Ch2_Euclidean_Space_Ex_2.4.ipynb

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    "<b> Question 1</b><br>\n",
    "\n",
    "Determine whether the following vectors span $\\mathbb{R}^2$: \n",
    "\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) $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$ <br>(c) $u = \\begin{pmatrix}\n",
    "2\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}\\qquad$(d) $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "10\\\\\n",
    "\\end{pmatrix}\\qquad$"
   ]
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    "<b> Solution</b> <br>\n",
    "\n",
    "<b>(a)</b> Lets take an arbitary vector $w = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$ and scalar $k$ and $c$ to make linear combination\n",
    "\n",
    "$ke_1 + ce_2 = w$\n",
    "\n",
    "$\\Rightarrow k\\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$ + $c\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} =\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$ + $\\begin{pmatrix}\n",
    "0\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix} =\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Here, $k=a$, $c=b$, which makes $w = ku + cv$. Therefore, $e_1$ and $e_2$ span $\\mathbb{R}^2$\n",
    "\n",
    "<b>(b)</b> Again, taking an arbitary vector $w = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$ which makes $w = ku + cv$<br>\n",
    "\n",
    "$\\Rightarrow k\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} + c\\begin{pmatrix}\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k-c\\\\\n",
    "k+c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives us: \n",
    "$k - c = a \\dots eq(1)$<br>\n",
    "$k + c = b \\dots eq(2)$<br>\n",
    "\n",
    "From both eq, we get $2k = a+b$ <br>\n",
    "$\\Rightarrow k = \\frac{a+b}{2}$ Therefore, $c = \\frac{b-a}{2}$ and span $\\mathbb{R}^2$\n",
    "\n",
    "<b>Solution (c)</b><br>\n",
    "\n",
    "Let $w = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$ satisfying $ku + cv = w$\n",
    "\n",
    "Then, $ku + cv = k\\begin{pmatrix}\n",
    "2\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} + c\\begin{pmatrix}\n",
    "-1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "2k -c\\\\\n",
    "2k -c\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives us:<br>\n",
    "2k -c = 0<br>\n",
    "2k -c = 0<br>\n",
    "$0 = a - b$ or $a = b$<br>\n",
    "\n",
    "$\\Rightarrow$ given vector $u$ and $v$ span $\\begin{pmatrix}\n",
    "a\\\\\n",
    "a\\\\\n",
    "\\end{pmatrix}$ and do not span $\\mathbb{R}^2$\n",
    "\n",
    "<b>Solution (d)</b><br>\n",
    "\n",
    "Let $w = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$ satisfying $ku + cv = w$\n",
    "\n",
    "Then, $ku + cv = k\\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} + c\\begin{pmatrix}\n",
    "-1\\\\\n",
    "10\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k -c\\\\\n",
    "2k +10c\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives Augmented Matrix as :\n",
    "$\\left(\\begin{array}{cc|c}\n",
    "1&-1&a\\\\\n",
    "2&10&b\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "After Gaussian elimination it is :\n",
    "$\\left(\\begin{array}{cc|c}\n",
    "1&-1&a\\\\\n",
    "0&12&b-2a\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "$\\Rightarrow 12c = b -2a$<br>\n",
    "$\\Rightarrow c = \\frac{b -2a}{12}$<br>\n",
    "\n",
    "Then, from the top row, we can deduce $k = a +c$<br>\n",
    "\n",
    "$\\Rightarrow k = a + \\frac{b - 2a}{12}$ or $k = \\frac{b + 10a}{12}$\n",
    "\n",
    "Here, with both $k$ and $c$ scalar, given vector $u$ and $v$ span $\\mathbb{R}^2$"
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    "------------------------------------------------------------"
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    "<b> Question 2</b><br>\n",
    "\n",
    "\n",
    "Determine whether the following vectors span $\\mathbb{R}^3$: \n",
    "\n",
    "(a) $ u = \\begin{pmatrix}\n",
    "2\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, $v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "2\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, and $w = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "(b) $ u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$, $v = \\begin{pmatrix}\n",
    "2\\\\\n",
