Ch2_Euclidean_Space_Ex_2.3.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "3b001b13",
   "metadata": {},
   "source": [
    "<b>Question 1</b><br>\n",
    "\n",
    "Determine whether the following vectors are linearly dependent in $\\mathbb{R}^2$ <br>\n",
    "(a) $e_1 = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, e_2 = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\qquad$ (b)$u = \\begin{pmatrix}\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "-6\\\\\n",
    "-8\\\\\n",
    "\\end{pmatrix}\\qquad$(c)$u = \\begin{pmatrix}\n",
    "6\\\\\n",
    "10\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "-3\\\\\n",
    "5\\\\\n",
    "\\end{pmatrix}\\qquad$<br>\n",
    "(d) $u = \\begin{pmatrix}\n",
    "\\pi\\\\\n",
    "-2\\pi\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}\\qquad$(e)$u = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, e_2 = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\qquad$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b82d5782",
   "metadata": {},
   "source": [
    "<b>Solution</b>\n",
    "\n",
    "\n",
    "<b>(a)</b> \n",
    "For linear combination: $k_1v_1 + k_2v_2 + ... + k_nv_n = 0$,\n",
    "\n",
    "If $k_1, k_2, ..., k_n = 0$, then independent<br>otherwise, dependent\n",
    "\n",
    "Now, using $k_1$ and $k_2$ scalar adn equating to zero, we have\n",
    "\n",
    "$k_1e_1 + k_2e_2 = 0$\n",
    "\n",
    "$\\Rightarrow k_1\\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}  + \\begin{pmatrix}\n",
    "0\\\\\n",
    "k_2\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "k_2\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, <b>independent</b>\n",
    "\n",
    "\n",
    "<b>Solution (b)</b>\n",
    "\n",
    "$k_1u + k_2v = k_1\\begin{pmatrix}\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "-6\\\\\n",
    "-8\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "3k_1\\\\\n",
    "4k_1\\\\\n",
    "\\end{pmatrix} + \\begin{pmatrix}\n",
    "-6k_2\n",
    "\\\\\n",
    "-8k_2\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "So, we have <br>\n",
    "\n",
    "$3k_1 -6k_2 = 0$<br>\n",
    "$4k_1 -8k_2 = 0$\n",
    "\n",
    "Here, if $k_2 = 1$, we get then $k_1 = 2$ <br>\n",
    "Therefore, $k_1 =2$, and $k_2 = 1$ and <b>linearly dependent</b>\n",
    "\n",
    "<b>Solution (c)</b> <br>\n",
    "\n",
    "Here, $u = -2v$\n",
    "\n",
    "$\\Rightarrow (1)u + 2v = 0$\n",
    "\n",
    "Therefore, $k_1 = 1$, and $k_2 = 2$, and <b>linearly dependent</b>\n",
    "\n",
    "\n",
    "<b>Solution (d)</b> <br>\n",
    "\n",
    "Here, $u = -\\pi v$\n",
    "\n",
    "$\\Rightarrow u + \\pi v = 0$ \n",
    "\n",
    "Therefore, $k_1 = 1$, and $k_2 \\pi$, and <b>linearly dependent</b>\n",
    "\n",
    "<b>Solution (e)</b><br>\n",
    "\n",
    "Here, one vector $u = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, therefore <b>linearly dependent</b>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ea09aedd",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ee36b254",
   "metadata": {},
   "source": [
    "<b>Question 2</b><br>\n",
    "\n",
    "Determine whether the following vectors are linearly dependent in $\\mathbb{R}^3$ <br>\n",
    "(a) $e_1 = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, e_2 = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, e_3 = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\qquad$\n",
    "(b) $u = \\begin{pmatrix}\n",
    "2\\\\\n",
    "2\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}, w =\\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\qquad$\n",
