Note - check if need revising
\begin{enumerate}
\item Positivity
Frobenius norm, sometimes also called the Euclidean norm
special case:
Tips: pretty sure think
\begin{equation}
\left.\begin{array}{l}
{\|f\|p=\left(∫ab|f(τ)|p d τ\right)\frac{1{p}}, \quad p ∈[1, ∞]}
{\|f\|∞=\displaystyle\sup \scriptscriptstyle a \leq t \leq b|f(t)| }\end{array} \quad,
\right\} \quad \mathscr{L}p-\text { norms }
\end{equation}
$f ∈ \mathscr{L}p ⇔\|f\|p$ is bounded, i.e. $∃ c:\|f\|p \leq c$
\begin{table}[h]
\begin{tabular}{cl}
rank & object
\hline
0 & scalar \
1 & vector \
2 & matrix (/Dyad) \
Example: $A=\left[\begin{array}{lll}{0} & {1} & {1} \ {\sqrt{2}} & {2} & {0} \ {0} & {1} & {1}\end{array}\right]$
The SVD is defined as $$ A=P Σ QT $$
Computing: $$ A AT=\left[\begin{array}{lll}{2} & {2} & {2} \ {2} & {6} & {2} \ {2} & {2} & {2}\end{array}\right] $$
\begin{equation}
\begin{aligned}
-λ3+10 λ2-16 λ &=-λ\left(λ2-10 λ+16\right)
&=-λ(λ-8)(λ-2)
\end{aligned}
\end{equation}
Eigenvals of $A AT$ are
Giving out the matrix $$ Σ= 03x3+σ= \left[\begin{array}{ccc}{2 \sqrt{2}} & {0} & {0} \ {0} & {\sqrt{2}} & {0} \ {0} & {0} & {0}\end{array}\right] $$
Finding the eigenvectors
Yeilding
$$
AT A=\left[\begin{array}{ccc}{2} & {2 \sqrt{2}} & {0} \ {2 \sqrt{2}} & {6} & {2} \ {0} & {2} & {2}\end{array}\right]
$$
With the eigenvals
We have then the SVD defined as $$ A=P Σ QT $$
\newpage
\begin{align*} λ&=1 \quad \operatorname{NUL}(A-1 I) = \text { SPAN }\left\{\left[\begin{array}{l}{1} \ {0} \ {0}\end{array}\right],\left[\begin{array}{c}{0} \ {1} \ {-1}\end{array}\right]\right\} \end{align*} 2 eigenvectors? :-d (for some reason we are calling these vectors the first and third)
\begin{align*}
A V1&=λ V1
A V2&=V1+λ V2
\end{align*}
\begin{equation*}
\begin{array}{l}{A V2-λ V2=V1} \ {(A-λ I) V2=V1}\end{array}
\end{equation*}
\left[\begin{array}{lll}{0} & {1} & {1} \ {0} & {0} & {0} \ {0} & {0} & {0}\end{array}\right] V2&=\left[\begin{array}{l}{1} \ {0} \ {0}\end{array}\right]
\end{align*}
\begin{equation} V2=\left[\begin{array}{l}{0} \ {0} \ {1}\end{array}\right]+\bcancel{x\left[\begin{array}{l}{1} \ {0} \ {0}\end{array}\right]+y\left[\begin{array}{c}{0} \ {1} \ {-1}\end{array}\right]} \end{equation} Let \begin{equation} V2=\left[\begin{array}{l}{0} \ {0} \ {1}\end{array}\right] \end{equation}
\begin{equation} V3=\left[\begin{array}{c}{0} \ {1} \ {-1}\end{array}\right] \end{equation}
\begin{equation} A \quad \text { has the from }\left[\begin{array}{lll}1 & 1 &0 \ 0 & 1 &0 \ 0 & 0 &1\end{array}\right] \end{equation}
\begin{equation} P=\left[\begin{array}{lll}{V1} & {V2} & {V3}\end{array}\right]= \left[\begin{array}{ccc}{1} & {0} & {0} \ {0} & {0} & {1} \ {0} & {1} & {-1}\end{array}\right] \end{equation} then \begin{equation} A=P J P-1 \quad J=\begin{equation} \left[\begin{array}{lll}{1} & {1} & {0} \ {0} & {1} & {0} \ {0} & {0} & {1}\end{array}\right] \end{equation} \end{equation}
A square matrix A is positive definite if there is a positive scalar α such that \begin{equation} xT A x \geq α xT x, \quad \text { for all } x ∈ \mathbf{R}n \end{equation} It is positive semidefinite if \begin{equation} xT A x \geq 0, \quad \text { for all } x ∈ \mathbb{R}n \end{equation} We can recognize that a symmetric matrix is positive definite by computing its eigenvalues and verifying that they are all positive, or by performing a Cholesky factorization.
