Ch5_Linear_Transform_Ex5.3.ipynb

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    "<b> Question 1 </b> \n",
    "\n",
    "Determine:<br>\n",
    "(i)ker$(T)\\qquad$(ii)nullity$(T)\\qquad$(ii)range$(T)\\qquad$(iv) rank$(T)$\n",
    "\n",
    "for the following linear transformation without using the dimension theorem:\n",
    "\n",
    "(a)$T: \\mathbb{R}^2 \\to \\mathbb{R}^2$ given by $T(v) = Av$ where $A = \\begin{pmatrix}\n",
    "1&2\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "(b)$T: \\mathbb{R}^3 \\to \\mathbb{R}^3$ given by $T(v) = Av$ where $A = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&1&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "(c)$T: \\mathbb{R}^3 \\to \\mathbb{R}^3$ given by $T(v) = Av$ where $A = \\begin{pmatrix}\n",
    "1&3&5\\\\\n",
    "2&6&10\\\\\n",
    "4&12&20\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "(d)$T: P_3 \\to P_3$ given by $T(p) = xp'$\n",
    "\n",
    "(e)$T: P_3 \\to P_2$ given by $T(p) = p'$\n",
    "\n",
    "(f)$T: P_3 \\to \\mathbb{R}$ given by $T(p) = \\int_0^1P(x) dx$\n",
    "\n",
    "(g)$T: M_{22} \\to P_1$ given by $T\\begin{pmatrix}\\begin{pmatrix}\n",
    "a&b\\\\\n",
    "c&d\\\\\n",
    "\\end{pmatrix}\\end{pmatrix} = (a+c)x + (b+d)$ "
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    "<b> Solution (a) </b>\n",
    "\n",
    "<b>(i)</b><br>\n",
    "Let $v = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix}$. For ker$(T)$, we need to find $x$ & $y$ such that $Tv = Av = O$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1&2\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix}\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = O$\n",
    "\n",
    "$\\Rightarrow x + 2y = 0$<br>\n",
    "and, $x + 2y = 0$\n",
    "\n",
    "If $y = r$, then $x = -2r$ \n",
    "\n",
    "$\\Rightarrow v = \\begin{pmatrix}\n",
    "-2r\\\\\n",
    "r\\\\\n",
    "\\end{pmatrix}$ or $r\\begin{pmatrix}\n",
    "-2\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Hence, ker$(T) = \\{ r\\begin{pmatrix}\n",
    "-2\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} | r \\in \\mathbb{R}^n \\}$\n",
    "\n",
    "\n",
    "<b>(ii)</b> \n",
    "\n",
    "$nullity(T) = dim(T)$.<br>\n",
    "where, $dim(T)$ is number of vectors in the basis for $ker(T)$.\n",
    "\n",
    "Here, basis = $\\{\\begin{pmatrix}\n",
    "-2\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\}$\n",
    "\n",
    "Therefore, $dim(ker(T)) =1$<br>\n",
    "$\\Rightarrow nullity(T) = 1$\n",
    "\n",
    "\n",
    "<b> (iii)</b>\n",
    "\n",
    "To get range, we need to solve $Tv = Av$ again, where $v = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$Tv = \\begin{pmatrix}\n",
    "1&2\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix}\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "x + 2y\\\\\n",
    "x + 2y\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Thus, $range(T) = \\{ \\begin{pmatrix}\n",
    "a\\\\\n",
    "a\\\\\n",
    "\\end{pmatrix} | a \\in \\mathbb{R} \\}$ for both entries being same\n",
    "\n",
    "<b>(iv)</b>\n",
    "\n",
    "$rank(T) = dim(range(T))$\n",
    "\n",
    "Dimension of range is the number of vectors in the basis for $range(T)$\n",