    "2\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$, and $w = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "(c) $ u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$, $v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, and $w = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "(d) $ u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$, $v = \\begin{pmatrix}\n",
    "2\\\\\n",
    "4\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, and $w = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "-2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$"
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    "<b>Solution(a)</b>\n",
    "\n",
    "Let $k_1, k_2, k_3$ satisfy $k_1u + k_2v + k_3w = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Then, we have <br>\n",
    "\n",
    "$k_1\\begin{pmatrix}\n",
    "2\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "0\\\\\n",
    "2\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "2k_1\\\\\n",
    "2k_2\\\\\n",
    "2k_3\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives us:<br>\n",
    "$2k_1 = a$<br>\n",
    "$2k_2 = b$<br>\n",
    "$2k_3 = c$<br>\n",
    "\n",
    "$\\Rightarrow k_1 = \\frac{a}{2}, k_2 = \\frac{b}{2}, k_3 = \\frac{c}{2}$\n",
    "\n",
    "Then, for any real scalar, $a, b, c$ vector $u, v, \\text{and}, w$ span $\\mathbb{R}^3$\n",
    "\n",
    "\n",
    "\n",
    "<b>Solution(b)</b>\n",
    "\n",
    "Let $k_1, k_2, k_3$ satisfy $k_1u + k_2v + k_3w = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Then, we have <br>\n",
    "\n",
    "$k_1\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "2\\\\\n",
    "2\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "k_1\\\\\n",
    "k_1\\\\\n",
    "\\end{pmatrix} + \\begin{pmatrix}\n",
    "2k_2\\\\\n",
    "2k_2\\\\\n",
    "2k_2\\\\\n",
    "\\end{pmatrix} + \\begin{pmatrix}\n",
    "k_3\\\\\n",
    "2k_3\\\\\n",
    "3k_3\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives Augmented Matrix as :\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&2&1&a\\\\\n",
    "2&1&3&b\\\\\n",
    "1&2&3&c\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Next step is as :<br>\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&2&1&a\\\\\n",
    "0&0&1&b-a\\\\\n",
    "0&0&1&c-b\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Then, <br>\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&2&1&a\\\\\n",
    "0&0&1&b-a\\\\\n",
    "0&0&0&c-b-(b-a)\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Here, it means $c = 2b - a$. Therefore, for any scalars, $u, v$, and $w$ do not span $\\mathbb{R}^3$\n",
    "\n",
    "\n",
    "<b>(c)</b><br>\n",
    "\n",
    "Let $k_1, k_2, k_3$ satisfy $k_1u + k_2v + k_3w = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Then, we have <br>\n",
    "\n",
    "$k_1\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k_1 + k_2 + k_3\\\\\n",
    "k_1 + k_2\\\\\n",
    "k_1\\\\\n",
    "\\end{pmatrix}\n",
    "= \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives us:<br>\n",
    "$k_1 = c$<br>\n",
    "$k_1 + k_2 = b$  $\\qquad \\Rightarrow k_2 = b -c$<br>\n",
    "$k_3 = a -c -(b+c) \\qquad \\Rightarrow k_3 = a -b$\n",
    "\n",
    "\n",
    "Therefore, any scalars, $u, v$ and $w$ span $\\mathbb{R}^3$\n",
    "\n",
    "\n",
    "<b>Solution (d)</b>"
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    "------------------------------------------------------------"
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    "<b> Question 3</b><br>\n",
    "\n",
    "\n",
    "Determine whether the following vectors form a basis for $\\mathbb{R}^2$: \n",
    "\n",
    "(a) $u = \\begin{pmatrix}\n",
    "1&2\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "0&1\\\\\n",
    "\\end{pmatrix}^T\\qquad$\n",
    "\n",
    "(b) $u = \\begin{pmatrix}\n",
    "-2&-4\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "1&2\\\\\n",
    "\\end{pmatrix}^T\\$\n",
    "\n",
    "(c) $u = \\begin{pmatrix}\n",
    "4&1\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "1&3\\\\\n",
    "\\end{pmatrix}^T\\qquad$\n",
    "\n",
    "(d) $u = \\begin{pmatrix}\n",
    "3&5\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "2&3\\\\\n",
    "\\end{pmatrix}^T\\qquad$"
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    "<b> Solution: <br>\n",
    "    Key thing</b> There are two conditions to be satisfied to establish if $n$ vectors form a basis. <br>\n",