3D point as 3-vector $$ \mathbf{X}=\left[\begin{array}{l}{X} \ {Y} \ {Z}\end{array}\right] $$
3D point using affine homogeneous
$$
\left[\begin{array}{l}{\mathbf{X}} \ {1}\end{array}\right]=\left[\begin{array}{l}{X} \ {Y} \ {Z} \ {1}\end{array}\right]
$$
Inverse of rotation matrix
\begin{equation}
\begin{aligned} \mathrm{X}′ &=\mathrm{RX}+\mathrm{t}
\mathrm{X}′-\mathrm{t} &=\mathrm{RX} \
\mathrm{R}\top\left(\mathrm{X}′-\mathrm{t}\right) &=\mathrm{X} \
\mathrm{R}\top \mathrm{X}′-\mathrm{R}\top \mathrm{t} &=\mathrm{X}
\end{aligned}
\end{equation}
This gives in homogenius
rotation matrix in vector form \begin{equation} \begin{array}{l} θ ← norm(r) \ r ← r/ θ \ R = cos{θ} I + (1- cos{θ} ) r r^T + sin{θ}\left[\begin{array}{ccc}{0} & {-rz} & {ry} \ {rz} & {0} & {-rx} \ {-ry} & {rx} & {0}\end{array}\right] \end{array} \end{equation}
Inverse transformation can be also done easily, since
\begin{equation} sin{θ}\left[\begin{array}{ccc}{0} & {-rz} & {ry} \ {rz} & {0} & {-rx} \ {-ry} & {rx} & {0}\end{array}\right]= \frac{R - R^T}{2} \end{equation}
A rotation vector is a convenient and most compact representation of a rotation matrix (since any rotation matrix has just 3 degrees of freedom)
\newpage
\begin{equation*} \begin{array}{|c|c|c|c|c|}\hline S & {N} & {S ⇒ N} & {S ⇐ N} & {S ⇔ N} \ \hline T & {T} & {T} & {T} & {T} \ \hline T & {F} & {F} & {T} & {F} \ \hline F & {T} & {T} & {F} & {F} \ \hline F & {F} & {T} & {T} & {T} \ \hline\end{array} \end{equation*}
\begin{center} \begin{tikzpicture} \def\radius{2cm} \def\mycolorbox#1{\textcolor{#1}{\rule{2ex}{2ex}}} \colorlet{colori}{blue!70} \colorlet{colorii}{red!70}
\coordinate (ceni); \coordinate[xshift=\radius] (cenii);
\draw[fill=colori,fill opacity=0.5] (ceni) circle (\radius); \draw[fill=colorii,fill opacity=0.5] (cenii) circle (\radius);
\draw ([xshift=-20pt,yshift=20pt]current bounding box.north west) rectangle ([xshift=20pt,yshift=-20pt]current bounding box.south east);
\node[yshift=10pt] at (current bounding box.north) {Venn diagram }; \node[xshift=-.5\radius] at (ceni) {$\mathbf{S}$}; \node[xshift=.5\radius] at (cenii) {$\mathbf{N}$}; \node[xshift=.9\radius] at (ceni) {$\mathbf{S}∩\mathbf{N}}$}; \node[xshift=10pt,yshift=10pt] at (current bounding box.south west) {$∅$}; \end{tikzpicture} \end{center}
Example based off one story from Aesop’s Fables:
\begin{table}[h!]
\begin{tabular}{|l|l|}
\hline
\makecell[l]{\textbf{True Positive (TP):}
Reality: A wolf threatened. \
Shepherd said: “Wolf.” \
Outcome: Shepherd is a hero. } &
\makecell[l]{\textbf{False Positive (FP):}\
Reality: No wolf threatened. \
Shepherd said: “Wolf.” \
Outcome: Villagers are \ angry at shepherd for waking them up.} \ \hline
\makecell[l]{\textbf{False Negative (FN):}\
Reality: A wolf threatened. \
Shepherd said: “No wolf.” \
Outcome: The wolf ate all the sheep.} &
\makecell[l]{\textbf{True Negative (TN):}\
Reality: No wolf threatened. \
Shepherd said: “No wolf.” \
Outcome: Everyone is fine. } \ \hline
\end{tabular}
\end{table}
## ranodm
- f(f)=c^t - exponential
- f(f)=t^n (n el natural num) - polymonial
- f(f)=t^c - ??