i)]
\item $S^{\perp} = \overline{\text{span}\left( \S \right)}^{\perp}$,
\item $S^{\perp\perp} = \overline{\text{span}\left( \S \right)}$,
\item $S^{\perp} = \left\{ 0 \right\}$ if and only if $\overline{\text{span}\left( \S \right)} = \X$.
\end{enumerate}
[/latex]

Front: [latex]
Let $\S$ be a subset of an inner product space $\X$ which is not complete. What is $\S \oplus \S^{\perp}$?
[/latex]
Back: [latex]
Nothing in particular. If the space is not complete, then it might be that \[\X \neq \S \oplus \S^{\perp}.\]
[/latex]

Front: [latex]
What does it mean for a subset $\S$ of a Hilbert space $\X$ to be orthogonal? What about orthonormal?
[/latex]
Back: [latex]
A subset $\S$ of nonzero vectors of a Hilbert space $\X$ is orthogonal if any two distinct elements in $\S$ are orthogonal.
An orthogonal set $\S$ is called orthonormal if also $\left\Vert x \right\Vert = 1$ for all $x \in \S$.
[/latex]

Front: [latex]
Let $\left\{ x_\alpha \in \X | \alpha \in \I \right\}$ be an indexed subset of a normed space $\X$, where $\I$ may be countable or not. What does it mean for the sum $\sum_{\alpha \in \I}^{} x_\alpha $ to converge unconditionally to $x \in \mathcal{X} $?
[/latex]
Back: [latex]
For each finite subset $\J$ of $\I$, define the partial sum
\[
S_\J = \sum_{\alpha \in \J}^{} x_\alpha .
\]
The unordered sum $\sum_{\alpha \in \I}^{} x_\alpha $ converges to $x \in \X$ if for every $\varepsilon > 0$ there exists a finite subset $\J_\varepsilon \subseteq \I$ such that $\left\Vert S_\J - x \right\Vert < \varepsilon $ for all finite subsets $\J \subseteq \I$ containing $\J_\varepsilon $.
[/latex]

Front: [latex]
What does it mean for an unordered sum $\sum_{\alpha \in \I}^{} x_\alpha$ to be Cauchy?
[/latex]
Back: [latex]
An unordered sum $\sum_{\alpha \in \I}^{} x_\alpha $ is Cauchy if for every $\varepsilon > 0$ there exists a finite subset $\J_\varepsilon \subseteq \I$ such that the partial sums have $\left\Vert S_\K \right\Vert < \varepsilon $ for all finite sets $\K \subseteq \I \setminus \J_\varepsilon $.
[/latex]

Front: [latex]
Consider an unordered sum in a Banach space. How its its convergence related to whether it is Cauchy or not?
[/latex]
Back: [latex]
The unordered sum converges if and only if it is Cauchy.
[/latex]

Front: [latex]
Suppose that the unordered sum $\sum_{\alpha \in \I}^{} x_\alpha $ in a Banach space converges. How many nonzero elements $x_\alpha $ are there?
[/latex]
Back: [latex]
There are at most countably many nonzero $x_\alpha$.
[/latex]

Front: [latex]
Consider an indexed orthogonal subset $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ of a Hilbert space $\X$. What is an equivalent condition to unconditional convergence of $\sum_{\alpha \in \I}^{} x_\alpha $ that uses the $l^{2}$ norm?
[/latex]
Back: [latex]
$\sum_{\alpha \in \I}^{} x_\alpha $ converges unconditionally if and only if
\[
\sum_{\alpha \in \I}^{} \left\Vert x_\alpha \right\Vert ^{2}< \infty.
\]
[/latex]

Front: [latex]
Consider an indexed orthogonal subset $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ of a Hilbert space $\X$. Suppose the sum $\sum_{\alpha \in \I}^{} x_\alpha $ converges unconditionally. What can you say about the value of
\[
\sum_{\alpha \in \I}^{} \left\Vert x_\alpha \right\Vert ^{2}?
\]
[/latex]
Back: [latex]
$\sum_{\alpha \in \I}^{} \left\Vert x_\alpha \right\Vert ^{2} < \infty$ and
\[
\sum_{\alpha \in \I}^{} \left\Vert x_\alpha \right\Vert ^{2} = \left\Vert \sum_{\alpha \in \I}^{} x_\alpha \right\Vert ^{2}.
\]
[/latex]

Front: [latex]
Consider an indexed orthogonal subset $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ of a Hilbert space $\X$. Show that $\sum_{\alpha \in \I}^{} x_\alpha $ converges unconditionally if and only if $\sum_{\alpha \in \I}^{} \left\Vert x_\alpha \right\Vert ^{2}< \infty,$ and in that case
\[
\sum_{\alpha \in \I}^{} \left\Vert x_\alpha \right\Vert ^{2} = \left\Vert \sum_{\alpha \in \I}^{} x_\alpha \right\Vert ^{2}.
\]
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item From orthogonality, for any finite set $\K \subseteq \I$,
\[
\left\Vert \sum_{\alpha \in \K}^{} x_\alpha \right\Vert ^{2} = \sum_{\alpha ,\beta \in \K}^{} \left\langle x_\alpha , x_\beta \right\rangle = \sum_{\alpha \in \K}^{} \left\Vert x_\alpha \right\Vert ^{2}.
\]
\item Therefore, $\sum_{\alpha \in \I}^{} x_\alpha $ is Cauchy if and only if $\sum_{\alpha \in \I}^{} \left\Vert x_\alpha \right\Vert ^{2}$ is Cauchy.
\item So, one of the sums converges unconditionally if and only if the other one does.
\item The equality given follows from the continuity of the norm, since the sum is the limit of a sequence of finite partial sums.
\end{enumerate}
[/latex]