    "\n",
    "here, basis $= \\left[\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\right]$ and number of vectors is 1.\n",
    "\n",
    "Thus, $dim(range(T) = 1$\n",
    "\n",
    "$\\Rightarrow rank(T) = 1$\n",
    "\n",
    "<b> Solution (b)</b>\n",
    "\n",
    "<b>(i)</b>\n",
    "\n",
    "Let $v = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix}$. For $ker(T)$ we need to solve for $x, y, z$ such that $Tv = Av = O$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&1&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = O$\n",
    "\n",
    "$\\Rightarrow x = 0, y = 0, z = 0$\n",
    "\n",
    "$\\Rightarrow ker(T) = 0$\n",
    "\n",
    "<b>(ii)</b> \n",
    "\n",
    "$nullity(T) = dim(ker(T)$\n",
    "\n",
    "Here, $ker(T) = O$. Therefore, $dim(ker(T)) = 0$\n",
    "\n",
    "$\\Rightarrow nullity(T) = 0$\n",
    "\n",
    "<b>(iii)</b>\n",
    "\n",
    "$Tv = Av$ linear transformation where $v = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix}$ gives:\n",
    "\n",
    "$Tv = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&1&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix}\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $range(T) = \\mathbb{R}^3$\n",
    "\n",
    "<b>(iv)</b>\n",
    "\n",
    "$rank(T) = dim(range(T)$\n",
    "\n",
    "$\\Rightarrow rank(T) = dim(range(T)) = 3$\n",
    "\n",
    "<b> Solution (c) </b> \n",
    "\n",
    "<b>(i)</b>\n",
    "\n",
    "Let $v = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix}$. For $ker(T)$, we need to solve $x, y, z $ such that $Tv = Av = O$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1& 3 & 5\\\\\n",
    "2 & 6 & 10\\\\\n",
    "4 & 12 & 20\\\\\n",
    "\\end{pmatrix}\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Applying Gaussian, we get:<br>\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&3&5&0\\\\\n",
    "2&6&10&0\\\\\n",
    "4&12&20&0\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Then, with $R_2 = R_2 - 2R_1$ and $R_3 = R_3 - 4R_1$, we get:<br>\n",
    "\n",
    "$\\left(\\begin{array}{ccc|c}\n",
    "1&3&5&0\\\\\n",
    "0&0&0&0\\\\\n",
    "0&0&0&0\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "Thus, we get, $x + 3y + 5z = 0$. If $y = s$, and $z = r$, we get, $x = -3s -5r$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "-3s - 5r\\\\\n",
    "s\\\\\n",
    "r\\\\\n",
    "\\end{pmatrix}$ or $s\\begin{pmatrix}\n",
    "-3\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + r\\begin{pmatrix}\n",
    "-5\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $ker(T) = \\left[s\\begin{pmatrix}\n",
    "-3\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + r\\begin{pmatrix}\n",
    "-5\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} | s \\in \\mathbb{R}, r \\in \\mathbb{R}\\right]$\n",
    "\n",
    "<b>(ii)</b>\n",
    "\n",
    "$nullity(T) = dim(ker(T))$\n",
    "\n",
    "Dimension is the number of vectors in the basis for $ker(T)$. Here, \n",
    "$ker(T) = \\left[\\begin{pmatrix}\n",
    "-3\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, \\begin{pmatrix}\n",
    "-5\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\right]$ has 2 basis vectors.\n",