    "    1. The vectors span $\\mathbb{R}^n$<br>\n",
    "    2. The vectors are linearly independent\n",
    "\n",
    "However, if we have $n$ vectors in $\\mathbb{R}^n$, then we only need to <b>check one of the conditions, either independence or span</b>\n",
    "\n",
    "<i>This section is about span. We will check span</i>\n",
    "\n",
    "<b>(a)</b> Here, we have, $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$ and $v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "To check span, lets take $k$ and $c$ such that $ku + cv = w$ where $w$ is an arbitary vector $\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now<br>\n",
    "$k\\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} + c\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} =\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k\\\\\n",
    "2k+c\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow k =a$ and $c = b - 2k \\quad\\text{or}\\quad c = b-2a$\n",
    "\n",
    "Therefore, $u$ and $v$ span $\\mathbb{R}^2$<br>\n",
    "Hence, $u$ and $v$ <b>form basis</b>\n",
    "\n",
    "\n",
    "<b>(b)</b> Here, we have, $u = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "-4\\\\\n",
    "\\end{pmatrix}$ and $v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "To check span, lets take $k$ and $c$ such that $ku + cv = w$ where $w$ is an arbitary vector $\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now<br>\n",
    "$k\\begin{pmatrix}\n",
    "-2\\\\\n",
    "-4\\\\\n",
    "\\end{pmatrix} + c\\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} =\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \n",
    "\\begin{pmatrix}\n",
    "-2k+ c\\\\\n",
    "-4k + 2c\\\\\n",
    "\\end{pmatrix} =\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives Augmented Matrix as :\n",
    "$\\left(\\begin{array}{cc|c}\n",
    "-2&1&a\\\\\n",
    "-4&2&b\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "After Gaussian elimination it is :\n",
    "$\\left(\\begin{array}{cc|c}\n",
    "-2&1&a\\\\\n",
    "0&0&b\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Here, $b = 0$ <br>\n",
    "Therefore, $u$ and $v$ do not span $\\mathbb{R}^2$<br>\n",
    "Hence, $u$ and $v$ <b>do not form basis</b>\n",
    "\n",
    "\n",
    "<b>(c)</b><br>\n",
    "\n",
    "Here, we have, $u = \\begin{pmatrix}\n",
    "4\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$ and $v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "To check span, lets take $k$ and $c$ such that $ku + cv = w$ where $w$ is an arbitary vector $\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now<br>\n",
    "$k\\begin{pmatrix}\n",
    "4\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} + c\\begin{pmatrix}\n",
    "1\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} =\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives us:\n",
    "\n",
    "\n",
    "$4k + c = a$<br>\n",
    "$k + 3c = b$<br>\n",
    "\n",
    "Solving above equations, we get $k = \\frac{3a - b}{11}$ and $c = \\frac{-(a+b)}{11}$\n",
    "\n",
    "Therefore, $u$ and $v$ span $\\mathbb{R}^2$<br>\n",
    "Hence, $u$ and $v$ <b>form basis</b>\n",
    "\n",
    "\n",
    "\n",
    "<b>(d)</b><br>\n",
    "\n",
    "Here, we have, $u = \\begin{pmatrix}\n",
    "3\\\\\n",
    "5\\\\\n",
    "\\end{pmatrix}$ and $v = \\begin{pmatrix}\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "To check span, lets take $k$ and $c$ such that $ku + cv = w$ where $w$ is an arbitary vector $\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now<br>\n",
    "$k\\begin{pmatrix}\n",
    "3\\\\\n",
    "5\\\\\n",
    "\\end{pmatrix} + c\\begin{pmatrix}\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} =\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives us:\n",
    "\n",
    "\n",
    "$3k + 2c = a$<br>\n",
    "$5k + 3c = b$<br>\n",
    "\n",
    "Solving above equations, we get $k = 2b - 3a$ and $c = 2a - b$\n",
    "\n",
    "Therefore, $u$ and $v$ span $\\mathbb{R}^2$<br>\n",
    "Hence, $u$ and $v$ <b>form basis</b>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8801831d",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0c88e113",
   "metadata": {},
   "source": [
    "<b> Question 4</b><br>\n",
    "\n",
    "\n",
    "Explain why the following vectors do not form a basis for the indicated Euclidean space : \n",
    "\n",
    "(a) $u = \\begin{pmatrix}\n",