Front: [latex]
State Bessel's inequality.
[/latex]
Back: [latex]
Let $\X$ be a Hilbert space, $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ an orthonormal subset of $\X$, and $x \in \X$. Then
\begin{enumerate}[(a)]
\item $\sum_{\alpha \in \I}^{} \left| \left\langle x, x_\alpha \right\rangle \right|^{2} \le \left\Vert x \right\Vert ^{2}$,
\item $x_\S \triangleq \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha $ converges unconditionally,
\item $x - x_\S \in \S^{\perp}$.
\end{enumerate}
[/latex]

Front: [latex]
What is the closed linear span $\left[ \S \right]$ of a subset $\S$ of a Hilbert space?
[/latex]
Back: [latex]
\[
\left[ \S \right] = \left\{ \sum_{x \in \S}^{} c_xx | c_x \in \F, \sum_{x \in \S}^{} c_xx \text{ converges unconditionally} \right\}.
\]
[/latex]

Front: [latex]
Rewrite the definition of a closed linear span
\[
\left[ \S \right] = \left\{ \sum_{x \in \S}^{} c_xx | c_x \in \F, \sum_{x \in \S}^{} c_xx \text{ converges unconditionally} \right\}.
\] with a statement equivalent to unconditional convergence, but easier to check.
[/latex]
Back: [latex]
Write $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$. Then, the closed linear span may be written as
\[
\left[ \S \right] = \left\{ \sum_{\alpha \in \I}^{} c_\alpha x_\alpha | c_\alpha \in \F, \sum_{\alpha \in \I}^{} \left| c_\alpha \right|^{2} < \infty \right\}.
\]
[/latex]

Front: [latex]
Suppose $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ is an orthonormal subset of a Hilbert space $\X$. What is the condition
\[
\left\langle x, x_\alpha \right\rangle = 0 \text{ for all } \alpha \in \I \text{ implies that } x = 0
\] equivalent to?
[/latex]
Back: [latex]
The following are equivalent:
\begin{enumerate}[(a)]
\item $\left\langle x, x_\alpha \right\rangle = 0$ for all $\alpha \in \I$ implies that $x = 0$,
\item $x = \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha $ for all $x \in \X$,
\item $\left\Vert x \right\Vert ^{2} = \sum_{\alpha \in \I}^{} \left| \left\langle x, x_\alpha \right\rangle \right|^{2}$ for all $x \in \X$,
\item $\left[ \S \right] = \X$,
\item $\S$ is a maximal orthonormal set, i.e. it has the property that if $\S' \supset \S$ is another orthonormal set, then necessarily $\S' = \S$.
\end{enumerate}
The orthonormal subset $\mathcal{S} $ that fulfils these criteria is called \emph{complete}.
[/latex]

Front: [latex]
Suppose $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ is an orthonormal subset of a Hilbert space $\X$. Show that
\[
\text{$\left\langle x, x_\alpha \right\rangle = 0$ for all $\alpha \in \I$ implies that $x = 0$}
\] implies
\[
\text{$x = \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha $ for all $x \in \X$}.
\]
[/latex]
Back: [latex]
The assumption states that $\S^{\perp} = \left\{ 0 \right\}$. Since $\S$ is orthonormal, we can use Bessel's inequality to get that for any $x \in \X$,
\begin{enumerate}[(1)]
\item $x_\S \triangleq \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha$ converges unconditionally, and
\item $x - x_\S \in \S^{\perp}$.
\end{enumerate}
Therefore, since $\S^{\perp} = \left\{ 0 \right\}$, it must be that $x - x_\S = 0$ so \[x = \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha.\]
[/latex]

Front: [latex]
Suppose $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ is an orthonormal subset of a Hilbert space $\X$. Show that
\[
\text{$x = \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha $ for all $x \in \X$}.
\] implies
\[
\text{$\left\Vert x \right\Vert ^{2} = \sum_{\alpha \in \I}^{} \left| \left\langle x, x_\alpha \right\rangle \right|^{2}$ for all $x \in \X$}.
\]
[/latex]
Back: [latex]
We know from a lemma that if $\left\{ x_\alpha \right\}_{\alpha \in \mathcal{I} }$ are orthogonal and the sum $\sum_{\alpha \in \I}^{} x_\alpha $ converges unconditionally, then
\[
\left\Vert \sum_{\alpha \in \I}^{ } x_\alpha \right\Vert ^{2} = \sum_{\alpha \in \I}^{} \left\Vert x_\alpha \right\Vert ^{2}.
\]
Define $y_\alpha := \left\langle x, x_\alpha \right\rangle x_\alpha $. Now we can write our assumptions as
\[x = \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha = \sum_{\alpha \in \I}^{} y_\alpha \]
so we know $\sum_{\alpha \in \I}^{} y_\alpha $ converges unconditionally. Therefore, we get that
\[\begin{aligned}
\left\Vert x \right\Vert ^{2} &= \left\Vert \sum_{\alpha \in \I }^{} y_\alpha \right\Vert^{2} &\text{(Assumption)} \\
&= \sum_{\alpha \in \I}^{} \left\Vert y_\alpha \right\Vert ^{2} &\text{(Lemma above)} \\
&= \sum_{\alpha \in \I}^{} \left\Vert \left\langle x, x_\alpha \right\rangle x_\alpha \right\Vert ^{2} &\text{(Def. of $y_\alpha $)} \\
&= \sum_{\alpha \in \I}^{} \left| \left\langle x, x_\alpha \right\rangle \right| \left\Vert x_\alpha \right\Vert ^{2} &\text{(Homogeneity of $\left\Vert \cdot \right\Vert $)} \\
&= \sum_{\alpha \in \I}^{} \left| \left\langle x, x_\alpha \right\rangle \right|^{2}. &\text{($\left\Vert x_\alpha \right\Vert = 1$)}
\end{aligned}\]
[/latex]

