From ec3613bf682de0025e4a111979dc1954b516b1cd Mon Sep 17 00:00:00 2001 From: Luc Date: Wed, 10 Jan 2024 22:03:23 +0100 Subject: [PATCH] Section determinants almost finished. --- .../linear-algebra/determinants.md | 109 +++++++++++++++++- 1 file changed, 108 insertions(+), 1 deletion(-) diff --git a/docs/en/mathematics/linear-algebra/determinants.md b/docs/en/mathematics/linear-algebra/determinants.md index 130f7e7..b192f78 100644 --- a/docs/en/mathematics/linear-algebra/determinants.md +++ b/docs/en/mathematics/linear-algebra/determinants.md @@ -81,7 +81,31 @@ $$ ??? note "*Proof*:" - Will be added later. + Let $A$ be a $n \times n$ triagular matrix with $n \in \mathbb{N}$ given by + + $$ + A = \begin{pmatrix} a_{11} & \cdots &a_{1n}\\ & \ddots & \vdots \\ & & a_{nn} \end{pmatrix}. + $$ + + We claim that $\det(A) = a_{11} \cdot a_{22} \cdots a_{nn}$. We first check the claim for $n=1$ which is given by $\det(A) = a_{11}$. + + Now suppose for some $k \in \mathbb{N}$, the determinant of a $k \times k$ triangular $A_{k}$ is given by + + $$ + \det(A_k) = a_1{11} \cdot a_{22} \cdots a_{kk} + $$ + + then by assumption + + $$ + \det(A_{k+1}) = \begin{pmatrix} A_k & a_{(k+1)1}\\& \vdots\\ 0 \cdots 0 & a_{(k+1)(k+1)}\end{pmatrix} = a_{(k+1)(k+1)} \det(A_k) + 0 = a_{11}a_1{11} \cdot a_{22} \cdots a_{kk} \cdot a_{(k+1)(k+1)}. + $$ + + Hence if the claim holds for some $k \in \mathbb{N}$ then it also holds for $k+1$. The principle of natural induction implies now that for all $n \in \mathbb{N}$ we have + + $$ + \det(A) = a_{11} \cdot a_{22} \cdots a_{nn}. + $$ > *Theorem*: let $A$ be an $n \times n$ matrix > @@ -92,3 +116,86 @@ $$ Will be added later. +> *Lemma*: let $A$ be an $n \times n$ matrix with $n \in \mathbb{N}$. If $A_{jk}$ denotes the cofactor of $a_{jk}$ for $k \in \mathbb{N}$ then +> +> $$ +> a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn} = \begin{cases} \det(A) &\text{ if } i = j,\\ 0 &\text{ if } i \neq j.\end{cases} +> $$ + +??? note "*Proof*:" + +If $i = j$ then we obtain the cofactor expansion of $\det(A)$ along the $i$th row of $A$. + +If $i \neq j$, let $A^*$ be the matrix obtained by replacing the $j$th row of $A$ by the $i$th row of $A$ + +$$ + A^* = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \begin{array}{ll} j\text{th row}\\ \\ \\ \\\end{array} +$$ + +since two rows of $A^*$ are the same its determinant must be zero. It follows from the cofactor expansion of $\det(A^*)$ along the $j$th row that + +$$ +\begin{align*} + 0 &= \det(A^*) = a_{i1} A_{j1}^* + a_{i2} A_{j2}^* + \dots + a_{in} A_{jn}^*, \\ + &= a_{i1} A_{j1} + a_{i2} A_{j2} + \dots + a_{in} A_{jn}. +\end{align*} +$$ + +> *Theorem*: let $E$ be an $n \times n$ elementary matrix and $A$ an $n \times n$ matrix with $n \in \mathbb{N}$ then we have +> +> $$ +> \det(E A) = \det(E) \det(A), +> $$ +> +> where +> +> $$ +> \det(E) = \begin{cases} -1 &\text{ if $E$ is of type I},\\ \alpha \in \mathbb{R}\backslash \{0\} &\text{ if $E$ is of type II},\\ 1 &\text{ if $E$ is of type III}. \end{cases} +> $$ + +??? note "*Proof*:" + + Will be added later. + +Similar results hold for column operations, since for the elementary matrix $E$, $E^T$ is also an elementary matrix and $\det(A E) = \det((AE)^T) = \det(E^T A^T) = \det(E^T) \det(A^T) = \det(E) \det(A)$. + +> *Theorem*: an $n \times n$ matrix A with $n \in \mathbb{N}$ is singular if and only if +> +> $$ +> \det(A) = 0 +> $$ + +??? note "*Proof*:" + + Let $A$ be an $n \times n$ matrixwith $n \in \mathbb{N}$. Matrix $A$ can be reduced to row echelon form with a finite number of row operations obtaining + + $$ + U = E_k E_{k-1} \cdots E_1 A, + $$ + + where $U$ is in $n \times n$ row echelon form and $E_i$ are $n \times n$ elementary matrices for $i \in \{1, \dots, k\}$. It follows then that + + $$ + \begin{align*} + \det(U) &= \det(E_k E_{k-1} \cdots E_1 A), \\ + &= \det(E_k) \det(E_{k-1}) \cdots \det(E_1) \det(A). + \end{align*} + $$ + + Since the determinants of the elementary matrices are all nonzero, it follows that $\det(A) = 0$ if and only if $\det(U) = 0$. If $A$ is singular then $U$ has a row consisting entirely of zeros and hence $\det(U) = 0$. If $A$ is nonsingular then $U$ is triangular with 1's along the diagonal and hence $\det(U) = 1$. + +From this theorem we may pose a method for computing $\det(A)$ by taking + +$$ +\det(A) = \Big(\det(E_k) \det(E_{k-1} \cdots \det(E_1)\Big)^{-1}. +$$ + +> *Theorem*: let $A$ and $B$ be $n \times n$ matrices with $n \in \mathbb{N}$ then +> +> $$ +> \det(AB) = \det(A) \det(B) +> $$ + +??? note "*Proof*:" + + Will be added later. \ No newline at end of file