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eth-summaries/semester3/analysis-ii/cheat-sheet-rb/parts/01_linalg.tex
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Relevant definitions used throughout Analysis II.\\
\subtext{$\textbf{A} \in \R^{m \times n},\quad x,y \in \R^n,\quad \alpha \in \R$}
\begin{footnotesize}
\definition \textbf{Scalar Product} $x \cdot y :=\sum_{i=0}^{n} (x_i \cdot y_i)$\\
\definition \textbf{Euclidian Norm} $||x|| := \displaystyle\sqrt{\sum_{i=1}^{n} x_i^2}$\\
\lemma \textbf{Properties of} $||x||$
\begin{center}
$
\begin{array}{ll}
(i) & ||x|| \geq 0 \\
(ii) & ||x|| \iff x = 0 \\
(iii) & ||\alpha x|| = \alpha \cdot ||x|| \\
(iv) & ||x + y|| \leq ||x|| + ||y||\quad \text{(Triangle Inequality)}
\end{array}
$
\end{center}
\definition \textbf{Definiteness}
\begin{center}
$
\begin{array}{lcl}
\text{Positive Definite} &\iffdef& x^\top \textbf{A} x > 0\ \forall x \in \R^n_{\neq 0} \\
\text{Negative Definite} &\iffdef& x^\top \textbf{A} x < 0\ \forall x \in \R^n_{\neq 0}
\end{array}
$
\end{center}
\color{gray}\scriptsize
If $0$ is allowed, $\textbf{A}$ is called positive/negative semi-definite.
\color{black}\footnotesize
\definition \textbf{Trace} $\text{Tr}(\textbf{A}) := \displaystyle\sum_{i=0}^{\text{min}(m,n)} (\textbf{A})_{i, i}$\\
\lemma \textbf{Inverse} of $\textbf{A} \in \R^{2\times2}$
$$
\textbf{A}^{-1}
=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}^{-1}
=
\frac{1}{\det(\textbf{A})}
\begin{bmatrix}
d & -b \\
-c & a
\end{bmatrix}
$$
\color{gray}\scriptsize
For $A \in \R^{n \times n}$: Gauss Algorithm.
\color{black}\footnotesize
\lemma \textbf{Determinant} of $\textbf{A} \in \R^{2\times2}$
$$
\det(\textbf{A})
=
\det\left(
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\right) = ad - bc
$$
\color{gray}\scriptsize
For $A \in \R^{n \times n}$: Cofactor method.
\color{black}\footnotesize
\lemma \textbf{Determinant} of $A \in \R^{3\times3}$ (Sarrus)
% https://tex.stackexchange.com/a/32981/184539
$$\begin{tikzpicture}[>=stealth]
\matrix [%
matrix of math nodes,
column sep=1em,
row sep=1em
] (sarrus) {%
a_{11} & a_{12} & a_{13} & a_{11} & a_{12} \\
a_{21} & a_{22} & a_{23} & a_{21} & a_{22} \\
a_{31} & a_{32} & a_{33} & a_{31} & a_{32} \\
};
\path ($(sarrus-1-1.north west)-(0.5em,0)$) edge ($(sarrus-3-1.south west)-(0.5em,0)$)
($(sarrus-1-3.north east)+(0.5em,0)$) edge ($(sarrus-3-3.south east)+(0.5em,0)$)
(sarrus-1-1) edge (sarrus-2-2)
(sarrus-2-2) edge[->] (sarrus-3-3)
(sarrus-1-2) edge (sarrus-2-3)
(sarrus-2-3) edge[->] (sarrus-3-4)
(sarrus-1-3) edge (sarrus-2-4)
(sarrus-2-4) edge[->] (sarrus-3-5)
(sarrus-3-1) edge[dashed] (sarrus-2-2)
(sarrus-2-2) edge[->,dashed] (sarrus-1-3)
(sarrus-3-2) edge[dashed] (sarrus-2-3)
(sarrus-2-3) edge[->,dashed] (sarrus-1-4)
(sarrus-3-3) edge[dashed] (sarrus-2-4)
(sarrus-2-4) edge[->,dashed] (sarrus-1-5);
\foreach \c in {1,2,3} {\node[anchor=south] at (sarrus-1-\c.north) {$+$};};
\foreach \c in {1,2,3} {\node[anchor=north] at (sarrus-3-\c.south) {$-$};};
\end{tikzpicture}$$
\lemma \textbf{Properties of Eigenvalues}
$$
\text{Tr}(\textbf{A}) = \sum_{i=0}^{n} \lambda_i \qquad\qquad \text{det}(\textbf{A}) = \prod_{i=0}^{n} \lambda_i
$$
\color{gray}\scriptsize
To find $\lambda_i$ solve $\det(\textbf{A} - \lambda \textbf{I}) = 0$.
\color{black}\footnotesize
\end{footnotesize}