diff --git a/chapters/linear_algebra.tex b/chapters/linear_algebra.tex index 0615dd9..5858818 100644 --- a/chapters/linear_algebra.tex +++ b/chapters/linear_algebra.tex @@ -7,4 +7,5 @@ \subfile{sections/vector_spaces.tex} \subfile{sections/matrices.tex} + \subfile{sections/determinant.tex} \end{document} \ No newline at end of file diff --git a/chapters/sections/determinant.tex b/chapters/sections/determinant.tex new file mode 100644 index 0000000..202c099 --- /dev/null +++ b/chapters/sections/determinant.tex @@ -0,0 +1,255 @@ +\documentclass[../../script.tex]{subfiles} + +% !TEX root = ../../script.tex + +\begin{document} +\section{The Determinant} + +In this section we always define $A \in \field^{n \times n}$ and $z_1, \cdots, z_n$ the row vectors of $A$. We declare the mapping +\[ + \det: \field^{n \times n} \longrightarrow \field +\] +and define +\[ + \det(A) := \det(z_1, z_2, \dots, z_n) +\] + +\begin{defi} + There exists exactly one mapping $\det$ such that + \begin{enumerate}[(i)] + \item It is linear in the first row, i.e. + \[ + \det(z_1 + \lambda\tilde{z_1}, z_2, \cdots, z_n) = \det(z_1, z_2, \cdots, z_n) + \lambda \det(\tilde{z_1}, z_2, \cdots, z_n) + \] + + \item If $\tilde{A}$ is obtained from $A$ by swapping two rows + \[ + \det(A) = -\det(\tilde{A}) + \] + + \item $\det(I) = 1$ + \end{enumerate} + This mapping is called the determinant, and we write + \[ + \det A = \begin{vmatrix} + a_{11} & \cdots & a_{1n} \\ + \vdots & \ddots & \vdots \\ + a_{n1} & \cdots & a_{nn} \\ + \end{vmatrix} + \] +\end{defi} + +\begin{eg} + \[ + \begin{vmatrix} + a_{11} & a_{12} \\ + a_{21} & a_{22} + \end{vmatrix} + = a_{11}a_{22} - a_{21}a_{12} + \] + \begin{align*} + \begin{vmatrix} + a_{11} & a_{12} & a_{13} \\ + a_{21} & a_{22} & a_{23} \\ + a_{31} & a_{32} & a_{33} \\ + \end{vmatrix} + = &a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} \\ + &- a_{31}a_{22}a_{13} - a_{32}a_{23}a_{11} - a_{33}a_{21}a_{12} + \end{align*} +\end{eg} + +\begin{rem} + \begin{enumerate}[(i)] + \item Every determinant is linear in every row + \item If two rows are equal then $\det(A) = 0$ + \item If one row (w.l.o.g. $z_1$) is a linear combination of the others, so + \[ + z_1 = \alpha_2z_2 + \alpha_3z_3 + \cdots + \alpha_nz_n, ~~\alpha_1, \cdots, \alpha_n \in \field + \] + then + \begin{align*} + \det(z_1, z_2, \cdots, z_n) = &\alpha_2 \underbrace{\det(z_2, z_2, z_3, \cdots, z_n)}_0 + \\ + &\alpha_3 \underbrace{\det(z_3, z_2, z_3, \cdots, z_n)}_0 + \\ + &\cdots + \\ + &\alpha_n \underbrace{\det(z_n, z_2, z_3, \cdots, z_n)}_0 \\ + &= 0 + \end{align*} + + \item Adding a multiple of a row to another doesn't change the determinant + + \item Define + \begin{align*} + T_{ij} && \text{ swaps rows } i \text{ and } j \\ + M_i(\lambda) && \text{ multiplies row } i \text{ with } \lambda \ne 0 \\ + L_{ij}(\lambda) && \text{ adds } \lambda \text{-times row } j \text{ to row } i \\ + \end{align*} + Then + \begin{align*} + \det(T_{ij} A) &= -\det(A) \\ + \det(L_{ij}(\lambda) A) &= \det(A) \\ + \det(M_i(\lambda) A) &= \lambda\det(A) + \end{align*} + \end{enumerate} +\end{rem} + +\begin{lem} + Let $\det$ be