Correction to notes in lecture 6

notes/lecture6.tex

@@ -23,7 +23,7 @@
 \maketitle
 
 \section*{2.5: Partial pivoting}
-Obviously, Algorithm 2.8 fils if one of the pivots becomes zero. In this case, we need to choose a different pivot element.
+Obviously, Algorithm 2.8 fails if one of the pivots becomes zero. In this case, we need to choose a different pivot element.
 
 \paragraph{Simple approach:} Column pivoting choose $|a^{(i)}_{k,i}|=\max_{i \leq l \leq n}|a^{(i)}_{l,i}|$
 In order to move the pivot element and the  corresponding row we switch row k and row i by a \textit{primitive matrix}:
@@ -107,18 +107,18 @@ output: pivoted LU decomposition $ \bar{L}\bar{U} = \bar{P}\bar{A}$
 \item For $i = 1,2,\ldots,n$ \begin{itemize}
 \item compute: k = $\argmax_{1 \leq j \leq n} | a^{(i)}_{p_j,i} |$ \% find pivot
 \item swap: $ p_i \gets\to p_k$
-\item $\bar{l}_i := \bar{a}^{(i)}_{:,i} / a^{(i)}_{p_i,p_i}$
-\item $\bar{u}_i := a_{p_i,:}$
+\item $\bar{l}_i := \bar{a}^{(i)}_{:,i} / a^{(i)}_{p_i,i}$
+\item $\bar{u}_i := a^{(i)}_{p_i,:}$
 \item compute: $\bar{A}_{i+1} = \bar{A}_i - \bar{l}_i \cdot \bar{u}_i$
 \end{itemize}
-\item set $\bar{P} := [\bar{l}_{p_1},\bar{l}_{p_2},\ldots, \bar{l}_{p_n}]^T$ \% $\bar{e}_i$ is i-th unit vector
-\item set $bar{L} = bar{P}[\bar{l}_1,\bar{l}_2,\cdots,\bar{l}_n]$
+\item set $\bar{P} := [\bar{e}_{p_1},\bar{e}_{p_2},\ldots, \bar{e}_{p_n}]^T$ \% $\bar{e}_i$ is i-th unit vector
+\item set $\bar{L} = \bar{P}[\bar{l}_1,\bar{l}_2,\cdots,\bar{l}_n]$
 \end{enumerate}
 
 \paragraph{Example 2.10} \textit{(omitted)}
 
 \section*{2.6: Cholesky decomposition}
-If $\bar{A}$ is symmetric and positive definite, i.e. all eigenvalues of $\bar{A}$ are bigger than zero or equivalently $bar{x}^T\bar{A}\bar{x} > 0$ for all $\bar{x} \neq 0$, we can compute a symmetric decomposition of $\bar{A}$.
+If $\bar{A}$ is symmetric and positive definite, i.e. all eigenvalues of $\bar{A}$ are bigger than zero or equivalently $\bar{x}^T\bar{A}\bar{x} > 0$ for all $\bar{x} \neq 0$, we can compute a symmetric decomposition of $\bar{A}$.
 
 \paragraph{Note:} if $\bar{A}$ is symmetric and positive definite, then the \textit{Schur complement} $\bar{S}:=\bar{A}_{2:n,2:n} - (\bar{a}_{2:n},1/a_{1,1})\bar{a}^T_{2:n,1}$, is symmetric and positive definite as well. In particular, it holds $s_{i,i} > 0$ and $a_{i,i} > 0$ !