    "\n",
    "Therefore, $nullity(T) = 2$\n",
    "\n",
    "<b>(iii)</b>\n",
    "\n",
    "$Tv = Av$ linear transformation where $v = \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix}$ gives:\n",
    "\n",
    "$Tv = \\begin{pmatrix}\n",
    "1 & 3 & 5\\\\\n",
    "0 & 0 & 0\\\\\n",
    "0 & 0 & 0\\\\\n",
    "\\end{pmatrix}\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "x + 3y + 5z\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Here, $x, y, z$ are any real numbers therefore, the $range(T) = \\mathbb{R}^1$\n",
    "\n",
    "<b>(iv)</b>\n",
    "\n",
    "$rank(T) = dim(range(T))$\n",
    "\n",
    "Here, dimension of $range(T) =1$.\n",
    "\n",
    "Therefore, $rank(T) =1$\n",
    "\n",
    "<b>Solution (d)</b>\n",
    "\n",
    "<b>(i)</b>\n",
    "\n",
    "Let $p = p(x) = ax^3 + bx^2 + cx + d$<br>\n",
    "\n",
    "For $ker(T)$, we need to solve, $T(p) = O$. We are given $T(p) = xp'$\n",
    "\n",
    "$\\Rightarrow T(p) = x(ax^3 + bx^2 + cx + d)'$\n",
    "\n",
    "$\\quad = x(3ax^2 + 2bx +c)$\n",
    "\n",
    "$\\quad = 3ax^3 + 2bx^2 + cx = 0$\n",
    "\n",
    "Here, only condition of $a = 0, b = 0, c = 0$ gives $T(p) = O$\n",
    "\n",
    "$\\Rightarrow ker(T) = \\{ax^3 + bx^2 + cx + d | a = b= c = 0\\}$\n",
    "\n",
    "$\\quad = \\{d | d \\in \\mathbb{R}\\} = P_0$\n",
    "\n",
    "$\\Rightarrow ker(T) = P_0$ i.e, set of constant polynomials.\n",
    "\n",
    "\n",
    "<b>(ii)</b> \n",
    "\n",
    "$nullity(T) = dim(ker(T))$\n",
    "\n",
    "Here, basis for $ker(T) = P_0$ is 1. Therefore, $dim(ker(T)) = 1$\n",
    "\n",
    "$\\Rightarrow nullity(T) = 1$\n",
    "\n",
    "<b>(iii)</b>\n",
    "\n",
    "$T(p) = xp'$ linear transformation where $p  = ax^3 + bx^2 + cx + d$ gives:\n",
    "\n",
    "$T(p) = 3ax^2 + 2bx^2 + cx$. A cubical polynomial where $3a = d, 2b =e, c = f$\n",
    "\n",
    "Therefore, $range(T) = \\{ dx^3 + ex^2 + fx | d, e, f \\in \\mathbb{R}\\}$\n",
    "\n",
    "<b>(iv)</b>\n",
    "\n",
    "$rank(T) = dim(range(T)$\n",
    "\n",
    "we have $range(T) = \\{ dx^3 + ex^2 + fx \\}$. Here, basis for $\\{dx^3 + ex^2 + fx \\}$ is $\\{x^3, x^2, x \\}$.\n",
    "\n",
    "Therefore, $dim(range(T) = 3$\n",
    "\n",
    "$\\Rightarrow rank(T) = 3$\n",
    "\n",
    "<b> Solution (e) </b><br>\n",
    "<b>(i)</b>\n",
    "\n",
    "\n",
    "Let $p = p(x) = ax^3 + bx^2 + cx + d$<br>\n",
    "\n",
    "For $ker(T)$, we need to solve, $T(p) = O$. We are given $T(p) = p'$\n",
    "\n",
    "$\\Rightarrow T(p) = x(ax^3 + bx^2 + cx + d)'$\n",
    "\n",
    "$\\quad = (3ax^2 + 2bx +c)$\n",
    "\n",
    "$\\quad = 3ax^3 + 2bx^ + c = 0$\n",
    "\n",
    "Here, only condition of $a = 0, b = 0$ gives $T(p) = O$\n",
    "\n",
    "$\\Rightarrow ker(T) = \\{ax^2 + bx + c | a = b = 0\\}$\n",
    "\n",
    "$\\quad = \\{c | c \\in \\mathbb{R}\\} = P_0$\n",
    "\n",
    "$\\Rightarrow ker(T) = P_0$ i.e, set of constant polynomials.\n",
    "\n",
    "<b>(ii)</b>\n",
    "\n",
    "$nullity(T) = dim(ker(T))$\n",
    "\n",
    "Here, basis for $ker(T) = P_0$ is 1. Therefore, $dim(ker(T)) = 1$\n",
    "\n",
    "$\\Rightarrow nullity(T) = 1$\n",
    "\n",
    "<b>(iii)</b>\n",