    "(c)$u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "-2\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix}$\n",
    "(d) $u = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "2\\\\\n",
    "-3\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "-4\\\\\n",
    "6\\\\\n",
    "\\end{pmatrix}, w =\\begin{pmatrix}\n",
    "2\\\\\n",
    "0\\\\\n",
    "6\\\\\n",
    "\\end{pmatrix}$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6c073988",
   "metadata": {},
   "source": [
    "<b>Solution(a)</b> <br>\n",
    "\n",
    "Lets take scalar $k_1, k_2, k_3$ and make it a equation<br>\n",
    "\n",
    "$k_1e_1 + k_2e_2 + k_3e_3 = 0$\n",
    "\n",
    "Then, we have <br>\n",
    "\n",
    "$k_1e_1 + k_2e_2 + k_3e_3 = k_1\\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + \\begin{pmatrix}\n",
    "0\\\\\n",
    "k_2\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "k_3\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "k_2\\\\\n",
    "k_3\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "Here, $k_1 = k_2 = k_3 = 0$ Therefore, <b>independent</b>\n",
    "\n",
    "<b>Solution (b)</b> <br>\n",
    "\n",
    "Taking scalars $k_1, k_2, k_3$ and then linear combination is as : <br>\n",
    "\n",
    "$k_1u + k_2v + k_3w = 0$\n",
    "\n",
    "Then, we have <br>\n",
    "\n",
    "$k_1u + k_2v + k_3w = k_1\\begin{pmatrix}\n",
    "2\\\\\n",
    "2\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, Augmented Matrix can be drawn as :\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "2&1&0&0\\\\\n",
    "2&2&0&0\\\\\n",
    "2&-1&1&0\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "After Gaussian elimination it is :\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "\\text{k_1} & \\text{k_2} & \\text{k_3} & \\text{}\\\\\n",
    "2&1&0&0\\\\\n",
    "0&1&0&0\\\\\n",
    "0&-2&1&0\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Now, we have following equations:<br>\n",
    "$2k_1 + k_2 = 0$<br>\n",
    "$k_2 = 0$<br>\n",
    "$-2k_2 + k_3 = 0$<br>\n",
    "\n",
    "$\\Rightarrow k_1 = 0, k_2 = 0$, and $k_3 = 0$\n",
    "\n",
    "Therefore, $u, v, w$ are linearly independent\n",
    "\n",
    "<b>Solution (c)</b><br>\n",
    "\n",
    "Here, we can observe $v = -2u$\n",
    "\n",
    "$v + 2u = 0$\n",
    "$\\Rightarrow 2u + 1v = 0$\n",
    "\n",
    "Therefore, scalars are not zero, and $u, v$ are linearly dependent\n",
    "\n",
    "<b>Solution(d)</b><br>\n",
    "\n",
    "Lets take scalar $k_1, k_2, k_3$ and make it a equation<br>\n",
    "\n",
    "$k_1u + k_2v + k_3w = 0$\n",
    "\n",
    "Then, we have <br>\n",
    "\n",
    "$k_1u + k_2v + k_3w = k_1\\begin{pmatrix}\n",
    "-1\\\\\n",
    "2\\\\\n",
    "-3\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "0\\\\\n",
    "4\\\\\n",
    "6\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "2\\\\\n",
    "0\\\\\n",
    "6\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, Augmented Matrix can be drawn as :\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "-1&0&2&0\\\\\n",
    "2&-4&0&0\\\\\n",
    "-3&6&6&0\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "After Guassian Elimination it would be :\n",
    "\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "-1&0&2&0\\\\\n",
    "0&-4&4&0\\\\\n",
    "0&0&-12&0\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "\n",
    "$\\Rightarrow k_3 =0$\n",
    "and susbequently $k_1 = 0$ and $k_2 = 0$\n",
    "\n",