    "1&2\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "0&1\\\\\n",
    "\\end{pmatrix}^T, w = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "\\end{pmatrix}^T$ for $\\mathbb{R}^2$\n",
    "\n",
    "(b) $u = \\begin{pmatrix}\n",
    "1&1&1\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "1&1&0\\\\\n",
    "\\end{pmatrix}^T, w = \\begin{pmatrix}\n",
    "0&0&0\\\\\n",
    "\\end{pmatrix}^T$ for $\\mathbb{R}^3$\n",
    "\n",
    "(c) $u = \\begin{pmatrix}\n",
    "1&1&1&1\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "1&1&0&-1\\\\\n",
    "\\end{pmatrix}^T, w = \\begin{pmatrix}\n",
    "-1&-2&-3&4\\\\\n",
    "\\end{pmatrix}^T, x = \\begin{pmatrix}\n",
    "2&2&2&2\\\\\n",
    "\\end{pmatrix}$ for $\\mathbb{R}^4$ \n",
    "\n",
    "(d) $u = \\begin{pmatrix}\n",
    "1&2&3&4&5\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "1&5&0&2&4\\\\\n",
    "\\end{pmatrix}^T, w = \\begin{pmatrix}\n",
    "3&2&-5&4&-9\\\\\n",
    "\\end{pmatrix}^T, x = \\begin{pmatrix}\n",
    "2&9&2&7&7\\\\\n",
    "\\end{pmatrix}$ for $\\mathbb{R}^5$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0e446ace",
   "metadata": {},
   "source": [
    "<b> Solution</b> \n",
    "\n",
    "<b>(a)</b> The basis of $\\mathbb{R}^n$ must have $n$ vectors, Therefore, $u, v, w$ cannot form a basis for $\\mathbb{R}^2$\n",
    "\n",
    "<b>(b)</b> We have $w = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$ Since zero vector makes vectors linearly dependent. $u, v, w$ do not form a basis\n",
    "\n",
    "\n",
    "<b>(c)</b> Here, we have vector $u = 2x$. Therefore, again linearly dependent, and cannot form a basis\n",
    "\n",
    "<b>(d)</b>The basis of $\\mathbb{R}^n$ must have $n$ vectors, Therefore, $u, v, w, x$ cannot form a basis for $\\mathbb{R}^5$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "211bcc6d",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a109615a",
   "metadata": {},
   "source": [
    "<b> Question 5</b><br>\n",
    "\n",
    "\n",
    "Determine whether vector $b$ is in the space spanned by the columns of matrix $A$: \n",
    "\n",
    "(a) $A = \\begin{pmatrix}\n",
    "1&1&0\\\\\n",
    "2&5&0\\\\\n",
    "3&4&9\\\\\n",
    "\\end{pmatrix}$, $b = \\begin{pmatrix}\n",
    "1\\\\\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix}\n",
    "\\qquad$ (b) $A = \\begin{pmatrix}\n",
    "1&2&3\\\\\n",
    "4&5&6\\\\\n",
    "7&8&9\\\\\n",
    "\\end{pmatrix}$, $b = \\begin{pmatrix}\n",
    "1\\\\\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix}\n",
    "$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "fb226431",
   "metadata": {},
   "source": [
    "If vector $b$ is in the space spanned by the columns of matrix $A$,<br>\n",
    "then, vector $x$ with any real scalar $k_1,\\dots K_n$ must satisfy $Ax = b$\n",
    "\n",
    "<b>Solution (a)</b><br>\n",
    "\n",
    "Let $x = \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "k_2\\\\\n",
    "k_3\\\\\n",
    "\\end{pmatrix}$, we need to find values of $x$ such that $Ax = b$ ie.,<br>\n",
    "$\\begin{pmatrix}\n",
    "1&1&0\\\\\n",
    "2&5&0\\\\\n",
    "3&4&5\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "k_1\\\\\n",
    "k_2\\\\\n",
    "k_3\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "1\\\\\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Creating Agumented Matrix \n",
    "\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&1&0&1\\\\\n",
    "2&5&0&3\\\\\n",
    "3&4&5&4\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "$\\Rightarrow\\left(\\begin{array}{ccc|c}\n",
    "1&1&0&1\\\\\n",
    "0&3&0&1\\\\\n",
    "0&1&9&1\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "This gives us:<br>\n",
    "$3k_2 = 1$<br>\n",
    "$\\Rightarrow k_2 = \\frac{1}{3}$\n",
    "\n",
    "Then, $k_1 + 1 - k_2 \\qquad \\Rightarrow k_1 = \\frac{2}{3}$<br>\n",
    "and $9k_3 = 1 - k_2 \\qquad \\Rightarrow k_3 = \\frac{2}{27}$\n",
    "\n",
    "Checking $Ax$<br>\n",
    "$\\frac{2}{3}\\begin{pmatrix}\n",
    "1\\\\\n",
    "2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} + \\frac{1}{3}\\begin{pmatrix}\n",
    "1\\\\\n",
    "5\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix} + \\frac{2}{27}\\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "9\\\\\n",
    "\\end{pmatrix}$ \n",
    "\n",
    "$= \\begin{pmatrix}\n",
    "2/3&1/3&0\\\\\n",
    "4/3&5/3&0\\\\\n",
