feat: 1522 lec 7

ma1522/1522 Notes.tex

@@ -54,6 +54,34 @@
 \hr
 \input{lec_06.tex}
 \newpage
+\section{Lecture 7}
+\hr
+\input{lec_07.tex}
+\newpage
+\section{Lecture 8}
+\hr
+\input{lec_08.tex}
+\newpage
+\section{Lecture 9}
+\hr
+\input{lec_09.tex}
+\newpage
+\section{Lecture 10}
+\hr
+\input{lec_10.tex}
+\newpage
+\section{Lecture 11}
+\hr
+\input{lec_11.tex}
+\newpage
+\section{Lecture 12}
+\hr
+\input{lec_12.tex}
+\newpage
+\section{Lecture 13}
+\hr
+\input{lec_13.tex}
+\newpage
 
 \section{Reference}
 

ma1522/lec_04.tex

@@ -63,6 +63,7 @@
     \item \textbf{Upper Triangular} if all entries \textbf{below} diagonal are zero.
       \subitem $A = (a_{ij})_{n \times n}$ is upper triangular $\iff a_{ij} = 0 \text{ if } i > j$
     \item \textbf{Lower Triangular} if all entries \textbf{above} diagonal are zero.
+      \label{def:ltm}
       \subitem $A = (a_{ij})_{n \times n}$ is lower triangular $\iff a_{ij} = 0 \text{ if } i < j$
       \subitem if Matrix is both Lower and Upper triangular, its a Diagonal Matrix.
   \end{itemize}

