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85 lines
3.4 KiB
TeX
85 lines
3.4 KiB
TeX
\documentclass[11pt]{amsart}
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\usepackage{geometry}
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\geometry{a4paper}
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\usepackage{graphicx}
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\usepackage{amssymb}
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\usepackage{amsmath,amsthm}
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\usepackage{epstopdf}
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\usepackage{hyperref}
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\usepackage{lmodern}
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% Reproducible builds
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\pdfinfoomitdate=1
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\pdftrailerid{}
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\pdfsuppressptexinfo=-1
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\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}
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\title{Proof that the determinant is multiplicative}
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\author{Patrick Stevens}
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\date{17th April 2014}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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\newenvironment{definition}[1][Definition]{\begin{trivlist}
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\item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}}
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\begin{document}
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\maketitle
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\tiny \begin{center} \url{https://www.patrickstevens.co.uk/misc/MultiplicativeDetProof/MultiplicativeDetProof.pdf} \end{center}
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\normalsize
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\
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This is a very concrete proof of the multiplicity of the determinant.
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It contains no cleverness at all, and is simply manipulation of expressions.
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\
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\begin{definition}
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The \emph{determinant} of a matrix $A$ is given by $$\sum_{\sigma \in S_n} \epsilon(\sigma) \prod_{i=1}^n A_{i, \sigma(i)}$$
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where $S_n$ is the symmetric group on $n$ elements, and $\epsilon$ is the signature of that element.
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\end{definition}
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\begin{lemma}
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Let $\rho \in S_n$, and let $A$ be a matrix. Then
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$$\sum_{\sigma \in S_n} \epsilon(\sigma) \prod_{i=1}^n A_{\rho(i) \sigma(i)} = \epsilon(\rho) \det(A)$$
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\end{lemma}
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\begin{proof}
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\begin{align*}
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\epsilon(\rho) \det(A) &= \epsilon(\rho) \sum_{\sigma \in S_n} \epsilon(\sigma) \prod_{i=1}^n A_{i, \sigma(i)}
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\\&= \sum_{\sigma \in S_n} \epsilon(\sigma \rho) \prod_{i=1}^n A_{i, \sigma(i)}
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\\&= \sum_{\sigma \in S_n} \epsilon(\sigma \rho) \prod_{i=1}^n A_{\rho(i), \sigma(\rho(i))}
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\\&= \sum_{\tau \in S_n} \epsilon(\tau) \prod_{i=1}^n A_{\rho(i), \tau(i)}
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\\&= \sum_{\tau \in S_n} \epsilon(\sigma) \prod_{i=1}^n A_{\rho(i), \sigma(i)}
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\end{align*}
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\end{proof}
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\begin{theorem}$$\det(AB) = \det(A) \det(B)$$
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\end{theorem}
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\begin{proof}
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We use summation convention throughout.
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\begin{align*}
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\det(AB) &= \sum_{\sigma \in S_n} \epsilon(\sigma) \prod_{i=1}^n (AB)_{i, \sigma(i)}
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\\ &= \sum_{\sigma \in S_n} \epsilon(\sigma) A_{1, k_1} B_{k_1, \sigma(1)} A_{2, k_2} B_{k_2, \sigma(2)} \dots A_{n, k_n} B_{k_n, \sigma(n)}
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\\ &= \sum_{\sigma \in S_n} \epsilon(\sigma) A_{1, k_1} A_{2, k_2} \dots A_{n, k_n} B_{k_1, \sigma(1)} \dots B_{k_n, \sigma(n)}
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\end{align*}
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But the $k_1, \dots, k_n$ only ever contribute when they are a permutation of $1, \dots, n$, because (assuming wlog $k_1 = k_2$) for each $\sigma^+$ there exists $\sigma^-$ such that $\sigma^+(1) = \sigma^-(2), \sigma^-(1) = \sigma^+(2), \sigma^-(k) = \sigma^+(k)$ for other $k$. Then we have
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$$A_{1, k_1} B_{k_1, \sigma^+(1)} A_{2 k_1} B_{k_1 \sigma^+(2)} \text{terms} = A_{1, k_1} B_{k_1, \sigma^-(1)} A_{2 k_1} B_{k_1 \sigma^-(2)} \text{terms}$$
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and because $\epsilon$ negates the sign, we have that these two terms cancel.
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Hence the sum over $k_i$ is in fact a sum over all $\rho$ such that $\rho(i) = k_i$ for all $i$:
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Then $$\det(AB) = \sum_{\rho \in S_n} \sum_{\sigma \in S_n} \epsilon(\sigma) A_{1, \rho(1)} A_{2, \rho(2)} \dots A_{n, \rho(n)} B_{\rho(1), \sigma(1)} \dots B_{\rho(n), \sigma(n)}$$
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Applying the lemma gives
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\begin{align*}
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\det(AB) &= \det(B) \sum_{\rho \in S_n} \epsilon(\rho) A_{1, \rho(1)} A_{2, \rho(2)} \dots A_{n, \rho(n)}
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\\&= \det(B) \det(A)
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\end{align*}
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\end{proof}
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\end{document} |