    "Therefore, $u, v, w$ are linearly independent"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ee0357bf",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "651e9ad1",
   "metadata": {},
   "source": [
    "<b>Question 3</b><br>\n",
    "\n",
    "Determine whether the following vectors are linearly dependent in $\\mathbb{R}^4$\n",
    "\n",
    "\n",
    "(a) $u = \\begin{pmatrix}\n",
    "0&-1&0&3\\\\\n",
    "\\end{pmatrix}^T\\quad, v = \\begin{pmatrix}\n",
    "1&0&5&0\\\\\n",
    "\\end{pmatrix}^T\\quad, w = \\begin{pmatrix}\n",
    "2&1&0&0\\\\\n",
    "\\end{pmatrix}^T\\quad, x = \\begin{pmatrix}\n",
    "0&1&0&-4\\\\\n",
    "\\end{pmatrix}^T\\quad$ \n",
    "\n",
    "(b) $u = \\begin{pmatrix}\n",
    "1&-1&3&3\\\\\n",
    "\\end{pmatrix}^T\\quad, v = \\begin{pmatrix}\n",
    "0&1&5&0\\\\\n",
    "\\end{pmatrix}^T\\quad, w = \\begin{pmatrix}\n",
    "-3&-6&-9&-4\\\\\n",
    "\\end{pmatrix}^T\\quad, x = \\begin{pmatrix}\n",
    "-5&5&-15&-15\\\\\n",
    "\\end{pmatrix}^T\\quad$\n",
    "\n",
    "(c) $u = \\begin{pmatrix}\n",
    "-2&2&3&4\\\\\n",
    "\\end{pmatrix}^T\\quad, v = \\begin{pmatrix}\n",
    "0&3&-2&-3\\\\\n",
    "\\end{pmatrix}^T\\quad, w = \\begin{pmatrix}\n",
    "2&-2&-1&0\\\\\n",
    "\\end{pmatrix}^T\\quad, x = \\begin{pmatrix}\n",
    "0&3&0&1\\\\\n",
    "\\end{pmatrix}^T\\quad$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7cc1b590",
   "metadata": {},
   "source": [
    "<b> Solution(a) </b><br>\n",
    "\n",
    "Here, we have $u = \\begin{pmatrix}\n",
    "0\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "5\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, w = \\begin{pmatrix}\n",
    "2\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}x = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "-4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Taking scalars $k_1, k_2, k_3, k_4$ and then linear combination is as : <br>\n",
    "\n",
    "$k_1u + k_2v + k_3w  + k_4x= 0$\n",
    "\n",
    "Then, we have <br>\n",
    "\n",
    "$k_1u + k_2v + k_3w +k_4x= k_1\\begin{pmatrix}\n",
    "0\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "5\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "2\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_4 \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "-4\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, Augmented Matrix can be drawn as :\n",
    "$\\left(\\begin{array}{cccc|c}\n",
    "1&1&2&0&0\\\\\n",
    "-1&0&1&1&0\\\\\n",
    "0&5&0&0&0\\\\\n",
    "3&0&0&-4&0\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "$\\Rightarrow 5K_2 = 0$ Therefore, $k_2 = 0$\n",
    "\n",
    "Subsequently, $k_3 = 0$, $k_1 = 0$, $k_4 = 0$\n",
    "\n",
    "Therefore, $u, v, w, x$ are linearly independent\n",
    "\n",
    "<b> Solution(b)</b><br>\n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "3\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "5\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, w = \\begin{pmatrix}\n",
    "-3\\\\\n",
    "-6\\\\\n",
    "-9\\\\\n",
    "-4\\\\\n",
    "\\end{pmatrix}, x = \\begin{pmatrix}\n",
    "-5\\\\\n",
    "5\\\\\n",
    "-15\\\\\n",
    "-15\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Here, we can see, $x = -5u$ <br>\n",
    "$\\Rightarrow 5u + x = 0$\n",
    "Therefore, linearly dependent\n",
    "\n",
    "\n",
    "<b> Solution(c)</b><br>\n",
    "\n",
    "Here, $u = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "3\\\\\n",
    "-2\\\\\n",
    "-3\\\\\n",
    "\\end{pmatrix}, w = \\begin{pmatrix}\n",
