\( \newcommand{\bb}{\mathbb} \newcommand{\bf}{\mathbf} \newcommand{\cal}{\mathcal} \newcommand{\L}{\left} \newcommand{\R}{\right} \newcommand{\C}{\mathbb{R}} \newcommand{\F}{\mathbb{F}} \newcommand{\mat}[1]{\begin{bmatrix} #1 \end{bmatrix}} \newcommand{\qed}{\quad\square} \newcommand{\opn}{\operatorname} \)

3.2 [R] Number Theory?

TipRemark

Remark. Suppose we have a matrix \(A\), two of whose rows are \(\bf{u}\) and \(\bf{v}\). How to construct a matrix \(A'\) with the opposite determinant?

The simplest way is to negate either row. However, we can also replace \(\bf{v}\) with \(\bf{u} - \bf{v}\). In number theory, we have \[-v \equiv u-v \pmod{u}\]

Quotient Spaces

Warning

Waiting for the author to learn.

Bézout’s Identity

TipRemark

Remark. Bézout’s identity states that for given \(a\) and \(b\), there exist \(x\) and \(y\) such that \(ax+by = \gcd(a,b)\).

Consider the determinant of \(\mat{m_1 & m_2 \\ x_1 & x_2}\). \[\det\mat{m_1 & m_2 \\ x_1 & x_2} = m_1x_2 - m_2x_1\]

If the determinant is \(1\), which implies the bounded area by \(\mat{m_1 & m_2}\) and \(\mat{x_1 & x_2}\) is \(1\), then \(m_1\) and \(m_2\) are coprime. Additionally, \[m_1x_2 \equiv 1 \pmod{m_2}\] \[-m_2x_1 \equiv 1 \pmod{m_1}\]

Thus, \(x_2\) is the inverse of \(m_1\) modulo \(m_2\), and \(-x_1\) is the inverse of \(m_2\) modulo \(m_1\).