Front: [latex]
Suppose $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ is an orthonormal subset of a Hilbert space $\X$. Show that
\[
\text{$\left\langle x, x_\alpha \right\rangle = 0$ for all $\alpha \in \I$ implies that $x = 0$}
\] implies
\[
\left[ \S \right] = \X.
\]
[/latex]
Back: [latex]
The assumption states that $\S^{\perp} = \left\{ 0 \right\}$, so we can write
\[\begin{aligned}
\left[ \S \right]^{\perp} &= \overline{\text{span}\left( \S \right)}^{\perp} &\text{(Def. of $\left[ \cdot \right]$)} \\
&= \S^{\perp} &\text{(Lemma)} \\
&= \left\{ 0 \right\}.
\end{aligned}\]
Hence, $\left[ \S \right] = \X$.
[/latex]

Front: [latex]
Suppose $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ is an orthonormal subset of a Hilbert space $\X$. Show that
\[
\left[ \S \right] = \X
\] implies that $\S$ is a maximal orthonormal set, i.e. it has the property that if $\S' \supset \S$ is another orthonormal set, then necessarily $\S' = \S$.
[/latex]
Back: [latex]
The assumption implies that $x = \sum_{\alpha \in \I}^{} c_\alpha x_\alpha $ for all $x \in \X$. Therefore, if $x \perp x_\alpha $ for all $\alpha \in \I$, then $x = 0$.
[/latex]

Front: [latex]
Suppose $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ is an orthonormal subset of a Hilbert space $\X$. Show that if $\S$ is a maximal orthogonal set, i.e. has the property
\[\text{if $\S' \supset \S$ is another orthonormal set, then necessarily $\S' = \S$} \]
then
\[
\left\langle x, x_\alpha \right\rangle = 0 \text{ for all } \alpha \in \I \text{ implies that } x = 0.
\]
[/latex]
Back: [latex]
Suppose for a contradiction that there exists an $x \in \X$ such that $\left\langle x, x_\alpha \right\rangle = 0$ for all $\alpha \in \I$, but $x \neq 0$. Then $S \cup \left\{ \frac{x}{ \left\Vert x \right\Vert} \right\}$ is an orthonormal set and is a strict superset of $\S$.
[/latex]

Front: [latex]
What does it mean for an orthonormal subset $S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ of a Hilbert space to be complete?
[/latex]
Back: [latex]
This means that $\S$ fulfils any of these equivalent criteria:
\begin{enumerate}[(a)]
\item $\left\langle x, x_\alpha \right\rangle = 0$ for all $\alpha \in \I$ implies that $x = 0$,
\item $x = \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha $ for all $x \in \X$,
\item $\left\Vert x \right\Vert ^{2} = \sum_{\alpha \in \I}^{} \left| \left\langle x, x_\alpha \right\rangle \right|^{2}$ for all $x \in \X$,
\item $\left[ \S \right] = \X$,
\item $\S$ is a maximal orthonormal set, i.e. it has the property that if $\S' \supset \S$ is another orthonormal set, then necessarily $\S' = \S$.
\end{enumerate}
[/latex]

Front: [latex]
What's an orthonormal basis of a Hilbert space $\X$?
[/latex]
Back: [latex]
An orthonormal basis of a Hilbert space $\X$ is an orthonormal subset of $\X$ which is complete.
[/latex]

Front: [latex]
Suppose $\S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ is an orthonormal basis for $\X$. State the Parseval's identity.
[/latex]
Back: [latex]
For any $x \in \X$,
\[
\left\Vert x \right\Vert ^{2} = \sum_{\alpha \in \I}^{} \left| \left\langle x, x_\alpha \right\rangle \right|^{2}
\]
[/latex]

Front: [latex]
What are separable Hilbert spaces?
[/latex]
Back: [latex]
Separable Hilbert spaces are those which have a countable orthonormal basis.
[/latex]

Front: [latex]
Let $\X, \Y$ be vector spaces over $\F$. What is a linear operator from $\X$ to $\Y$? What are its range, rank, null space, and kernel?
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item
A linear operator from $\X$ to $\Y$ is a function $\T : \X \to \Y$ that satisfies
\[
\T \left( \alpha x + \beta y \right) = \alpha \T x + \beta \T y
\] for all $x, y \in \X$ and $\alpha , \beta \in \F$.
\item
The range of $\T$ is the set $\Rc\left( \T \right) = \left\{ \T x : x \in \X \right\}$.
\item The rank of $\T$ is the dimension of $\Rc\left( \T \right)$.
\item The null space of $\T$ is the set $\No\left( \T \right) = \left\{ x \in \X : \T x = 0 \right\}$.
\item The kernel is the same as the null space.
\end{enumerate}
[/latex]