    "2\\\\\n",
    "-2\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, x = \\begin{pmatrix}\n",
    "0\\\\\n",
    "3\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "Taking scalars $k_1, k_2, k_3, k_4$ and then linear combination is as : <br>\n",
    "\n",
    "$k_1u + k_2v + k_3w  + k_4x= 0$\n",
    "\n",
    "Then, we have <br>\n",
    "\n",
    "$k_1u + k_2v + k_3w +k_4x= k_1\\begin{pmatrix}\n",
    "-2\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "0\\\\\n",
    "3\\\\\n",
    "-2\\\\\n",
    "-3\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "2\\\\\n",
    "-2\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_4 \\begin{pmatrix}\n",
    "0\\\\\n",
    "3\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "Equations from this can be drawn as :\n",
    "\n",
    "$-2k_1 +\\quad + 2k_3+\\quad = 0$<br>\n",
    "$2k_1 + 3k_2 - 2k_3 + 3k_4 = 0$<br>\n",
    "$3k_1 + -2k_2 - k_3 +\\quad = 0$<br>\n",
    "$4k_1 - 3k_2 + \\quad + k_4 = 0$\n",
    "\n",
    "Here, we can see solving first equation, we get $k_1 = k_3$. If we take $k_1 = 0$, every other scalar, $k_3, k_2, k_4$ would be $0$.\n",
    "\n",
    "But the trick is $k_1u + k_2v + k_3w + k_4x = 0$ only holds if $k_1 = k_2 = k_3 = K_4 = 0$ then independent\n",
    "\n",
    "If we take, $k_1 = 1$, then subsquently we will have $k_2 =1, k_3 =1$ and $k_4 = -1$ satisfying $k_1u + k_2v + k_3w + k_4x = 0$.\n",
    "\n",
    "Therefore, as non-zero scalars satisfy equations, given vectors, $u, v, w, x$ are <b>linearly dependent</b>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c0f4cab4",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6ea7e536",
   "metadata": {},
   "source": [
    "<b>Question4 </b><br>\n",
    "\n",
    "Let $u$ and $v$ be non-zero vectors in $\\mathbb{R}^n$. Prove that if $u = kv$ where $k$ is a real scalar then the vectors $u$ and $v$ are linearly dependent (Vectors $u$ and $v$ are scalar multiples of each other. "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9905e960",
   "metadata": {},
   "source": [
    "<b> Solution</b><br>\n",
    "\n",
    "Here, we have $u = kv$\n",
    "\n",
    "$\\Rightarrow (1)u - kv = 0$\n",
    "\n",
    "Since, non-zero scalars satisfy equation, vector $u$ and $v$ are linearly dependent"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6c823746",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "33dc0610",
   "metadata": {},
   "source": [
    "<b>Question 5</b><br>\n",
    "\n",
    "Let $u$, $v$ and $w$ be any vectors in $\\mathbb{R}^n$. Prove that the vectors $u + v$, $v + w$, and $u - w$ are linearly dependent"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f281621a",
   "metadata": {},
   "source": []
  },
  {
   "cell_type": "markdown",
   "id": "bf1879cc",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8ce46f8c",
   "metadata": {},
   "source": [
    "<b>Question 6</b><br>\n",
    "\n",
    "Let $e_1$ and $e_2$ be the standard unit vectors in $\\mathbb{R}^2$. Prove that $e_1$, $e_1 + e_2$ are linearly independent"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1a064526",
   "metadata": {},
   "source": [
    "<b> Solution</b><br>\n",
    "\n",
    "Lets take scalars as $k_1, k_2$ and the linear combination as :\n",
    "\n",
    "$k_1(e_1) + k_2(e_1 + e_2) = 0$\n",
    "\n",
    "$\\Rightarrow k_1e_1 + k_2e_1 + k_2e_2 = 0$\n",
    "\n",
    "$\\Rightarrow (k_1 + k_2)e_1 + k_2e_2 = 0$\n",
    "\n",
