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[Analysis] More Linear Algebra
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RobinB27
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Relevant definitions used throughout Analysis II.
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Relevant definitions used throughout Analysis II.\\
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\subtext{$\textbf{A} \in \R^{m \times n},\quad x,y \in \R^n,\quad \alpha \in \R$}
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\definition \textbf{Scalar Product} $x \cdot y :=\sum_{i=0}^{n} (x_i \cdot y_i)$
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\definition \textbf{Euclidian Norm} $||x|| := \displaystyle\sqrt{\sum_{i=1}^{n} x_i^2}$\\
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\subtext{Used to generalize $|x|$ in many Analysis I definitions}
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\lemma \textbf{Properties of} $||x||$
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\begin{center}
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$
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\begin{array}{ll}
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(i) & ||x|| \geq 0 \\
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(ii) & ||x|| \iff x = 0 \\
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(iii) & ||\alpha x|| = \alpha \cdot ||x|| \\
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(iv) & ||x + y|| \leq ||x|| + ||y||\quad \text{(Triangle Inequality)}
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\end{array}
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$
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\end{center}
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\definition \textbf{Definiteness}
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\begin{center}
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$
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\begin{array}{lcl}
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\text{Positive Definite} &\iffdef& x^\top \textbf{A} x > 0\ \forall x \in \R^n_{\neq 0} \\
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\text{Negative Definite} &\iffdef& x^\top \textbf{A} x < 0\ \forall x \in \R^n_{\neq 0}
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\end{array}
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$
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\end{center}
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\smalltext{If $0$ is allowed, $\textbf{A}$ is called positive/negative semi-definite.}
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\definition \textbf{Trace} $\text{Tr}(\textbf{A}) := \displaystyle\sum_{i=0}^{\text{min}(m,n)} (\textbf{A})_{i, i}$
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\begin{footnotesize}
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\lemma \textbf{Determinant} of $\textbf{A} \in \R^{2\times2}$
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$$
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\det(\textbf{A})
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=
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\det\left(
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\begin{bmatrix}
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a & b \\
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c & d
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\end{bmatrix}
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\right) = ad - bc
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$$
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\definition \textbf{Scalar Product} $x \cdot y :=\sum_{i=0}^{n} (x_i \cdot y_i)$\\
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\definition \textbf{Euclidian Norm} $||x|| := \displaystyle\sqrt{\sum_{i=1}^{n} x_i^2}$\\
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\lemma \textbf{Properties of} $||x||$
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\begin{center}
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$
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\begin{array}{ll}
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(i) & ||x|| \geq 0 \\
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(ii) & ||x|| \iff x = 0 \\
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(iii) & ||\alpha x|| = \alpha \cdot ||x|| \\
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(iv) & ||x + y|| \leq ||x|| + ||y||\quad \text{(Triangle Inequality)}
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\end{array}
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$
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\end{center}
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\definition \textbf{Definiteness}
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\begin{center}
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$
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\begin{array}{lcl}
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\text{Positive Definite} &\iffdef& x^\top \textbf{A} x > 0\ \forall x \in \R^n_{\neq 0} \\
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\text{Negative Definite} &\iffdef& x^\top \textbf{A} x < 0\ \forall x \in \R^n_{\neq 0}
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\end{array}
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$
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\end{center}
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\color{gray}\scriptsize
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If $0$ is allowed, $\textbf{A}$ is called positive/negative semi-definite.
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\color{black}\footnotesize
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\definition \textbf{Trace} $\text{Tr}(\textbf{A}) := \displaystyle\sum_{i=0}^{\text{min}(m,n)} (\textbf{A})_{i, i}$\\
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\lemma \textbf{Inverse} of $\textbf{A} \in \R^{2\times2}$
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$$
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\textbf{A}^{-1}
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@@ -61,4 +46,61 @@ Relevant definitions used throughout Analysis II.
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-c & a
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\end{bmatrix}
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$$
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\end{footnotesize}
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\color{gray}\scriptsize
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For $A \in \R^{n \times n}$: Gauss Algorithm.
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\color{black}\footnotesize
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\lemma \textbf{Determinant} of $\textbf{A} \in \R^{2\times2}$
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$$
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\det(\textbf{A})
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=
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\det\left(
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\begin{bmatrix}
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a & b \\
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c & d
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\end{bmatrix}
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\right) = ad - bc
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$$
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\color{gray}\scriptsize
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For $A \in \R^{n \times n}$: Cofactor method.
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\color{black}\footnotesize
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\lemma \textbf{Determinant} of $A \in \R^{3\times3}$ (Sarrus)
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% https://tex.stackexchange.com/a/32981/184539
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$$\begin{tikzpicture}[>=stealth]
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\matrix [%
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matrix of math nodes,
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column sep=1em,
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row sep=1em
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] (sarrus) {%
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a_{11} & a_{12} & a_{13} & a_{11} & a_{12} \\
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a_{21} & a_{22} & a_{23} & a_{21} & a_{22} \\
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a_{31} & a_{32} & a_{33} & a_{31} & a_{32} \\
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};
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\path ($(sarrus-1-1.north west)-(0.5em,0)$) edge ($(sarrus-3-1.south west)-(0.5em,0)$)
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($(sarrus-1-3.north east)+(0.5em,0)$) edge ($(sarrus-3-3.south east)+(0.5em,0)$)
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(sarrus-1-1) edge (sarrus-2-2)
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(sarrus-2-2) edge[->] (sarrus-3-3)
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(sarrus-1-2) edge (sarrus-2-3)
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(sarrus-2-3) edge[->] (sarrus-3-4)
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(sarrus-1-3) edge (sarrus-2-4)
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(sarrus-2-4) edge[->] (sarrus-3-5)
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(sarrus-3-1) edge[dashed] (sarrus-2-2)
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(sarrus-2-2) edge[->,dashed] (sarrus-1-3)
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(sarrus-3-2) edge[dashed] (sarrus-2-3)
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(sarrus-2-3) edge[->,dashed] (sarrus-1-4)
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(sarrus-3-3) edge[dashed] (sarrus-2-4)
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(sarrus-2-4) edge[->,dashed] (sarrus-1-5);
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\foreach \c in {1,2,3} {\node[anchor=south] at (sarrus-1-\c.north) {$+$};};
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\foreach \c in {1,2,3} {\node[anchor=north] at (sarrus-3-\c.south) {$-$};};
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\end{tikzpicture}$$
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\lemma \textbf{Properties of Eigenvalues}
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$$
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\text{Tr}(\textbf{A}) = \sum_{i=0}^{n} \lambda_i \qquad\qquad \text{det}(\textbf{A}) = \prod_{i=0}^{n} \lambda_i
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$$
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\color{gray}\scriptsize
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To find $\lambda_i$ solve $\det(\textbf{A} - \lambda \textbf{I}) = 0$.
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\color{black}\footnotesize
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\end{footnotesize}
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