    "2&4/3&2/3\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "1\\\\\n",
    "3\\\\\n",
    "4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "With these values of $k_1, k_2, \\text{and} k_3$ we can get $b$. Therefore, vector $b$ spans columns of matrix $A$\n",
    "\n",
    "\n",
    "<b>Solution(b)</b>\n",
    "\n",
    "Lets take $x = \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "k_2\\\\\n",
    "k_3\\\\\n",
    "\\end{pmatrix}$, if we get to find values of $k_1, k_2, K_3$ such that $Ax = b$ then vector $b$ spans $A$\n",
    "\n",
    "\n",
    "Creating Agumented Matrix $(A | b)$ \n",
    "\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&2&3&1\\\\\n",
    "4&5&6&3\\\\\n",
    "7&8&9&4\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Solving this with Gaussian Elimination, we get\n",
    "\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&2&3&1\\\\\n",
    "3&3&3&2\\\\\n",
    "0&0&0&-1\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Third row, suggest $0k_1 + 0k_2 + 0k_3 = 0 $ impliying $0 = -1$\n",
    "which is inconsistent. Therefore, any values of $k_1, K_2, k_3$ doesn't satisfy $Ax = b$.\n",
    "\n",
    "Hence, vector $b$ is not in the space spanned by the columns of matrix $A$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3a509676",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "30927a9f",
   "metadata": {},
   "source": [
    "<b> Question 6</b><br>\n",
    "\n",
    "\n",
    "Show that the following vectors form a basis for $\\mathbb{R}^3$ : \n",
    "\n",
    "(a) $u = \\begin{pmatrix}\n",
    "5&0&0\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "0&6&0\\\\\n",
    "\\end{pmatrix}^T, w = \\begin{pmatrix}\n",
    "0&0&7\\\\\n",
    "\\end{pmatrix}^T$\n",
    "\n",
    "(b) $u = \\begin{pmatrix}\n",
    "\\alpha&1&1\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "1&\\beta&0\\\\\n",
    "\\end{pmatrix}^T, w = \\begin{pmatrix}\n",
    "0&0&\\gamma\\\\\n",
    "\\end{pmatrix}^T$, where $\\alpha$, $\\beta$, and $\\gamma$ are any non-zero real numbers."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "665b6404",
   "metadata": {},
   "source": [
    "<b> Solution: <br>\n",
    "    Key thing</b> There are two conditions to be satisfied to establish if $n$ vectors form a basis. <br>\n",
    "    1. The vectors span $\\mathbb{R}^n$<br>\n",
    "    2. The vectors are linearly independent\n",
    "\n",
    "However, if we have $n$ vectors in $\\mathbb{R}^n$, then we only need to <b>check one of the conditions, either independence or span</b>\n",
    "\n",
    "<b>(a)</b> Here, we have, $u = \\begin{pmatrix}\n",
    "5\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "6\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$ and $w = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "7\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "To check span, lets take $k_1, k_2$ and $k_3$ such that $k_1u + k_2v + k_3 = \\sigma$ where $\\sigma$ is an arbitary vector $\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now<br>\n",
    "$k_1\\begin{pmatrix}\n",
    "5\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "0\\\\\n",
    "6\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "7\\\\\n",
    "\\end{pmatrix}=\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "5k_1\\\\\n",
    "6k_2\\\\\n",
    "7k_3\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow k_1 =\\frac{a}{5}, k_2 = \\frac{b}{6}$ and $k_3 = \\frac{c}{7}$\n",
    "\n",
    "Therefore, $u, v$ and $w$ span $\\mathbb{R}^2$<br>\n",
    "\n",
    "Hence, $u, v$ and $w$ <b>form a basis</b> of $\\mathbb{R}^3$\n",
    "\n",
    "\n",
    "<b>Solution (b)</b> \n",
    "\n",
    "Here, we have, $u = \\begin{pmatrix}\n",
    "\\alpha\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "\\beta\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$ and $w = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\gamma\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This problem is same as previous one and evetually we will get\n",
    "\n",
    "$k_1 =\\frac{a}{\\alpha}, k_2 = \\frac{b}{\\beta}$ and $k_3 = \\frac{c}{\\gamma}$\n",
    "\n",
    "Since, $\\alpha, \\beta$ and $\\gamma$ are non-zero real numbers, $u, v$ and $w$ span $\\mathbb{R}^2$<br>\n",
    "\n",
    "Hence, $u, v$ and $w$ <b>form a basis</b> of $\\mathbb{R}^3$\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7dc53bca",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d26e6c4a",