    "\n",
    "$T(p) = p'$ linear transformation where $p  = ax^3 + bx^2 + cx + d$ gives:\n",
    "\n",
    "$T(p) = 3ax^2 + 2bx + c$. A polynomial where $3a = d, 2b =e, c = f$\n",
    "\n",
    "Therefore, $range(T) = \\{ dx^2 + ex + f | d, e, f \\in \\mathbb{R}\\}$ a vector space of set of polynomials of degree 2 or less. <br>\n",
    "Thus, $range(T) = P_2$\n",
    "\n",
    "<b>(iv)</b>\n",
    "\n",
    "$rank(T) = dim(range(T)$\n",
    "\n",
    "we know that $dim(P_k) = k +1$. So, $dim(P_2) = 3$ \n",
    "\n",
    "Therefore, $dim(range(T) = 3$\n",
    "\n",
    "$\\Rightarrow rank(T) = 3$\n",
    "\n",
    "<b>Solution(f)</b>\n",
    "\n",
    "<b>(i)</b><br>\n",
    "Let $p = ax^3 + bx^2 + cx + d$. For $ker(T)$, we need to solve $T(p) = \\int_0^1 p(x) dx$ such that $T(p) = O$\n",
    "\n",
    "$\\Rightarrow T(p) = \\int_0^1 [ax^3 + bx^2 + cx + d] dx$\n",
    "\n",
    "$\\quad = \\left[ \\frac{ax^4}{4} + \\frac{bx^3}{3} + \\frac{cx^2}{2} + dx \\right]_0^1$\n",
    "\n",
    "$\\quad = \\frac{a}{4} + \\frac{b}{3} + \\frac{c}{2} + d = 0$\n",
    "\n",
    "$\\quad = 3a + 4b + 6c + 12d = 0$\n",
    "\n",
    "$\\quad a = -\\frac{4}{3}b - 2c -4d$\n",
    "\n",
    "Here, $ker(T) = \\{ ax^3 + bx^2 + cx + d | a = -\\frac{4}{3}b - 2c -4d \\}$\n",
    "\n",
    "<b>(ii)</b><br>\n",
    "\n",
    "$nullity(T) = dim(ker(T))$\n",
    "\n",
    "Dimension is the number of vectos in the basis of kernel.\n",
    "\n",
    "Therefore, solving $p(x)$ with $a = -\\frac{4}{3}b - 2c -4d$, we get: \n",
    "\n",
    "$P(x) = ax^3 + bx^2 + cx + d$\n",
    "\n",
    "$\\quad = (-\\frac{4}{3}b - 2c -4d)x^3 + bx^2 + cx +d$\n",
    "\n",
    "$\\quad = -\\frac{4}{3}bx^3 - 2cx^3 -4dx^3 + bx^2 + cx + d$\n",
    "\n",
    "$\\quad = (x^2 - \\frac{4}{3}x^3)b + (x - 2x^3)c + (1 - 4x^3)d$\n",
    "\n",
    "Thus, basis for $ker(T) = \\left[(x^2 - \\frac{4}{3}x^3), (x - 2x^3), (1 - 4x^3) \\right]$\n",
    "\n",
    "Therefore, $dim(ker(T)) = 3$\n",
    "\n",
    "$\\Rightarrow nullity(T) = 3$\n",
    "\n",
    "<b>(iii)</b><br>\n",
    "\n",
    "Given linear transformation where $p(x) = ax^3 + bx^2 + cx + d$ gives:\n",
    "\n",
    "$T(p) = \\frac{a}{4} + \\frac{b}{3} + \\frac{c}{2} +d $ \n",
    "\n",
    "Here, $a, b, c, d$ are any real numbers.<br>\n",
    "Therefore, $range(T) = \\mathbb{R}$\n",
    "\n",
    "\n",
    "<b>(iv)</b><br>\n",
    "$rank(T) = dim(range(T)) = 1$\n",
    "\n",
    "<b>Solution (g) </b>\n",
    "\n",
    "Let $A = \\begin{pmatrix}\n",
    "a & b\\\\\n",
    "c & d\\\\\n",
    "\\end{pmatrix}$ where transformation $T\\begin{pmatrix}\\begin{pmatrix}\n",
    "a&b\\\\\n",
    "c&d\\\\\n",
    "\\end{pmatrix}\\end{pmatrix} = (a+c)x + (b+d)$ gives zero vector. i.e, $T(A) = O$\n",
    "\n",
    "$\\Rightarrow T\\begin{pmatrix}\\begin{pmatrix}\n",
    "a&b\\\\\n",
    "c&d\\\\\n",
    "\\end{pmatrix}\\end{pmatrix} = (a+c)x + (b+d) = 0$\n",
    "\n",
    "This gives $a + c = 0$ and $b+d = 0$.<br>\n",
    "Therefore, $c = -a, d = -b$.\n",
    "\n",
    "Hence, the $ker(T) = \\left[ \\begin{pmatrix}\n",
    "a & b\\\\\n",
    "-a & -b\\\\\n",
    "\\end{pmatrix} \\right]$\n",
    "\n",
    "<b>(ii)</b><br>\n",
    "$nullity(T) = dim(ker(T))$\n",
