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