   "metadata": {},
   "source": [
    "<b> Question 7</b><br>\n",
    "\n",
    "\n",
    "Show that the vectors $u = \\begin{pmatrix}\n",
    "1&1&2\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "-1&1&-2\\\\\n",
    "\\end{pmatrix}^T, w = \\begin{pmatrix}\n",
    "1&-5&2\\\\\n",
    "\\end{pmatrix}^T$ do not form a basis for $\\mathbb{R}^3$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d66d8609",
   "metadata": {},
   "source": [
    "<b> Solution </b>\n",
    "\n",
    "\n",
    "As we have 3 vectors in $\\mathbb{R}^3$ we need to check either it is linearly dependent or do not span, to prove $u, v, w$ do not form a basis for $\\mathbb{R}^3$\n",
    "\n",
    "\n",
    "Here, we have, $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "1\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix}$ and $w = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-5\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Let $k_1, k_2$ and $k_3$ satisfy $k_1u + k_2v + k_3w = \\sigma$ where $\\sigma$ is an arbitary vector $\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow k_1\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "-1\\\\\n",
    "1\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix} + k_3\\begin{pmatrix}\n",
    "1\\\\\n",
    "-5\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "This gives us Augmented Matrix as:\n",
    "\n",
    "\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&-1&1&a\\\\\n",
    "1&1&-5&b\\\\\n",
    "2&-2&2&c\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Solving this with Gaussian Elimination, we get\n",
    "\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&-1&1&a\\\\\n",
    "0&2&6&b-a\\\\\n",
    "0&0&0&c-2a\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "From the last row, we get<br>\n",
    "$ c - 2a = 0$<br>\n",
    "$\\Rightarrow c = 2a$\n",
    "\n",
    "Therefore, vector span $\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "2a\\\\\n",
    "\\end{pmatrix}$ and do not span $\\begin{pmatrix}\n",
    "a\\\\\n",
    "b\\\\\n",
    "c\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Hence, do not form a basis for $\\mathbb{R}^3$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ce0a659b",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a93cffc5",
   "metadata": {},
   "source": [
    "<b> Question 8</b><br>\n",
    "\n",
    "\n",
    "Show that the vectors \n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "1&1&1&1\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "0&1&1&1\\\\\n",
    "\\end{pmatrix}^T, w = \\begin{pmatrix}\n",
    "0&0&1&1\\\\\n",
    "\\end{pmatrix}^T, x = \\begin{pmatrix}\n",
    "0&0&0&1\\\\\n",
    "\\end{pmatrix}^T$ form a basis for $\\mathbb{R}^4$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5babe048",
   "metadata": {},
   "source": [
    "<b> Solution: <br>\n",
    "    Key thing</b> There are two conditions to be satisfied to establish if $n$ vectors form a basis. <br>\n",
    "    1. The vectors span $\\mathbb{R}^n$<br>\n",
    "    2. The vectors are linearly independent\n",
    "\n",
    "However, if we have $n$ vectors in $\\mathbb{R}^n$, then we only need to <b>check one of the conditions, either independence or span</b>\n",
    "\n",
    "Here, we have\n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}, w = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}, \\text{and}\\quad x = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "For linear combination <br>\n",
    "$k_1u + k_2v +k_3w + k_4x = 0$\n",
    "\n",
    "If $k_1 = k_2 = k_3 = k_4 = 0$, then <b>independent</b>\n",
    "\n",
    "Now, $k_1u + k_2v +k_3w + k_4x =k_1 \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} + k_2\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} + k_3 \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} + k_4\\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "k_1 + k_2\\\\\n",
    "k_1 + k_2 + k_3\\\\\n",
    "k_1 + k_2 + k_3 + k_4\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Solving above equations we get, $k_1 = k_2 = k_3 = k_4 = 0$\n",
    "\n",
    "$\\Rightarrow u, v, w, x$ are linearly independent\n",
    "\n",
    "Hence, $u, v, w, x$ form a basis for $\\mathbb{R}^4$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6b82b959",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d5014c4a",