    "\n",
    "Here, $ker(T) = \\begin{pmatrix}\n",
    "a & b\\\\\n",
    "-a & -b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "and using row operations, it is:<br>\n",
    "$ker(T) = \\begin{pmatrix}\n",
    "a & b\\\\\n",
    "0 & 0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Next, basis for $\\begin{pmatrix}\n",
    "a & b\\\\\n",
    "0 & 0\\\\\n",
    "\\end{pmatrix}$ is $\\left[ \\begin{pmatrix}\n",
    "1 & 0\\\\\n",
    "0 & 0\\\\\n",
    "\\end{pmatrix}, \\begin{pmatrix}\n",
    "0 & 1\\\\\n",
    "0 & 0\\\\\n",
    "\\end{pmatrix}\n",
    "\\right]$\n",
    "\n",
    "Therefore, $dim(ker(T)) = 2$.<br>\n",
    "Hence, $nullity(T) =24 $\n",
    "\n",
    "<b>(iii)</b>\n",
    "\n",
    "With given linear transformation, we get, <br>\n",
    "\n",
    "$T(v) = T\\begin{pmatrix}\\begin{pmatrix}\n",
    "a&b\\\\\n",
    "c&d\\\\\n",
    "\\end{pmatrix}\\end{pmatrix} = (a+c)x + (b+d)$\n",
    "\n",
    "$\\Rightarrow range = \\{ (a+c)x + (b+d)\\}$\n",
    "\n",
    "Therefore, $range(T) = P_1$ i.e, set of linear polynomials\n",
    "\n",
    "<b>(iv)</b>\n",
    "$rank(T) = dim(range(T))$\n",
    "\n",
    "we know that , $dim(P_k) = k + 1$\n",
    "\n",
    "Thus, $rank(T) = 1+1 = 2$"
   ]
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    "--------------"
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   "source": [
    "<b>Question 2 </b> \n",
    "\n",
    "For each of the examples in parts(a) to parts(g) in question 1 verify the dimension theorem 5.12 which says nullity$(T)$ + rank$(T) = n$ where $n$ is the dimension of the domain of the linear transformation $T$."
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    "<b> Solution (a)</b>\n",
    "\n",
    "We are given $T: \\mathbb{R}^2 \\to \\mathbb{R}^2$\n",
    "\n",
    "$\\Rightarrow$ dimension of $\\mathbb{R}^2 = 2$<br>\n",
    "$\\Rightarrow n = 2$\n",
    "\n",
    "And, we have deduced from the prior question, <br>\n",
    "$nullity(T) =1$ and $rank(T) = 1$\n",
    "\n",
    "$\\Rightarrow nullity(T) + rank(T) = 1 + 1 =2 $\n",
    "\n",
    "Thus, <b>$nullity(T) + rank(T) =n$ proved</b>\n",
    "\n",
    "\n",
    "<b> Solution (b) </b>\n",
    "\n",
    "We are given $T: \\mathbb{R}^3 \\to \\mathbb{R}^3$\n",
    "\n",
    "$\\Rightarrow$ dimension of $\\mathbb{R}^3 = 3$<br>\n",
    "$\\Rightarrow n = 3$\n",
    "\n",
    "And, we have deduced from the prior question, <br>\n",
    "$nullity(T) =0$ and $rank(T) = 3$\n",
    "\n",
    "$\\Rightarrow nullity(T) + rank(T) = 0 + 3 =3 $\n",
    "\n",
    "Thus, <b>$nullity(T) + rank(T) =n$ proved</b>\n",
    "\n",
    "\n",
    "<b> Solution (c) </b>\n",
    "\n",
    "We are given $T: \\mathbb{R}^3 \\to \\mathbb{R}^3$\n",
    "\n",
    "$\\Rightarrow$ dimension of $\\mathbb{R}^3 = 3$<br>\n",
    "$\\Rightarrow n = 3$\n",
    "\n",
    "And, we have deduced from the prior question, <br>\n",
    "$nullity(T) =2$ and $rank(T) = 1$\n",
    "\n",
    "$\\Rightarrow nullity(T) + rank(T) = 2 + 1 =3 $\n",
    "\n",
    "Thus, <b>$nullity(T) + rank(T) =n$ proved</b>\n",
    "\n",
    "\n",
    "<b>Solution (d) </b>\n",
    "\n",
    "We are given $T: P_3 \\to P_3$ such as $T(p) = xp'$\n",
    "\n",
