Exercises and solutions

Author: bvanderlei

Inner_Products.ipynb

@@ -46,7 +46,7 @@
     "\\end{equation}\n",
     "$$\n",
     "\n",
-    "We see that $(AB)_{23} = U\\cdot V$, the dot product of the second row of $A$ with the third row of $B$.  In general the entry $(AB)_{ij}$ will be the dot product of the $i$th row of $A$ with the $j$th column of $B$.  \n",
+    "We see that $(AB)_{23} = U\\cdot V$, the dot product of the second row of $A$ with the third column of $B$.  In general the entry $(AB)_{ij}$ will be the dot product of the $i$th row of $A$ with the $j$th column of $B$.  \n",
     "\n",
     "Taking advantage of this connection, we could also write the definition of the dot product by viewing the vectors as $n\\times 1$ matrices, and making use of matrix transposes.\n",
     "\n",
@@ -529,6 +529,106 @@
    "source": [
     "## Code solution here."
    ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**Exercise 4:** Create a Python function named $\\texttt{DifferenceMagnitude}$ which takes two $n \\times 1$ vectors as arguments and returns the magnitude of their difference. Use this function to determine $|| X - Y||$ where vectors $X$ and $Y$ are given below.\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "X = \\left[ \\begin{array}{r} 3 \\\\ -1 \\\\ 2  \\end{array}\\right] \\hspace{1cm} \n",
+    "Y = \\left[ \\begin{array}{r} 4 \\\\ 2 \\\\ 0  \\end{array}\\right] \n",
+    "\\end{equation}\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 36,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## Code solution here"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**Exercise 5:** Create a Python function which takes an $n \\times 1$ vector as its argument and returns the unit vector of the argument vector. Use the NumPy $\\texttt{random}$ module to generate a random $3 \\times 1 $ vector, and test this function on that vector."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 37,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## Code solution here"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**Exercise 6:** Find a vector $Y$ in $\\mathbb{R}^2$ such that $X\\cdot Y = ||Y - X||$ where the vector $X$ is given below. Is the vector $Y$ unique ? Verify your answer through a computation.\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "X = \\left[ \\begin{array}{r} 1 \\\\ 1  \\end{array}\\right] \n",
+    "\\end{equation}\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 38,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## Code solution here"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**Exercise 7:** Create a Python function named $\\texttt{Angle}$ which takes two vectors as arguments and returns $\\cos{\\theta}$ where ${\\theta}$ is the angle between the two vectors. Use this function to show that vectors $U$ and $W$ are orthogonal.\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "U = \\left[ \\begin{array}{r} 1 \\\\ -1 \\\\ 2  \\end{array}\\right] \\hspace{1cm} \n",
+    "W = \\left[ \\begin{array}{r} 2 \\\\ 0 \\\\ -1  \\end{array}\\right] \n",
+    "\\end{equation}\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 39,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## Code solution here"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "**Exercise 8:** Given that $||X+Y|| = 3$, $X\\cdot Y = 2$, find $||X-Y||$."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "## Code solution here"
+   ]
   }
  ],
  "metadata": {