    "Since, $e_1, e_2$ are standard unit vectors(or linearly independent as proven earlier), therefore, scalars must be 0 ie. $k_1 + k_2 = 0$ and $k_2 = 0$ \n",
    "\n",
    "$\\Rightarrow k_1 = 2$ and $k_2 = 0$\n",
    "\n",
    "Therefore, $e_1$ and $e_1 + e_2$ linearly independent"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b1e219b2",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "abb158f0",
   "metadata": {},
   "source": [
    "<b>Question 7</b><br>\n",
    "\n",
    "Let $e_1$, $e_2$ and $e_3$ be the standard unit vectors in $\\mathbb{R}^2$. Prove that $e_1$, $e_1 + e_2$ and $e_1 + e_2 + e_3$ are linearly independent"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0e8894ff",
   "metadata": {},
   "source": [
    "<b> Solution</b><br>\n",
    "\n",
    "Lets take scalars as $k_1, k_2, k_3$ and the linear combination as :\n",
    "\n",
    "$k_1e_1 + k_2(e_1 + e_2) + k_3(e_1 + e_2 + e_3) =0$\n",
    "\n",
    "$\\Rightarrow k_1e_1 + k_2e_1 + k_2e_2 + k_3e_1 + k_3e_2 + k_3e_3 = 0$\n",
    "\n",
    "$\\Rightarrow (k_1 + k_2 + k_3)e_1 + (k_2 + k_3)e_2 + k_3e_3 = 0$\n",
    "\n",
    "Since, $e_1, e_2, e_3$ are standard unit vectors(or linearly independent), therefore, its scalars must be 0 ie. $k_1 + k_2 +k_3 = 0$, $k_2 + k_3 =0$ and $k_3 = 0$ \n",
    "\n",
    "$\\Rightarrow k_1 = 2$, $k_2 = 0$, and $k_2 = 0$\n",
    "\n",
    "Therefore, $e_1$, $e_1 + e_2$ and $e_1 + e_2 + e_3$ linearly independent"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "474052de",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "94e29d42",
   "metadata": {},
   "source": [
    "<b>Question 8</b><br>\n",
    "\n",
    "Let $u$, $v$, $w$ and $x$ be linearly independent vectors in $\\mathbb{R}^n$. Prove that \n",
    "\n",
    "$\\qquad \\qquad u + v$, $v + w$, $w + x$, and $u + x$\n",
    "\n",
    "are linearly dependent vectors in $\\mathbb{R}^n$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d4e50f18",
   "metadata": {},
   "source": [
    "<b>Solution</b> <br>\n",
    "\n",
    "Lets take scalars $k_1, k_2, k_3, \\text{and}, k_4$ and creating linear combination equating to 0 such that:\n",
    "\n",
    "$ k_1(u + v) + k_2(v+w) + k_3(w+x) + k_4(u +x) = 0 .................eq(1)$\n",
    "\n",
    "$\\Rightarrow k_1u + k_1v + k_2v + k_2w + k_3w + k_3x + k_4u + k_4x = 0$\n",
    "\n",
    "$\\Rightarrow (k_1 + k_4)u + (k_1 + k_2)v + (k_2 + k_3)w + (k_3 + k_4)x = 0$\n",
    "\n",
    "Since, $u, v, w, x$ are linearly independent<br>\n",
    "Therefore, <br>\n",
    "$k_1 + k_4 = 0 \\qquad \\Rightarrow k_1 = - k_4$,<br>\n",
    "$k_1 + k_2 = 0 \\qquad \\Rightarrow k_2 = -k_1$,<br>\n",
    "$k_2 + k_3 = 0 \\qquad \\Rightarrow k_3 = -k_2$,<br>\n",
    "$k_3 + k_4 = 0 \\qquad \\Rightarrow k_4 = -k_3$\n",
    "\n",
    "If $k_1 = 1$, then \n",
    "\n",
    "$k_2 = -1$ <br>\n",
    "$k_3 = 1$<br>\n",
    "$k_4 = -1$\n",
    "\n",
    "This gives us scalar values of $eq(1)$ are not zeroes.\n",
    "Therefore, $u + v$, $v+w$, $w+x$ and $u + x$ vectors are <b>Linearly Dependent</b>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cfc1bba1",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "63b8a23c",
   "metadata": {},
   "source": [
    "<b>Question 9</b><br>\n",
    "\n",
    "Let $u$, $v$, $w$ be linearly independent vectors in $\\mathbb{R}^n$. Let the vector $x$ be in $\\mathbb{R}^n$ such that $x = k_1 u + k_2 v + k_3 w$, where the $k's$ are real scalars. Prove that this representation of the vector $x$ is unique."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "dad98471",
   "metadata": {},
   "source": [