    "we know that $dim(P_k) = k+1$.<br>\n",
    "$\\Rightarrow n $ for $P_3 = 3 + 1$<br>\n",
    "$\\Rightarrow n =4$\n",
    "\n",
    "And, we have calculated $nullity(T) = 1$ and $rank(T) =3$\n",
    "\n",
    "$\\Rightarrow nullity(T) + rank(T) = 1 + 3 = 4$\n",
    "\n",
    "Hence, $nullity(T) + rank(T) = n$\n",
    "\n",
    "\n",
    "<b>Solution (e) </b>\n",
    "\n",
    "We are given $T: P_3 \\to P_2$ such that $T(p) =p'$\n",
    "\n",
    "we know that $dim(P_k) = k+1$<br>\n",
    "$\\Rightarrow n $ of the domain $P_3 = 3 + 1$<br>\n",
    "$\\Rightarrow n =4$\n",
    "\n",
    "And, we have calculated $nullity(T) = 1$ and $rank(T) =3$\n",
    "\n",
    "$\\Rightarrow nullity(T) + rank(T) = 1 + 3 = 4$\n",
    "\n",
    "Hence, $nullity(T) + rank(T) = n$\n",
    "\n",
    "\n",
    "<b> Solution (f) </b>\n",
    "\n",
    "We are given $T: P_3 \\to \\mathbb{R}$ such that $T(p) = \\int_0^1 p(x) dx$.\n",
    "\n",
    "we know that $dim(P_k) = k+1$.<br>\n",
    "$\\Rightarrow n $ of the domain $P_3 = 3 + 1$<br>\n",
    "$\\Rightarrow n =4$\n",
    "\n",
    "And, we have calculated $nullity(T) = 3$ and $rank(T) =1$\n",
    "\n",
    "$\\Rightarrow nullity(T) + rank(T) = 3 + 1 = 4$\n",
    "\n",
    "Hence, $nullity(T) + rank(T) = n$\n",
    "\n",
    "\n",
    "<b>Solution (g) </b>\n",
    "\n",
    "We are given $T: M_{22} \\to P_1$ \n",
    "\n",
    "Here, the dimension of the domain is 2+2  = 4\n",
    "\n",
    "And, we have deduced $nullity(T) =2$ and $rank(T) =2$\n",
    "\n",
    "Thus, $nullity(T) + rank(T) = 2 + 2 =4$\n",
    "\n",
    "Hence, $nullity(T) + rank(T) = n$"
   ]
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   "id": "7cd495bb",
   "metadata": {},
   "source": [
    "--------------"
   ]
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   "source": [
    "<b> Question 3 </b> \n",
    "\n",
    "Let $T: \\mathbb{R}^5 \\to \\mathbb{R}^3$ be given by $T(u) = Au$ where $u$ is in $\\mathbb{R}^5$ and <br> \n",
    "\n",
    "$\\begin{pmatrix}\n",
    "1&3&4&2&1\\\\\n",
    "2&6&-7&-2&5\\\\\n",
    "4&12&6&4&6\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Find a basis for (i)ker$(T)\\qquad$ (ii)range$(T)$"
   ]
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   "id": "fb989715",
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    "<b> Solution (i)</b> \n",
    "\n",
    "$Kernel$ is the set of vectors in the start vector, here $\\mathbb{R}^5$ that are transformed to the zero vector.\n",
    "\n",
    "Here, we would be finding basis vectors for $ker(T)$ with $Rx = O$, after getting reduced row echelon vector, $R$\n",
    "\n",
    "$R = \\begin{pmatrix}\n",
    "1&3&4&2&1\\\\\n",
    "2&6&-7&-2&5\\\\\n",
    "4&12&6&4&6\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "After reduced row echelon:\n",
    "\n",
    "$R = \\begin{pmatrix}\n",
    "1 & 3 & 9 & 0.4 & 1.8\\\\\n",
    "0 & 0 & 1 & 0.4 & -0.2\\\\\n",
    "0 & 0 & 0 & 0 & 0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "With $Rx = O$, we get following equations: \n",
    "\n",
    "$x_1 + 3x_2 + 9x_3 + 0.4x_4 + 1.8x_5 = 0$<br>\n",
    "$\\qquad\\qquad x_3 + 0.4x_4 -0.2x_5 = 0$\n",
    "\n",
    "Now, if $x_2 = r, x_4 = s$ and $x_5 =t$, we get:\n",