Front: [latex]
Consider a linear operator $\T : \X \to \Y$. What kind of sets are the range and null space of $\T$? What are they subsets of?
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item The range $\Rc\left( \T \right)$ is a subspace of $\Y$, whereas
\item the null space $\No \left( \T \right)$ is a subspace of $\X$.
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\T$ is a linear operator. What can you say about the openness/closedness of its null space $\No\left( \T \right)$?
[/latex]
Back: [latex]
If $\T$ is continuous, then $\No\left( \T \right)$ is closed.
[/latex]

Front: [latex]
What does it mean for a linear operator $\T : \X \to \Y$ to be bounded?
[/latex]
Back: [latex]
A linear operator $\T : \X \to \Y$ is bounded if there exists a constant $K > 0$ such that for all $x \in \X$,
\[
\left\Vert \T x \right\Vert _\Y \le K \left\Vert x \right\Vert _\X.
\]
[/latex]

Front: [latex]
Suppose $\X$ and $\Y$ are vector spaces. What does $\B \left( \X, \Y \right)$ stand for?
[/latex]
Back: [latex]
$\B \left( \X, \Y \right)$ is the set of bounded linear operators $\T : \X \to \Y$.
[/latex]

Front: [latex]
Suppose $\X, \Y$ are normed spaces. What kind of space is $\B \left( \X, \Y \right)$?
[/latex]
Back: [latex]
We know that in this case, $\B\left( \X, \Y \right)$ is a normed space with norm
\[
\left\Vert \T \right\Vert = \sup_{\left\Vert x \right\Vert _\X = 1} \left\Vert \T x \right\Vert _{\Y}.
\] This is called the operator norm of an operator between two normed vector spaces.
Furthermore, if $\Y$ is a Banach space, then $\B\left( \X, \Y \right)$ is a Banach space.
[/latex]

Front: [latex]
Suppose $\X, \Y$ are normed spaces, and $\mathbb{T} $ is a bounded linear operator from $\X$ to $\Y$. What is $\left\Vert \mathbb{T} \right\Vert $?
[/latex]
Back: [latex]
It's the operator norm
\[
\left\Vert \T \right\Vert = \sup_{\left\Vert x \right\Vert _\X = 1} \left\Vert \T x \right\Vert _{\Y}.
\]
[/latex]

Front: [latex]
Suppose $\X, \Y$ are normed spaces, and $\Y : \X \to \Y$ is a linear operator. What does $\T$ being continuous imply about $\left\Vert \mathbb{T} \right\Vert $?
[/latex]
Back: [latex]
The following are equivalent:
\begin{enumerate}[(a)]
\item $\T$ is continuous,
\item $\left\Vert \T \right\Vert < \infty$,
\item $\left\Vert \T \right\Vert $ is bounded.
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\X, \Y$ are normed spaces, and $\Y : \X \to \Y$ is a linear operator. Show that if $\T$ is continuous, then $\left\Vert \T \right\Vert < \infty$.
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item
Suppose that $\left\Vert \T \right\Vert = \infty$.
\item Then, there exists a sequence $\left\{ x_n \right\}_n^{\infty}$ of elements $x_n \in \X$ with $\left\Vert x_n \right\Vert _\X = 1$ such that $\left\Vert \T x_n \right\Vert _\Y > n$ for $n$ sufficiently large.
\item Set $y_n = x_n / n$. Now $y_n \to 0$, but $\left\Vert \T y_n \right\Vert \ge 1$.
\item Therefore, $\T$ is not continuous as $\lim_{n \to \infty} \T y_n \neq \T 0 = 0$.
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\X, \Y$ are normed spaces, and $\Y : \X \to \Y$ is a linear operator. Show that if $\left\Vert \T \right\Vert < \infty$, then $\T$ is bounded.
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item Take any $x \in \X$.
\item For $x = 0$, $\left\Vert \T x \right\Vert _\Y = 0$ to satisfy linearity, so $\left\Vert \T x \right\Vert _\Y$ is obviously bounded.
\item For $x \neq 0$,
\[\begin{aligned}
\left\Vert \T x \right\Vert _\Y &= \left\Vert x \right\Vert _\X \left\Vert \T \frac{x}{\left\Vert x \right\Vert _\X} \right\Vert _\Y &\text{(Linearity)} \\
&\le \left\Vert \T \right\Vert \left\Vert x \right\Vert _\X &\text{(Def. of $\left\Vert \T \right\Vert$ )}\\
& < \infty.&\text{(Assumption)}
\end{aligned}\]
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\X, \Y$ are normed spaces, and $\Y : \X \to \Y$ is a linear operator. Show that if $\T$ is bounded, then $\T$ is continuous.
[/latex]
Back: [latex]
If $\T$ is bounded, then it is uniformly continuous, and hence also just continuous.
[/latex]