the determinent, and $A, B \in \field^{n \times n}$. Let $A$ be in row echelon form, then + \[ + \det(AB) = a_{11} \cdot a_{22} \cdot \cdots \cdot a_{nn} \cdot \det(B) + \] +\end{lem} +\begin{proof} + First consider the case of $A$ not being invertible. This means that the last row of $A$ is a zero row, which in turn means that $\det(A) = 0$. + This also means that the last row of $AB$ is a zero row and therefore $\det(AB) = 0$. + + Now let $A$ be invertible. This means that all the diagonal entries are non-zero. It is possible to bring $A$ into diagonal form without changing + the diagonal entries themselves. So, w.l.o.g. let $A$ be in diagonal form: + \begin{equation} + A = M_n(a_{nn}) \cdot \cdots \cdot M_2(a_{22})M_1(a_{11}) I + \end{equation} + and thus + \begin{equation} + \begin{split} + \det(AB) &= \det(M_n(a_{nn}) \cdot \cdots \cdot M_2(a_{22})M_1(a_{11}) B) \\ + &= a_{nn} \cdot \cdots \cdot a_{22} \cdot a_{11} \det(B) + \end{split} + \end{equation} +\end{proof} + +\begin{rem} + For $B = I$ this results in + \[ + \det(A) = a_{11} a_{22} \cdots a_{nn} + \] +\end{rem} + +\begin{thm} + Let $A, B \in \field^{n \times n}$. Then + \[ + \det AB = \det A \cdot \det B + \] +\end{thm} +\begin{proof} + Let $i, j \in \set{1, \cdots, n}$ and $\lambda \ne 0$. Then + \begin{subequations} + \begin{equation} + \det(T_{ij} AB) = -\det(AB) + \end{equation} + \begin{equation} + \det(L_{ij}(\lambda) AB) = \det(AB) + \end{equation} + \end{subequations} + Bring $A$ with $T_{ij}$ and $L_{ij}(\lambda)$ operations into row echelon form. Then + \begin{equation} + \det(AB) = a_{11}a_{22} \cdots a_{nn} \cdot \det(B) + \end{equation} + and therefore + \begin{equation} + \det(AB) = \det A \cdot \det B + \end{equation} +\end{proof} + +\begin{cor} + \[ + A \in \field^{n \times n} \text{ invertible } \iff \det A \ne 0 + \] +\end{cor} +\begin{proof} + Row operations don't effect the invertibility or the determinant (except for the sign) of a matrix. Therefore we can limit ourselves to matrices in + row echelon form (w.l.o.g.). Let $A$ be in row echelon form, then + \begin{equation} + \begin{split} + \det A \ne 0 &\iff a_{11} a_{22} \cdots a_{nn} \ne 0 \\ + &\iff a_{11} \ne 0, a_{22} \ne 0, \cdots, a_{nn} \ne 0 \\ + &\iff A \text{ invertible since diagonal entries are non-zero} + \end{split} + \end{equation} +\end{proof} + +\begin{thm} + \[ + \det A = \det A^T + \] +\end{thm} +\begin{proof} + First consider the explicit representation of row operations: + \begin{subequations} + \begin{equation} + T_{ij} = \kbordermatrix{ + & & j & & i & \\ + & 1 & & & & \\ + i & & 0 & & 1 & \\ + & & & 1 & & \\ + j & & 1 & & 0 & \\ + & & & & & 1 \\ + } + \end{equation} + \begin{equation} + L_{ij}(\lambda) = \kbordermatrix{ + & & & & j & \\ + & 1 & & & & \\ + i & & 1 & & \lambda & \\ + & & & 1 & & \\ + & & & & 1 & \\ + & & & & & 1 \\ + } + \end{equation} + \end{subequations} + Thus we can see + \begin{subequations} + \begin{equation} + \det(T_{ij}) = \det(T_{ij}^T) = -1 + \end{equation} + \begin{equation} + \det(L_{ij}(\lambda)) = \det(L_{ij}(\lambda)^T) = 1 + \end{equation} + \end{subequations} + Let $T$ be one of