    "\n",
    "$x_3 = -0.4s + 0.2t$\n",
    "\n",
    "and $x_1 + 3r + 9(-0.4s + 0.2t) + 0.4s + 1.8t = 0$\n",
    "\n",
    "$\\Rightarrow x_1 + 3r -3.6s + 0.4s + 1.8t + 1.8t  =0$\n",
    "\n",
    "$\\Rightarrow x_1 + 3r -3.2s + 3.6t = 0$\n",
    "\n",
    "$\\Rightarrow x_1 = -3r + 3.2s -3.6t$\n",
    "\n",
    "Thus, $x = \\begin{pmatrix}\n",
    "x_1\\\\\n",
    "x_2\\\\\n",
    "x_3\\\\\n",
    "x_4\\\\\n",
    "x_5\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "-3r + 3.2s -3.6t\\\\\n",
    "r\\\\\n",
    "-0.4s + 0.2t\\\\\n",
    "s\\\\\n",
    "t\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow x = r\\begin{pmatrix}\n",
    "-3\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + s\\begin{pmatrix}\n",
    "3.2\\\\\n",
    "0\\\\\n",
    "-0.4\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} + t\\begin{pmatrix}\n",
    "-3.6\\\\\n",
    "0\\\\\n",
    "0.2\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, basis for $ker(T) = \\left[\\begin{pmatrix}\n",
    "-3\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, \\begin{pmatrix}\n",
    "3.2\\\\\n",
    "0\\\\\n",
    "-0.4\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, \\begin{pmatrix}\n",
    "-3.6\\\\\n",
    "0\\\\\n",
    "0.2\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}\\right]$\n",
    "\n",
    "<b> Solution (ii)</b>\n",
    "\n",
    "A basis for the $range(T)$ can be determined from the reduced row echelon form of $A^T$\n",
    "\n",
    "Here, $A = \\begin{pmatrix}\n",
    "1&3&4&2&1\\\\\n",
    "2&6&-7&-2&5\\\\\n",
    "4&12&6&4&6\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Therefore, $A^T = \\begin{pmatrix} \n",
    "1 & 2 & 4\\\\\n",
    "3 & 6 & 12\\\\\n",
    "4 & -7 & 6\\\\\n",
    "2 & -2 & 4\\\\\n",
    "1 & 5 & 6\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "which after RREF changes to:\n",
    "\n",
    "$RREF(A^T) = \\begin{pmatrix}\n",
    "1 & 0 & 8/3\\\\\n",
    "0 & 1 & 2/3\\\\\n",
    "0 & 0 & 0\\\\\n",
    "0 & 0 & 0\\\\\n",
    "0 & 0 & 0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Here, the first two rows forms the basis for the $range(T)$\n",
    "\n",
    "$\\Rightarrow$ basis for $range(T) = \\left[\\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "8/3\\\\\n",
    "\\end{pmatrix}, \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "2/3\\\\\n",
    "\\end{pmatrix}\\right]$"
   ]
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   "id": "e8929e5a",
   "metadata": {},
   "source": [
    "--------------------------"
   ]
  },
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   "id": "92b7b799",
   "metadata": {},
   "source": [
    "<b> Question 4 </b>\n",
    "\n",
    "Prove proposition 5.13 <br>\n",
    "\n",
    "Let $T: \\mathbb{R}^n \\to \\mathbb{R}^m$ be a linear transformation given by $T(x) = Ax$. Then range$(T)$ is the column space of $A$."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "607215ea",
   "metadata": {},
   "source": [
    "<b> Solution </b> \n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "6c3486c0",
   "metadata": {},
   "outputs": [],
   "source": []
  }
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