Front: [latex]
Let $\X, \Y$ be normed spaces, and $\T : \X \to \Y$ a linear operator. What's an upper bound on $\left\Vert \T x \right\Vert _\Y$ using $\left\Vert x \right\Vert _{\X}$?
[/latex]
Back: [latex]
For all $x \in \X$,
\[
\left\Vert \T x \right\Vert _\Y \le \left\Vert \T \right\Vert \left\Vert x \right\Vert _\X.
\]
[/latex]

Front: [latex]
Let $\X, \Y$ be normed spaces, and $\T : \X \to \Y$ a linear operator. Show that for all $x \in \X$,
\[
\left\Vert \T x \right\Vert _\Y \le \left\Vert \T \right\Vert \left\Vert x \right\Vert _\X.
\]
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item Take any $x \in \X$.
\item For $x = 0$, $\left\Vert \T x \right\Vert _\Y = 0$ to satisfy linearity, so $\left\Vert \T x \right\Vert _\Y$ is obviously bounded.
\item For $x \neq 0$,
\[\begin{aligned}
\left\Vert \T x \right\Vert _\Y &= \left\Vert x \right\Vert _\X \left\Vert \T \frac{x}{\left\Vert x \right\Vert _\X} \right\Vert _\Y &\text{(Linearity)} \\
&\le \left\Vert \T \right\Vert \left\Vert x \right\Vert _\X. &\text{(Def. of $\left\Vert \T \right\Vert$ )}\\
\end{aligned}\]
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\T_1$ and $\T_2$ are linear operators between normed spaces for which the range of $\T_2$ is contained in the domain of $\T_1$, so we can compose them. Suppose they are both bounded. What is an upper bound on $\left\Vert \T_1 \T_2 \right\Vert $?
[/latex]
Back: [latex]
\[
\left\Vert \T_1 \T_2 \right\Vert \le \left\Vert \T_1 \right\Vert \left\Vert \T_2 \right\Vert
\]
[/latex]

Front: [latex]
Suppose $\T_1$ and $\T_2$ are linear operators between normed spaces for which the range of $\T_2$ is contained in the domain of $\T_1$, so we can compose them. Suppose they are both bounded. Show that
\[
\left\Vert \T_1 \T_2 \right\Vert \le \left\Vert \T_1 \right\Vert \left\Vert \T_2 \right\Vert.
\]
[/latex]
Back: [latex]
\[\begin{aligned}
\left\Vert \T_1 \T_2 \right\Vert &= \sup_{\left\Vert x \right\Vert = 1} \left\{ \left\Vert \T_1 \T_2 x \right\Vert \right\}\\
&\le \sup_{\left\Vert x \right\Vert = 1} \left\{ \left\Vert \T_1 \right\Vert \left\Vert \T_2 x \right\Vert \right\}\\
&=\left\Vert \T_1 \right\Vert \left\Vert \T_2 \right\Vert.
\end{aligned}\]
[/latex]

Front: [latex]
Let $\X, \Y$ be normed spaces, and let $\T : \X \to \Y$ be a linear operator. What does it mean for $\T$ to be a linear isometry?
[/latex]
Back: [latex]
$\T$ is a linear isometry if $\left\Vert \T x \right\Vert _\Y = \left\Vert x \right\Vert _\X$ for all $x \in \X$.
[/latex]

Front: [latex]
What is a projection on a vector space $\X$?
[/latex]
Back: [latex]
A projection on a vector space $\X$ is a linear operator $\P : \X \to \X$ such that $\P^{2}=\P$.
[/latex]

Front: [latex]
Suppose $\P$ is a projection on $\X$. What can you say about $\X$ with respect to the range and null space of $\P$?
[/latex]
Back: [latex]
\[
\X = \Rc \left( \P \right) \oplus \No\left( \P \right)
\]
[/latex]

Front: [latex]
Suppose $\X$ is a vector space which is a direct sum $\S \oplus \Tc$. How many projections $\P$ are there such that $\S = \Rc\left( \P \right)$ and $\Tc = \No\left( \P \right)$?
[/latex]
Back: [latex]
There exists precisely one such projection.
[/latex]

Front: [latex]
Suppose $\X$ is a vector space which is a direct sum $\S \oplus \Tc$. What is the projection onto $\S$ along $\Tc$?
[/latex]
Back: [latex]
The projection onto $\S$ along $\Tc$ is the unique projection $\P : \X \to \X$ such that $\S = \Rc \left( \P \right)$ and $\Tc = \No\left( \P \right)$.
[/latex]

Front: [latex]
Suppose $\X$ is a vector space. Show that if $\P : \X \to \X$ is a projection, then $\X = \Rc\left( \P \right) \oplus \No\left( \P \right)$.
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item Note that $x \in \Rc\left( \P \right)$ if and only if $x = \P x$. By definition, $\P x \in \Rc\left( \P \right)$. Suppose that $x \in \Rc\left( \P \right)$. Then $x = \P z$ for some $z \in \X$ and, therefore, $\P x = \P \P z = \P z = x$.
\item Let $x \in \No\left( \P \right)\cap \Rc\left( \P \right)$. Then $x = \P x$ and $\P x = 0$, which implies $x = 0$. Hence, $\No\left( \P \right) \cap \Rc\left( \P \right) = \left\{ 0 \right\}$.
\item Now take any $x \in \X$. We can write it as $x = \P x + \left( \Id - \P \right)x$ with $\P x \in \Rc \left( \P \right)$ and $\left( \Id - \P \right)x \in \No\left( \P \right)$.
\item Thus, $\X = \Rc\left( \P \right) \oplus \No\left( \P \right)$.
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\X$ is a vector space. Show that if $\X = \S \oplus \Tc$, then there exists a unique projection $\P : \X \to \X$ such that $\S = \Rc\left( \P \right)$ and $\Tc = \No\left( \P \right)$.
[/latex]
Back: [latex]
By definition of the direct sum, each $x \in \X$ has a unique decomposition $x = s + t$ with $s \in \S$ and $t \in \Tc$. We can then define the projection by $\P x = s$.
[/latex]