    "<b>Solution</b> <br>\n",
    "\n",
    "Lets take another real scalars $c's$ as $c_1$, $c_2$, and $c_3$ such that\n",
    "\n",
    "$x = c_1u + c_2v + c_3w$\n",
    "\n",
    "We have, <br>\n",
    "$x = k_1 u + k_2 v + k_3 w$\n",
    "\n",
    "Therefore, <br>\n",
    "$c_1u + c_2v + c_3w = k_1 u + k_2 v + k_3$\n",
    "\n",
    "$\\Rightarrow c_1 - k_1u + c_2v - k_2v + c_3w - k_3w = 0$\n",
    "\n",
    "$\\Rightarrow (c_1 - k_1)u + (c_2 - k_2)v + (c_3 - k_3)w = 0$\n",
    "\n",
    "As, $u, v, w$ are linearly independent, <br>\n",
    "$c_1 - k_1 = 0\\qquad \\Rightarrow c_1 = k_1$<br>\n",
    "$c_2 - k_2 = 0\\qquad \\Rightarrow c_2 = k_2$<br>\n",
    "$c_3 - k_3 = 0\\qquad \\Rightarrow c_3 = k_3$\n",
    "\n",
    "Hence, <b>unique</b>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ac667888",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e329304a",
   "metadata": {},
   "source": [
    "<b>Question 10</b><br>\n",
    "\n",
    "Let $S = \\{v_1, v_2, v_3, ..., v_n\\}$ be a set of linear independent vectors in $\\mathbb{R}^n$. <br>Prove that  $T = \\{c_1v_1, c_2v_2, c_3v_3, ..., c_nv_n\\}$, where the $c's$ are real non-zero scalars, is also a set of linear independent vectors."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c9b6496f",
   "metadata": {},
   "source": [
    "<b> Solution</b>\n",
    "\n",
    "Lets take scalars $k_1, k_2, \\dots k_n$ and linear combination of $T$ set of vectors as: \n",
    "\n",
    "$$k_1(c_1v_1) + k_2(c_2v_2) + k_3(c_3v_3) + \\cdots k_n(c_nv_n) = 0$$\n",
    "\n",
    "We know that,<br>\n",
    "$k(u\\cdot v) = (ku)\\cdot v$\n",
    "\n",
    "Threfore, above equation can be written as :\n",
    "\n",
    "$(k_1c_1)v_1 + (k_2c_2)v_2 + (k_3c_3)v_3 + \\dots + (k_nc_n)v_n = 0$\n",
    "\n",
    "We are given that, set of vectors $S = \\{v_1, v_2, v_3, ..., v_n\\}$ are linearly independent. Therefore, their scalars must equate to zero.\n",
    "\n",
    "$\\Rightarrow k_1c_1 = k_2c_2 = \\dots =k_nc_n = 0$\n",
    "\n",
    "$\\Rightarrow k_1 = k_2 = k_3 = 0$<br>\n",
    "Therefore, T set of vectors must also be linearly independent\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "788f8b98",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7af093d3",
   "metadata": {},
   "source": [
    "<b>Question 11</b><br>\n",
    "\n",
    "Let $S = \\{v_1, v_2, v_3, ..., v_n\\}$ is linearly independent then any subset $T = \\{v_1, v_2, v_3, ..., v_m\\}$, where $ m \\le n$, is also linearly independent."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6e138913",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9fce01b6",
   "metadata": {},
   "source": [
    "<b>Question 12</b><br>\n",
    "\n",
    "Determine the real values of <i>t</i> in the following vectors which form linear independence in $\\mathbb{R}^3$ :\n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "t\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "t\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$, and $w = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "t\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "94a967d0",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6ee47533",
   "metadata": {},
   "source": [
    "<b>Question 13</b><br>\n",
    "\n",
    "Prove proposition 2.23"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "9c3cd583",
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3 (ipykernel)",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.9.9"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 5
}