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 <[email protected]>
Signed-off-by: Teodor Dutu <[email protected]>
Signed-off-by: Razvan Deaconescu <[email protected]>

chapters/vectors-matrix-ops/ranks-determinants/reading/read.md

@@ -0,0 +1,379 @@
+# 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
+
+$$a_1=\mathbf{c_1} = -\displaystyle\frac{k_2}{k_1} a_2 - \ldots - \displaystyle\frac{k_2}{k_1} a_n$$
+
+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
+
+$$
+A^{-1} = \displaystyle\frac{1}{ad-bc}
+  \begin{bmatrix}
+    2 & 3 \\
+    3 & 1
+  \end{bmatrix}
+$$
+
+### 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