   "metadata": {},
   "source": [
    "<b>Question 9</b><br>\n",
    "\n",
    "\n",
    "Let $u = \\begin{pmatrix}\n",
    "1&5&0\\\\\n",
    "\\end{pmatrix}^T, v = \\begin{pmatrix}\n",
    "0&1&0\\\\\n",
    "\\end{pmatrix}^T$ \n",
    "\n",
    "(a) Determine the dependency of $u$ and $v$ <br>\n",
    "(b) Find the space spanned by $\\{u, v\\}$ <br>\n",
    "(c) Write down a general vector $w$ such that $\\{u, v, w\\}$ form a basis for $\\mathbb{R}^3$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d67cb36c",
   "metadata": {},
   "source": [
    "<b>Solution (a)</b>\n",
    "\n",
    "For linear combination $k_1u + k_2v = 0$ \n",
    "if $k_1 = k_2 = 0$ then independent, otherwise dependent\n",
    "\n",
    "Here,<br>\n",
    "$k_1\\begin{pmatrix}\n",
    "1\\\\\n",
    "5\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}+ k_2\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "5k_1 + k_2\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$k_1 = 0$<br>\n",
    "$k_2 = 0$\n",
    "\n",
    "Hence, vector $u$ and $v$ are <b>linearly independent</b>\n",
    "\n",
    "<b>Solution (b)</b>\n",
    "\n",
    "The space spanned by given vector $u$ and $v$ is linear combination of $u$ and $v$ with any real $k_1, k_2$ and as:<br>\n",
    "$k_1\\begin{pmatrix}\n",
    "1\\\\\n",
    "5\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}+ k_2\\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "k_1\\\\\n",
    "5k_1 + k_2\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Thus, we get vector $u, v$ span $x-y plane$\n",
    "\n",
    "<b>Solution(c)</b>\n",
    "\n",
    "To get a general vector 𝑤 such that {𝑢,𝑣,𝑤} form a basis for $\\mathbb{R}^3$, we need $w$ linearly independent of $u =\\begin{pmatrix}\n",
    "1\\\\\n",
    "5\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, and $v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Since, $u$ and $v$ span $\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, $w$ can be $\\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix}$ where $z$ is any non-zero real number."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "aaa67108",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7af940dd",
   "metadata": {},
   "source": [
    "<b> Question 10</b><br>\n",
    "\n",
    "\n",
    "Let $v_1, v_2, v_3, \\dots v_n$ span $\\mathbb{R}^n$. Show that $v_1, v_2, v_3, \\dots v_n$ and $w$ also span $\\mathbb{R}^n$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e15ee64a",
   "metadata": {},
   "source": [
    "<b>Solution</b> <br>\n",
    "\n",
    "Since, $v_1, v_2, v_3, \\dots v_n$ span $\\mathbb{R}^n$ <br>\n",
    "\n",
    "So, for any vector $u$ in $\\mathbb{R}^n$, we can write:\n",
    "\n",
    "\n",
    "$u = k_1v_1 + k_2v_2 + \\dots + k_nv_n$\n",
    "\n",
    "\n",
    "$\\Rightarrow u =  k_1v_1 + k_2v_2 + \\dots + k_nv_n + 0w$\n",
    "\n",
    "Hence, $v_1, v_2, v_3, \\dots v_n$ and $w$ also span $\\mathbb{R}^n$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "bc8bbb70",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "022e97f4",
   "metadata": {},
   "source": [
    "<b> Question 11</b><br>\n",
    "\n",
    "\n",
    "Let $S = \\{v_1, v_2, v_3 \\dots v_n\\}$ be a basis for $\\mathbb{R}^n$. Prove that $T = \\{k_1v_1, k_2v_2, k_3v_3 \\dots k_nv_n\\}$ <br>\n",
    "where none of the $k's$ are zero, is also a basis for $\\mathbb{R}^n$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6d51aee4",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "46d4f9e2",
   "metadata": {},
   "source": [
    "<b> Question 12</b><br>\n",
    "\n",
    "Let $A = (v_1, v_2 \\dots v_n)$. Prove that $A$ is invertible $\\iff \\{v_1, \\dots, v_n\\}$ form a basis for $\\mathbb{R}^n$ "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "08a327ba",
   "metadata": {},
   "source": [
    "<b>Solution</b> <br>\n",
    "We know that, if $A$ be the $n$ by $n$ matrix whose columns are given by the vectors $v_1, v_2, \\dots v_n$:\n",
    "\n",
    "$A = (v_1, v_2, \\dots, v_n)$\n",
    "\n",
    "Then, vectors $(v_1, v_2, \\dots, v_n$ are linearly independent $\\iff$ matrix $A$ is invertible\n",
    "\n",
    "$\\Rightarrow$ matrix $A$ is invertible $\\iff$ vectors $(v_1, v_2, \\dots, v_n$ are linearly independent\n",
    "\n",
    "We also know that, <b>\"Any $n$ linearly independent vectors in $\\mathbb{R}^n$\" form a basis for $\\mathbb{R}^n$</b>\n",