ma1522/lec_07.tex

@@ -0,0 +1,107 @@
+\begin{theorem} Main Theorem for Invertible Matrices \\
+  Let $A$ be a square matrix. Then the following are equivalent
+  \begin{enumerate}
+    \item $A$ is an invertible matrix.
+    \item Linear System $Ax = b$ has a unique solution
+    \item Linear System $Ax = 0$ has only the trivial solution
+    \item RREF of $A$ is $I$
+    \item A is the product of elementary matrices
+  \end{enumerate}
+\end{theorem}
+
+\begin{theorem} Find Inverse
+  \begin{itemize}
+    \item Let $A$ be an invertible Matrix.
+    \item RREF of $(A | I)$ is $(I | A^{-1})$
+  \end{itemize}
+
+  How to identify if Square Matrix is invertible?
+
+  \begin{itemize}
+    \item Square matrix is invertible
+      \subitem $\iff$ RREF is $I$
+      \subitem $\iff$ All columns in its REF are pivot
+      \subitem $\iff$ All rows in REF are nonzero
+    \item Square matrix is singular
+      \subitem $\iff$ RREF is \textbf{NOT} $I$
+      \subitem $\iff$ Some columns in its REF are non-pivot
+      \subitem $\iff$ Some rows in REF are zero.
+    \item $A$ and $B$ are square matrices such that $AB = I$
+      \subitem then $A$ and $B$ are invertible
+  \end{itemize}
+\end{theorem}
+
+\begin{defn}[LU Decomposition with Type 3 Operations]\ \\
+  \begin{itemize}
+    \item Type 3 Operations: $(R_i + cR_j, i > j)$
+    \item Let $A$ be a $m \times n$ matrix. Consider Gaussian Elimination $A \dashrightarrow R$
+    \item Let $R \dashrightarrow A$ be the operations in reverse
+    \item Apply the same operations to $I_m \dashrightarrow L$. Then $A = LR$
+    \item $L$ is a \hyperref[def:ltm]{lower triangular matrix} with 1 along diagonal
+    \item If $A$ is square matrix, $R = U$
+  \end{itemize}
+
+  Application: 
+  \begin{itemize}
+    \item $A$ has LU decomposition $A = LU$, $Ax = b$ i.e., $LUx = b$
+    \item Let $y = Ux$, then it is reduced to $Ly = b$
+    \item $Ly = b$ can be solved with forward substitution. 
+    \item $Ux = y$ is the REF of A.
+    \item $Ux = y$ can be solved using backward substitution.
+
+  \end{itemize}
+\end{defn}
+
+\begin{defn}[LU Decomposition with Type II Operations]\ \\
+  \begin{itemize}
+    \item Type 2 Operations: $(R_i \leftrightarrow R_j)$, where 2 rows are swapped
+    \item $A \xrightarrow[]{E_1} \bullet \xrightarrow[]{E_2}\bullet \xrightarrow[E_3]{R_i \iff R_j}\bullet \xrightarrow[]{E_4}\bullet \xrightarrow[]{E_5} R$
+    \item $A = E^{-1}_1E^{-1}_2E^{}_3E^{-1}_4E^{-1}_5R$
+    \item $E_3A = (E_3E^{-1}_1E^{-1}_2E_3)E^{-1}_4E^{-1}_5R$
+    \item $P = E_3, L = (E_3E^{-1}_1E^{-1}_2E_3)E^{-1}_4E^{-1}_5, R = U$, $PA = LU$
+  
+  \end{itemize}
+\end{defn}
+
+\begin{defn}[Column Operations]\ \\
+  \begin{itemize}
+    \item Pre-multiplication of Elementary matrix $\iff$ Elementary row operation
+      \subitem $A \to B \iff B = E_1E_2...E_kA$
+    \item Post-Multiplication of Elementary matrix $\iff$ Elementary Column Operation
+      \subitem $A \to B \iff B = AE_1E_2...E_k$
+    \item If $E$ is obtained from $I_n$ by single elementary column operation, then
+      \subitem $I \xrightarrow[]{kC_i}E \iff I \xrightarrow[]{kR_i}E$
+      \subitem $I \xrightarrow[]{C_i \leftrightarrow C_j}E \iff I \xrightarrow[]{R_i \leftrightarrow  R_j}E$
+      \subitem $I \xrightarrow[]{C_i + kC_j}E \iff I \xrightarrow[]{R_j + kR_i}E$
+  \end{itemize}
+\end{defn}
+
+\subsection{Determinants}
+
+\begin{defn}[Determinants of $2 \times 2$ Matrix]\ \\
+  \begin{itemize}
+    \item Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$
+    \item $\det(A) = |A| = ad - bc$
+    \item $\det(I_2) = 1$
+    \item $A \xrightarrow{cR_i} B \implies \det(B) = c\det(A)$
+    \item $A \xrightarrow{R_1 \leftrightarrow R_2} B \implies \det(B) = -\det(A)$
+    \item $A \xrightarrow{R_i + cR_j} B \implies \det(B) = \det(A), i \neq j$
+  \end{itemize}
+  Solving Linear equations with determinants for $2 \times 2$
+  \begin{itemize}
+    \item $x_1 = 
+      \dfrac{\begin{vmatrix} b_1 & a_{12} \\ b_2 & a_{22} \end{vmatrix}}
+      {\begin{vmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}}$, $x_2 = 
+      \dfrac{\begin{vmatrix} a_{11} & b_1 \\ a_{21} & b_2 \end{vmatrix}}
+      {\begin{vmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}}$
+  \end{itemize}
+\end{defn}
+
+\begin{defn}[Determinants of $3 \times 3$ Matrix]\ \\
+  \begin{itemize}
+    \item Suppose $A$ is invertible, then there exists EROs such that
+    \item $A \xrightarrow{ero_1} A_1 \rightarrow ... \rightarrow A_{k-1} \xrightarrow{ero_k}A_k = I$
+    \item Then $\det(A)$ can be evaluated backwards.
+      \subitem E.g. $A \xrightarrow{R_1 \leftrightarrow R_3} \bullet \xrightarrow{3R_2} \bullet \xrightarrow{R_2 + 2R_4} I \implies det(A) = 1 \to 1 \to \frac{1}{3} \to -\frac{1}{3}$
+  \end{itemize}
+\end{defn}