Front: [latex]
What is an orthogonal projection on a Hilbert space $\X$?
[/latex]
Back: [latex]
An orthogonal projection on a Hilbert space $\X$ is a linear operator $\P : \X \to \X$ such that $\P^{2} = \P$ and
\[
\left\langle \P x, y \right\rangle = \left\langle x, \P y \right\rangle \text{ for $x, y \in \X$.}
\]
[/latex]

Front: [latex]
Suppose $\X$ is a Hilbert space, and consider an orthogonal projection $\P: \X \to \X$. Knowing that we can decompose $\X = \Rc\left( \P \right) \oplus \No\left( \P \right)$, what can you say about the sets $\Rc\left( \P \right)$ and $\No\left( \P \right)$ with respect to each other?
[/latex]
Back: [latex]
\begin{enumerate}[(a)]
\item $\No\left( \P \right)\cap \Rc\left( \P \right) = \left\{ 0 \right\}$ from definition of direct sum, and
\item $\Rc\left( \P \right)$ is closed with $\Rc\left( \P \right)^{\perp} = \No\left( \P \right)$ since $\P$ is orthogonal.
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\X$ is a Hilbert space, and consider an orthogonal projection $\P: \X \to \X$. Knowing that we can decompose $\X = \Rc\left( \P \right) \oplus \No\left( \P \right)$, show that $\Rc\left( \P \right)$ is closed with $\Rc\left( \P \right)^{\perp} = \No\left( \P \right)$.
[/latex]
Back: [latex]
If $x = \P z \in \Rc\left( \P \right)$ and $y \in \No\left( \P \right)$, then since $\P$ is an orthogonal projection we have
\[
\left\langle x, y \right\rangle = \left\langle \P z, y \right\rangle = \left\langle z, \P y \right\rangle = 0.
\]
Therefore, $\Rc\left( \P \right) = \No\left( \P \right)^{\perp}$ and, since $\No\left( \P \right)$ is non-empty, $\Rc\left( \P \right)$ is closed.
[/latex]

Front: [latex]
Suppose $\X$ is a Hilbert space, and consider a closed subspace $\S$. Knowing that $\X = \S \oplus \S^{\perp}$, what can you say about the projection $\P$ such that $\S = \Rc\left( \P \right)$ and $\S^{\perp} = \No\left( \P \right)$?
[/latex]
Back: [latex]
It is unique and orthogonal.
[/latex]

Front: [latex]
Suppose $\X$ is a Hilbert space, and consider a closed subspace $\S$. Knowing that $\X = \S \oplus \S^{\perp}$, show that the projection $\P$ such that $\S = \Rc\left( \P \right)$ and $\S^{\perp} = \No\left( \P \right)$ is unique and orthogonal.
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item Define the projection $\P$ by
\[
\P x = s, \text{ where $x = s + t$ with $s \in \S$, $t \in \S^{\perp}$}.
\]
Then, $\Rc\left( \P \right) = \S$ and $\No\left( \P \right) = \S^{\perp}$.
\item Let $x_1, x_2 \in \X$ and $x_i = s_i + t_i$ be the unique decomposition of $x_i$ with $s_i \in \S$ and $t_i \in \S^{\perp}$, $i = 1, 2.$ Now, we have
\[\begin{aligned}
\left\langle x_1, \P x_2 \right\rangle &= \left\langle x_1, \P \left( s_2 + t_2 \right) \right\rangle \\
&= \left\langle x_1, s_2 \right\rangle &\text{($s_2 \in \Rc\left( \P \right)$, $t_2 \in \No\left( \P \right))$} \\
&= \left\langle s_1 + t_1, s_2 \right\rangle \\
&= \left\langle s_1, s_2 \right\rangle \\
&= \left\langle s_1, s_2 + t_2 \right\rangle \\
&= \left\langle \P x_1, s_2 + t_2 \right\rangle\\
&=\left\langle \P x_1, x_2 \right\rangle
\end{aligned}\]
so $\P$ is an orthogonal projection.
\item Uniqueness of $\P$ follows from $\X = \S \oplus \S^{\perp}$.
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\P$ is a non-zero orthogonal projection. What can you say about $\left\Vert \P \right\Vert$?
[/latex]
Back: [latex]
\[
\left\Vert \P \right\Vert = 1
\]
[/latex]