those matrices. Then + \begin{equation} + \begin{split} + \det((TA)^T) &= \det(A^T \cdot T^T) \\ + &= \det A^T \cdot \det T^T \\ + &= \det A^T \cdot \det T \\ + \end{split} + \end{equation} + and + \begin{equation} + \det TA = \det A \cdot \det T + \end{equation} + And therefore + \begin{equation} + \det((TA)^T) = \det(TA) \iff \det A^T = \det A + \end{equation} + Now w.l.o.g. let $A$ be in row echelon form. Let $A$ be non-invertible, i.e. the last row is a zero row. Thus $\det A = 0$. + This implies that $A^T$ has a zero column. Row operations that bring $A^T$ into row echelon form (w.l.o.g.) perserve this zero column. Therefore the + resulting matrix must also have a zero column, and thus $\det(A^T) = 0$. + + Now assume $A$ is invertible, and use row operations to bring $A$ into a diagonalised form (w.l.o.g.). For diagonalised matrices we know that + \begin{equation} + A = A^T \implies \det A = \det A^T + \end{equation} +\end{proof} + +\begin{rem} + Let $A_{ij}$ be the matrix you get by removing the $i$-th row and the $j$-th column from $A$. + \[ + \det A = \sum_{i=1}^n (-1)^{i+j} \cdot a_{ij} \cdot \det(A_{ij}), ~~j \in \set{1, \cdots, n} + \] +\end{rem} + +\begin{rem}[Leibniz formula] + Let $n \in \natn$, and let there be a bijective mapping + \[ + \sigma: \set{1, \cdots, n} \longrightarrow \set{1, \cdots, n} + \] + $\sigma$ is a permutation. The set of all permutations is labeled $S_n$, and it contains $n!$ elements. Then + \[ + \det A = \sum_{\sigma \in S_n} \sgn(\sigma) \prod_{i=1}^n a_{i, \sigma(i)} + \] + A permutation that swaps exactly two elements is called elementary permutation. Every permutation can be written as a number of consecutively executed elementary permutations. + \[ + \sgn(\sigma) = (-1)^k + \] + where $\sigma$ is the permutation in question and $k$ is the number of elementary permutations it consists of. +\end{rem} +\end{document} \ No newline at end of file diff --git a/script.pdf b/script.pdf index 14020df..e69de29 100644 Binary files a/script.pdf and b/script.pdf differ diff --git a/script.synctex(busy) b/script.synctex(busy) new file mode 100644 index 0000000..e69de29 diff --git a/script.tex b/script.tex index 90c7371..67ab224 100644 --- a/script.tex +++ b/script.tex @@ -7,6 +7,7 @@ \usepackage[shortlabels]{enumitem} \usepackage{multicol} \usepackage{tikz} +\usepackage{kbordermatrix} \usetikzlibrary{calc,trees,positioning,arrows,fit,shapes,angles} \usepackage{color} @@ -23,6 +24,7 @@ \DeclareMathOperator{\spn}{span} %\DeclareMathOperator{\dim}{dim} +\DeclareMathOperator{\sgn}{sgn} \newcommand{\natn}{\mathbb{N}} \newcommand{\intn}{\mathbb{Z}} @@ -54,6 +56,9 @@ \newcommand{\conv}[1]{\xrightarrow{\makebox[2em][c]{$\scriptstyle#1$}}} \newcommand{\convinf}{\conv{n \rightarrow \infty}} +\renewcommand{\kbldelim}{(} +\renewcommand{\kbrdelim}{)} + \newcommand{\reader}{Left as an exercise for the reader.} \newcommand{\setvert}{\,\vert\,} \newcommand{\set}[2][]{% @@ -99,7 +104,7 @@ \begin{document} -\title{Mathemacs for Physicists} +\title{Mathematics for Physicists} \author{https://www.github.com/Lauchmelder23/Mathematics} \maketitle \tableofcontents