[LinAlg] Fixes #3

semester1/linear-algebra/linAlg-janishutz.tex

@@ -44,7 +44,7 @@
         ``\textit{Lineare Algebra ist Kulturgut}''
     \end{Large}
 
-    \hspace{3cm} - Robert Weissmantel, 2024
+    \hspace{3cm} - Robert Weismantel, 2024
 \end{center}
 
 \vspace{3cm}

semester1/linear-algebra/parts/solving-sle.tex

@@ -2,9 +2,12 @@
 \vspace{-0.5pc}
 \section{Solving Linear Equations}
 \label{sec:sle}
-Put the system of linear equation's factors (i.e. for a linear equation $ax + by + cz = u$, we would put $a, b, c$ into the matrix and $u$ into the vector) into a matrix, where each row is an equation and the result into a vector $b$. Then, we solve $Ax = b$ by Gauss elimination.
+Put the system of linear equation's factors (i.e. for a linear equation $ax + by + cz = u$, we would put $a, b, c$ into the matrix and $u$ into the vector) into a matrix,
+where each row is an equation and the result into a vector $b$. Then, we solve $Ax = b$ by Gauss elimination.
 
-\textbf{Gauss elimination}: Transform matrix $A$ into upper triangle matrix by performing row transformations (adding, adding scalar multiples, multiplying by scalar) on it. All operations performed on $A$ have to also be performed on $b$. Typically, write down both as a matrix with a dividing line between $A$ and $b$. Then solve by back-substitution. Gauss elimination succeeds iff $A \in \R^{m \times m}$ and $A$'s columns are linearly independent. (Runtime: \tco{m^3})
+\textbf{Gauss elimination}: Transform matrix $A$ into upper triangle matrix by performing row transformations (adding, adding scalar multiples, multiplying by scalar) on it.
+All operations performed on $A$ have to also be performed on $b$. Typically, write down both as a matrix with a dividing line between $A$ and $b$.
+Then solve by back-substitution. Gauss elimination succeeds iff $A \in \R^{m \times m}$ and $A$'s columns are linearly independent. (Runtime: $\tco{m^3}$)
 
 \vspace{-0.5pc}
 \subsection{Inverse}
@@ -20,43 +23,46 @@ Put the system of linear equation's factors (i.e. for a linear equation $ax + by
 \subsection{LU-Decomposition}
 \label{sec:lu-decomp}
 \setcounter{all}{13}\shorttheorem \textbf{LU-Decomposition}: $A = LU$. $U$ upper triangle, result of Gauss elimination, $L$ lower triangle, $(E_1 \times E_2 \times \ldots \times E_n)^{-1}$.
-Transformation matrices $E$ ($E \cdot A = A_1$): transformation is a single entry in lower triangle, where the $i$ and $j$ are the two rows involved and the value of $e_{ij}$ is the operation performed on the two. $L^{-1}$ for size $3$: Diagonal, $L^{-1}_{1,2} = -L_{1,2}$, $L^{-1}_{2,3} = -L_{2,3}$, $L^{-1}_{1,3} = -L_{1,3}$.
+Transformation matrices $E$ ($E \cdot A = A_1$): transformation is a single entry in lower triangle, where the $i$ and $j$ are the two rows involved and the value of
+$e_{ij}$ is the operation performed on the two. $L^{-1}$ for size $3$: Diagonal, $L^{-1}_{1,2} = -L_{1,2}$, $L^{-1}_{2,3} = -L_{2,3}$, $L^{-1}_{1,3} = -L_{1,3}$.
 When multiplying up to three different ones, copies values to multiplied matrix.
-If it is impossible to decompose $A$ into $LU$ without row exchanges, we get $PA = LU$, where $P$ is a permutation matrix (indicating which rows have been swapped). Time complexity is improved significantly with this \tco{m^2}.
+If it is impossible to decompose $A$ into $LU$ without row exchanges, we get $PA = LU$, where $P$ is a permutation matrix (indicating which rows have been swapped).
+Time complexity is improved significantly with this $\tco{m^2}$.
 
-\shortdef \textbf{Permutations}: bijective function $\pi$ on matrix; Reorders the input structure (i.e. vector or matrix); 
+\shortdef \textbf{Permutations}: bijective function $\pi$ on matrix; Reorders the input structure (i.e. vector or matrix);
 \shortdef \textbf{Permutation matrix}: $p_{ij} = \begin{cases}
-    1 & \text{if } j = \pi(i)\\
-    0 & \text{else}
-\end{cases}$
-\shortlemma $P^{-1} = P^{\top}$. 
+        1 & \text{if } j = \pi(i) \\
+        0 & \text{else}
+    \end{cases}$
+\shortlemma $P^{-1} = P^{\top}$.
 
 \setcounter{all}{18}\shorttheorem \textbf{LUP-Decomposition}: $PA = LU$, where $U = P\times L A$. $P_j = I$ if no row swaps are performed at each step and $P = P_m \times P_{m - 1} \times \ldots \times P_1$. Rewriting this as $A = P^{\top}LU$, we can simply solve an SLE using LUP-Decomposition
 
 \vspace{-0.5pc}
 \subsection{Gauss-Jordan-Elimination}
 \label{sec:gauss-jordan}
-\textbf{Gauss-Jordan-Elimination}: Generalization of Gauss elimination to $m \times n$ matrices, still works similarly to Gauss elimination. We aim to find REF or RREF (see \ref{sec:matrices}). To describe, we say REF$(j_1, j_2, \ldots, j_r)$ or equivalently with RREF, where $j_r$ is the $r$-th pivot.
+\textbf{Gauss-Jordan-Elimination}: Generalization of Gauss elimination to $m \times n$ matrices, still works similarly to Gauss elimination.
+We aim to find REF or RREF (see \ref{sec:matrices}). To describe, we say REF$(j_1, j_2, \ldots, j_r)$ or equivalently with RREF, where $j_r$ is the $r$-th pivot.
 
 The solution is then in a vector, whose components are either $0$ or the $r$-th component of $b$. Example:
 \[
     \begin{bmatrix}
-        0 & 1 & 0 & 0 & 2 & 0\\
-        0 & 0 & 1 & 0 & 3 & 0\\
-        0 & 0 & 0 & 1 & 2 & 0\\
-        0 & 0 & 0 & 0 & 0 & 1\\
+        0 & 1 & 0 & 0 & 2 & 0 \\
+        0 & 0 & 1 & 0 & 3 & 0 \\
+        0 & 0 & 0 & 1 & 2 & 0 \\
+        0 & 0 & 0 & 0 & 0 & 1 \\
         0 & 0 & 0 & 0 & 0 & 0
     \end{bmatrix}
     \cdot
     \begin{bmatrix}
-        0\\b_1\\b_2\\b_3\\0\\b_4
+        0 \\b_1\\b_2\\b_3\\0\\b_4
     \end{bmatrix}
     =
     \begin{bmatrix}
-        b_1\\b_2\\b_3\\b_4\\\textcolor{Green}{0}
+        b_1 \\b_2\\b_3\\b_4\\\textcolor{Green}{0}
     \end{bmatrix}
 \]
-If the green marked entry in $b$ were not to be $0$, then the SLE would not have a solution. 
+If the green marked entry in $b$ were not to be $0$, then the SLE would not have a solution.
 
 
 \textbf{CR-Decomposition}: see \ref{sec:matrices} for exaplanation. \setcounter{all}{24}\shorttheorem is described there