Front: [latex]
Suppose $\P$ is a non-zero orthogonal projection. Show that $\left\Vert \P \right\Vert = 1$.
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item If $x \neq 0$, we get
\[\begin{aligned}
\left\Vert \P x \right\Vert &= \frac{\left\langle \P x, \P x \right\rangle }{\left\Vert \P x \right\Vert } \\
&= \frac{\left\langle x, \P x \right\rangle }{\left\Vert \P x \right\Vert } &\text{($\P$ is orthogonal)} \\
&\le \left\Vert x \right\Vert \cdot \frac{\left\Vert \P x \right\Vert }{\left\Vert \P x \right\Vert } &\text{(Cauchy-Schwarz)} \\
&= \left\Vert x \right\Vert .
\end{aligned}\]
\item Since $\left\Vert \P \right\Vert $ is defined as
\[
\left\Vert \P \right\Vert = \sup_{\left\Vert x \right\Vert = 1} \left\{ \left\Vert \P x \right\Vert \right\},
\] we have that $\left\Vert \P \right\Vert \le 1$.
\item For $\P x \neq 0$, we have $\left\Vert \P \left( \P x \right) \right\Vert = \left\Vert \P x \right\Vert $, which implies that \[\left\Vert \P \right\Vert = \sup_{\left\Vert x \right\Vert = 1} \left\{ \left\Vert \P x \right\Vert \right\} \ge 1.\]
\item Thus, $\left\Vert \P \right\Vert = 1$.
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\S = \left\{ x_\alpha : \alpha \in \I \right\}$ is an orthonormal system in a Hilbert space $\X$. Let $\V = \left[ \S \right]$ be the closed linear span of $\S$. What is an explicit formula for the orthogonal projection $\P_\V$ onto $\V$?
[/latex]
Back: [latex]
We can write $\P_\V$ as
\[
\P_\V x = \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha.
\]
[/latex]

Front: [latex]
Suppose $\S = \left\{ x_\alpha : \alpha \in \I \right\}$ is an orthonormal system in a Hilbert space $\X$. Let $\V = \left[ \S \right]$ be the closed linear span of $\S$. Show that the orthogonal projection $\P_\V$ onto $\V$ is given explicitly as
\[
\P_\V x = \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha.
\]
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item The set $\S$ can be extended to an orthonormal basis $\B$ of $\X$. This can be done by Gram-Schmidt orthogonalization if $\X$ is separable, or using Zorn's lemma generally.
\item So, we can enumerate $\B = \left\{ x_\alpha : \alpha \in \J \right\}$ by an index set $\J$ containing $\I$.
\item Since $\B$ is an orthonormal basis for $\X$, we know that we can write any $x \in \X$ uniquely as
\[\begin{aligned}
x &= \sum_{a \in \J}^{} \left\langle x, x_\alpha \right\rangle x_\alpha \\
&=\underbrace{\sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha}_{=:s} + \underbrace{\sum_{\alpha \in \J \setminus \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha}_{=:t} \\
&=s + t
\end{aligned}\] so the mapping $\P x = s$ is a projection.
\item Moreover, any such $s$ and $t$ are orthogonal since
\[\begin{aligned}
\left\langle s, t \right\rangle &=\left\langle \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha , \sum_{\beta \in \J \setminus \I}^{} \left\langle x, x_\beta \right\rangle x_\beta \right\rangle \\
&= \sum_{\substack{\alpha \in \I \\ \beta \in \J \setminus \I} }^{} \left\langle x, x_\alpha \right\rangle \left\langle x, x_\beta \right\rangle \underbrace{\left\langle x_\alpha , x_\beta \right\rangle}_{=0, \text{ as } \alpha \neq \beta } = 0,
\end{aligned}\]
so the projection $\P$ is orthogonal.
\item Since $\P$ is an orthogonal projection, we know that its range $\Rc\left( \P \right)$ is closed. Since we defined $\P$ as
\[
\P x = s = \sum_{\alpha \in \I}^{} \left\langle x, x_\alpha \right\rangle x_\alpha,
\] we have that $\Rc\left( \P \right) = \text{span}\left( \S \right)$. Finally, since the closure of a closed set is the set itself, we get that $\Rc\left( \P \right) = \left[ \S \right] = \V$. Hence, $\P$ is the desired projection onto $\V$.
\end{enumerate}
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Front: [latex]
Consider a Hilbert space $\X$ over $\F$. What do you call an element of $\B\left( \X, \F \right)$?
[/latex]
Back: [latex]
An element of $\B\left( \X, \F \right)$ is called a bounded linear functional.
[/latex]

Front: [latex]
State the Riesz representation theorem.
[/latex]
Back: [latex]
Let $f : \X \to \F$ be a bounded linear functional on a Hilbert space $\X$. Then there exists a unique $y_f \in \X$ such that $f(x) = \left\langle x, y_f \right\rangle $ for all $x \in \X$.
[/latex]

Front: [latex]
Let $\X, \Y$ be Hilbert spaces and $\T \in \B\left( \X, \Y \right)$ a bounded linear operator. What is the adjoint $\T^{*}$ of $\T$? What are its basic properties?
[/latex]
Back: [latex]
The adjoint $\T^{*} \in \B\left( \Y, \X \right)$ is the unique operator such that
\[
\left\langle \T x, y \right\rangle = \left\langle x, \T^{*}y \right\rangle , \text{ $x \in \X$, $y \in \Y$}
\] with $\left\Vert \T^{*} \right\Vert = \left\Vert \T \right\Vert $. Furthermore, $\T^{**} = \T$.
[/latex]

