From b4228a283852a428515d5ce1eabde0731869b464 Mon Sep 17 00:00:00 2001 From: Razvan Deaconescu Date: Wed, 13 Sep 2023 05:45:20 +0300 Subject: [PATCH] Add content for vectors-matrix-ops/ranks-determinants Add actual content to the skeleton of the `vectors-matrix-ops/ranks-determinants` section. Signed-off-by: Eggert Karl Hafsteinsson Signed-off-by: Teodor Dutu Signed-off-by: Razvan Deaconescu --- .../ranks-determinants/reading/README.md | 212 ++++++++++++++++++ 1 file changed, 212 insertions(+) diff --git a/chapters/vectors-matrix-ops/ranks-determinants/reading/README.md b/chapters/vectors-matrix-ops/ranks-determinants/reading/README.md index 46d1103..ce8f7ab 100644 --- a/chapters/vectors-matrix-ops/ranks-determinants/reading/README.md +++ b/chapters/vectors-matrix-ops/ranks-determinants/reading/README.md @@ -1 +1,213 @@ # Ranks and Determinants + +## The Rank of a Matrix + +The rank of an $n \times p$ matrix $A$, denoted by $\text{rank}(A)$, is the largest number of columns of $A$, which are not linearly dependent (i.e. the number of linearly independent columns). + +### Details + +Vectors $a_1, a_2, \ldots, a_n$ are said to be linearly dependent if there exist constants $k_1, \ldots, k_n$ that are not all zero, such that + +$$k_1 a_1 + k_2 a_2 + \ldots + k_n a_n = 0.$$ + +Note that if such constants exist, then we can write one of the $a$ 's as a linear combination of the rest, e.g. if $k_1 \neq 0$ then + +$$a_1=\mathbf{c_1} = -\frac{k_2}{k_1} a_2 - \ldots - \frac{k_2}{k_1} a_n$$ + +It can be shown that the rank of $A$, is the same as the rank of $A'$ + +i.e. the maximum number of linearly independent rows of $A$. + +:::note Note + +Note that if $\text{rank}(A) = p$, then the columns are linearly independent. + +::: + +### Examples + +:::info Example + +If + +$$A= \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right]$$ + +then $\text{rank}(A)$ = 2, since + +$$k_1 \left( \begin{array}{cc} 1 \\ 0 \\ \end{array} \right) + k_2 \left( \begin{array}{cc} 0 \\ 1 \\ \end{array} \right) = \left( \begin{array}{cc} 0 \\ 0 \\ \end{array} \right)$$ + +if and only if + +$$\left( \begin{array}{cc} k_1 \\ k_2 \\ \end{array} \right) = \left( \begin{array}{cc} 0 \\ 0 \\ \end{array} \right)$$ + +so the columns are linearly independent. + +::: + +:::info Example + +If + +$$A = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1\\ 0 & 0 & 0 \\ \end{array} \right]$$ + +then $\text{rank}(A)$ = 2. + +::: + +:::info Example + +If + +$$A = \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \\ \end{array} \right]$$ + +then $\text{rank}(A)$ = 2. since + +$$1 \left( \begin{array}{ccc} 1 \\ 0 \\ 0 \\ \end{array} \right) + 0 \left( \begin{array}{ccc} 0 \\ 1 \\ 1 \\ \end{array} \right) + (-1) \left( \begin{array}{ccc} 1 \\ 0 \\ 0 \\ \end{array} \right) = 0$$ + +(and hence the rank cannot be more than 2) but + +$$k_1 \left( \begin{array}{ccc} 1 \\ 0 \\ 0 \\ \end{array} \right) + k_2 \left( \begin{array}{ccc} 0 \\ 1 \\ 1 \\ \end{array} \right)$$ + +if and only if $k_1=k_2=0$ (and hence the rank must be at least 2). + +::: + +## The Determinant + +Recall that for a $2 \times 2$ matrix, + +$$A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ + +the inverse of $A$ is + +$$A^{-1}= \frac{1}{ad-bc} \begin{bmatrix} 2 & 3 \\ 3 & 1 \end{bmatrix}$$ + +### Details + +:::note Definition + +The number $ad-bc$ is called the **determinant** of the $2 \times 2$ matrix $A$. + +::: + +:::note Definition + +Now suppose $A$ is an $n \times n$ matrix. +An **elementary product** from the matrix is a product of $n$ terms based on taking exactly one term from each column of row $x$. +Each such term can be written in the form $a_{1j_1} \cdot a_{2j_2} \cdot a_{3j_3} \cdot \ldots \cdot a_{nj_n}$ where $j_1, \ldots, j_n$ is a permutation of the integers $1,2, \ldots, n$. +Each permutation $\sigma$ of the integers $1,2,\ldots,n$ can be performed by repeatedly interchanging two numbers. + +::: + +:::note Definition + +A **signed elementary product** is an elementary product with a positive sign if the number of interchanges in the permutation is even but negative otherwise. + +::: + +The determinant of $A$, $\det(A)$ or $\vert A \vert$, is the sum of all signed elementary products. + +### Examples + +:::info Example + +$$A= \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$ + +then + +$\vert A \vert = a_{1\underline{1}} a_{2\underline{2}} - a_{1\underline{2}}a_{2\underline{1}}$. + +::: + +:::info Example + +If + +$$A= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix},$$ + +Then + +$\vert A \vert$ + += $a_{11} a_{22} a_{33}$ This is the identity permutation and has positive sign + +$-a_{11} a_{23} a_{32}$ This is the permutation that only interchanges $2$ and $3$ + +$-a_{12} a_{21} a_{33}$ Only one interchange + +$+a_{12} a_{23} a_{31}$ Two interchanges + +$+a_{13} a_{21} a_{32}$ Two interchanges + +$-a_{13} a_{22} a_{31}$ Three interchanges + +::: + +:::info Example + +$$A= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$$ + +$\vert A \vert = -1$ + +::: + +:::info Example + +$$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$ + +$\vert A \vert = 1 \cdot 2 \cdot 3 = 6$ + +::: + +:::info Example + +$$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 3 & 0 \end{bmatrix}$$ + +$\vert A \vert = 0$ + +::: + +:::info Example + +$$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 3 & 0 \end{bmatrix}$$ + +$\vert A \vert = -6$ + +::: + +:::info Example + +$$A= \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}$$ + +$\vert A \vert = 0$ + +::: + +:::info Example + +$$A= \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}$$ + +$\vert A \vert = 0$ + +::: + +## Ranks, Inverses and Determinants + +The following statements are true for an $n\times n$ matrix $A$ : + +- $\text{rank} (A)= n$ + +- $\det(A)\neq 0$ + +- $A$ has an inverse + +### Details + +Suppose $A$ is an $n\times n$ matrix. +Then the following are truths: + +- $\text{rank} (A)= n$ + +- $\det(A)\neq 0$ + +- $A$ has an inverse