    "\n",
    "Therefore, $A$ is invertible $\\iff \\{v_1, \\dots, v_n\\}$ form a basis for $\\mathbb{R}^n$ "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3562ec31",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f9f42df6",
   "metadata": {},
   "source": [
    "<b> Question 13</b><br>\n",
    "\n",
    "\n",
    "Prove proposition 2.27<br>\n",
    "\"Any $n$ linearly independent vectors in $\\mathbb{R}^n$\" form a basis for $\\mathbb{R}^n$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "18b5c802",
   "metadata": {},
   "source": [
    "<b>Solution</b>\n",
    "\n",
    "We know that there is a proposition that states\n",
    "\n",
    "\n",
    "<b>\"If we have $v_1, v_2, v_3,\\dots v_m$ as different vectors in $\\mathbb{R}^n$, where $n \\lt m$, then the vectors $v_1, v_2, v_3, \\dots v_m$ are linearly dependent\"</b>\n",
    "\n",
    "Now, suppose if we have a set of $n$ independent vectors, S i.e, $S = \\{v_1, v_2, v_3, \\dots v_n\\}$ that does not span $\\mathbb{R}^n$.\n",
    "\n",
    "Then, there must be another $u$ that is not a linear combination of $S$.<br>\n",
    "\n",
    "Lets call this new set as $T$ which consists of now $n+1$ vectors in $\\mathbb{R}^n$ and written as $T = \\{v_1, v_2, v_3, \\dots v_n, u\\}$\n",
    "\n",
    "From earlier proposition, this set $T = \\{v_1, v_2, v_3, \\dots v_n, u\\}$ must be linearly dependent\n",
    "\n",
    "$\\Rightarrow c_1v_1 + c_2v_2 + \\dots c_nv_n + c_{n+1}u = 0$ where all c's are not zero.\n",
    "\n",
    "$\\Rightarrow u = (-c_1/c_{n+1})v_1 + (-c_2/c_{n+1})v_2 + ... + (-c_n/c_{n+1})v_n$\n",
    "\n",
    "We can see $u$ can be written as linear combination of $v_1, v_2, v_3, \\dots v_n$, the set of vectors $S$.\n",
    "\n",
    "Therefore, $u$ is in the span of $S$\n",
    "\n",
    "This contradiction our prior assumption that $u$ cannot be written as linear combination of $S$ and $S$, as set of linearly independent vectors does not span $\\mathbb{R}^n$\n",
    "\n",
    "<b>Hence, $S = \\{v_1, v_2, v_3, \\dots v_n\\}$  is a set of $n$ linearly independent vectors in $\\mathbb{R}^n$ so it a basis for $\\mathbb{R}^n$</b>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3bf34026",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1ed95758",
   "metadata": {},
   "source": [
    "<b> Question 14</b><br>\n",
    "\n",
    "Prove proposition 2.28<br>\n",
    "\"Any $n$ vectors which span $\\mathbb{R}^n$ form a basis for $\\mathbb{R}^n$\""
   ]
  },
  {
   "cell_type": "markdown",
   "id": "27445c65",
   "metadata": {},
   "source": [
    "<b>Solution</b>\n",
    "\n",
    "Let $S = \\{v_1, v_2, v_3, \\dots v_n\\}$ be set of $n$ vectors which span $\\mathbb{R}^n$\n",
    "\n",
    "As we know from Question 13, <b>\"Any $n$ linearly independent vectors in $\\mathbb{R}^n$ form a basis for $\\mathbb{R}^n$</b>\n",
    "\n",
    "We need to prove $S = \\{v_1, v_2, v_3, \\dots v_n\\}$ are linearly independent to be a basis for  $\\mathbb{R}^n$\n",
    "\n",
    "Suppose the vectors in $S$ are linearly dependent.<br>\n",
    "Then one of the vectors can be written as liner combination of others.\n",
    "\n",
    "If that vector is $v_n$ and the remaining vectors $T = {v_1, v_2, \\dots, v_n}$ still spans $\\mathbb{R}^n$\n",
    "\n",
    "If we keep repeating this process, the very last set of vectors would still span $\\mathbb{R}^n$ and linearly independent to form a basis for $\\mathbb{R}^n$\n",
    "\n",
    "This contradicts, one proposition <b>\"Every basis of $\\mathbb{R}^n$ contains exactly $n$ vectors\". Therefore our supposition $S = \\{v_1, v_2, v_3, \\dots v_n\\}$ to be linearly dependent must be wrong.\n",
    "    \n",
    "Hence, any $n$ vectors which span $\\mathbb{R}^n$ form a basis for $\\mathbb{R}^n$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6efdd541",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b56fd023",
   "metadata": {},
   "source": [
    "<b> Question 15</b><br>\n",
    "\n",
    "\n",
    "Let $\\{v_1, v_2, v_3, \\dots v_n \\}$ be a basis for $\\mathbb{R}^n$ and $A$ be an invertible matrix. Prove that $\\{Av_1, Av_2, Av_3,\\dots Av_n\\}$ is also a  basis for $\\mathbb{R}^n$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1c4f010d",
   "metadata": {},
   "source": [
    "------------------------------------------------------------"
   ]
  }
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