Front: [latex]
Let $\X$ be a Hilbert space and $\T \in \B\left( \X, \X \right)$. What does it mean for $\T$ to be self-adjoint?
[/latex]
Back: [latex]
$\T$ is self-adjoint if $\T = \T^{*}$.
[/latex]

Front: [latex]
Let $\X$ be a Hilbert space and $\T \in \B\left( \X, \X \right)$. What does it mean for $\T$ to be unitary?
[/latex]
Back: [latex]
$\T$ is unitary if it is invertible and if $\T^{-1} = \T^{*}$.
[/latex]

Front: [latex]
Let $\X$ be a Hilbert space and $\T \in \B\left( \X, \X \right)$. What does it mean for $\T$ to be normal?
[/latex]
Back: [latex]
$\T$ is normal if $\T\T^{*} = \T^{*}\T$.
[/latex]

Front: [latex]
Show that every orthogonal projection $\P$ is self-adjoint.
[/latex]
Back: [latex]
This follows directly from the definition. A projection $\P : \X \to \X$ is orthogonal if
\[
\left\langle \P x, y \right\rangle = \left\langle x , \P y \right\rangle \text{ for all $x, y \in \X$}
\] whereas an operator $\T^{*} \in \B\left( \Y, \X \right)$ is adjoint to $\T \in \B\left( \X, \Y \right)$ if
\[
\left\langle \T x, y \right\rangle = \left\langle x, \T^{*} y \right\rangle \text{ for all $x \in \X$, $y \in \Y$}.
\] Since $\left\Vert \P \right\Vert = 1$ so $\P \in \B\left( \X, \X \right)$, we can clearly see that $\P = \P^{*}$.
[/latex]

Front: [latex]
Show that if $\T : \X \to \X$ is a unitary operator then it is a linear isometry.
[/latex]
Back: [latex]
\begin{enumerate}[1.]
\item $\T$ is a linear isometry if $\left\Vert \T x \right\Vert _\Y = \left\Vert x \right\Vert _\X$ for all $x \in \X$.
\item If $\T$ is unitary, we can write
\[\begin{aligned}
\left\Vert \T x \right\Vert ^{2} &= \left\langle \T x, \T x \right\rangle \\
&= \left\langle \T^{*}\T x, x \right\rangle \\
&= \left\langle \text{Id }x, x \right\rangle \\
&= \left\Vert x \right\Vert ^{2} \text{ for all $x \in \X$.}
\end{aligned}\]
\end{enumerate}
[/latex]

Front: [latex]
Suppose $\X$ is a Hilbert space, and $\T$ is self-adjoint. How can you find $\left\Vert \T \right\Vert $?
[/latex]
Back: [latex]
\begin{enumerate}[(1)]
\item Since $\T$ is self-adjoint, we can use \[\left\Vert \T \right\Vert = \sup_{\left\Vert x \right\Vert = 1} \left\{ \left| \left\langle \T x, x \right\rangle \right| \right\}.\]
\item We can also use the definition of $\left\Vert \T \right\Vert $ directly, so \[\left\Vert \T \right\Vert = \sup_{\left\Vert x \right\Vert = 1} \left\{ \left\Vert \T x \right\Vert \right\}.\]
\end{enumerate}
[/latex]

Front: [latex]
Let $\X$ be a Hilbert space. When is a self-adjoint operator $\T \in \B\left( \X, \X \right)$
\begin{enumerate}[(1)]
\item positive semidefinite?
\item positive definite?
\end{enumerate}
[/latex]
Back: [latex]
The self-adjoint operator $\T$ is
\begin{enumerate}[(1)]
\item positive semidefinite if $\left\langle \T x, x \right\rangle \ge 0$ for all $x \in \X$. In this case, we write $\T \ge 0$.
\item positive definite if it is positive semidefinite and $\left\langle \T x, x \right\rangle = 0$ implies that $x = 0$. In this case, we write $\T > 0$.
\end{enumerate}
[/latex]

Front: [latex]
Consider a Hilbert space $\X$ and a semidefinite linear bounded operator $\T \in \B\left( \X, \X \right)$. What does it mean for an operator $\Ab \in \B\left( \X, \X \right)$ to be a square root of $\T$?
[/latex]
Back: [latex]
$\Ab$ is a square root of $\T$ if $\T = \Ab^{2}$.
[/latex]

Front: [latex]
Suppose $\X$ is a Hilbert space, $\T \in \B\left( \X, \X \right)$, $\T \ge 0$, and a square root $\A$ of $\T$. When does $\A$ exist? Is it unique? What are its basic properties?
[/latex]
Back: [latex]
If $\T \in \B\left( \X, \X \right)$ and $\T \ge 0$, then the square root exists and is unique. Furthermore, it is positive definite.
[/latex]

Front: [latex]
When can we say that an operator $\T : \X \to \Y$ is invertible?
[/latex]
Back: [latex]
An operator $\T$ is invertible if it is surjective and injective.
[/latex]

Front: [latex]
Suppose that an orthonormal subset $S = \left\{ x_\alpha \in \X | \alpha \in \I \right\}$ of a Hilbert space $\mathcal{H} $ is complete. How can you use it to represent $x \in \mathcal{H}$ as sum of orthogonal vectors?
[/latex]
Back: [latex]
We can write
\[
x = \sum_{\alpha \in \mathcal{I}}^{} \left\langle x, x_\alpha \right\rangle x_\alpha.
\]
[/latex]