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