Solutions updated

Author: bvanderlei

Inner_Products_Solutions.ipynb

@@ -1846,6 +1846,20 @@
     "**Exercise 7:** If $A$ is an $n \\times n$ invertible matrix, explain why $A^TA$ is also invertible. "
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "e9d5b7d6",
+   "metadata": {},
+   "source": [
+    "**Solution:**\n",
+    "\n",
+    "If we assume that the matrix $A$ is invertible, then it has a pivot in each column.  This means that the columns of $A$ are linearly independent and that $A$ has a $QR$ factorization.\n",
+    "\n",
+    "So, $ A^TA = (QR)^T(QR)$ = $ R^TQ^TQR = R^T(Q^TQ)R = R^TR $.\n",
+    "\n",
+    "Since $R$ is an upper-triangular matrix with non-zero enteries on the diagonal we can reduce it to the identity matrix $I$ by performing row operations.  This means that $R$ is invertible. The matrix $R^T$ is lower triangular with non-zero entries on the diagonal and is invertible for the same reason.  Since $A^TA$ is the product of two invertible matrices, it must also be invertible.  (*See exercises in [Inverse Matrices](Inverse_Matrices.ipynb).)*"
+   ]
+  },
   {
    "cell_type": "markdown",
    "id": "71cfd633",
@@ -3599,42 +3613,65 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 12,
-   "id": "a9b1fa21",
+   "execution_count": 14,
+   "id": "a547e7e3",
    "metadata": {},
    "outputs": [
     {
      "name": "stdout",
      "output_type": "stream",
      "text": [
-      "Pivot could not be found in column 2 .\n",
-      "Zero entry found in U pivot position 2 .\n",
-      "X_cap: \n",
-      " [[0.16666667]\n",
-      " [0.66666667]\n",
-      " [0.        ]] \n",
+      "A_aug_red: \n",
+      " [[1.         0.         2.         0.16666667]\n",
+      " [0.         1.         0.         0.66666667]\n",
+      " [0.         0.         0.         0.        ]] \n",
       "\n"
      ]
     }
    ],
    "source": [
-    "X_cap = lag.SolveSystem(A, B_cap)\n",
+    "A_aug = np.array([[1,2,2,1.5],[2,1,4,1],[1,2,2,1.5]])\n",
+    "A_aug_red = lag.FullRowReduction(A_aug)\n",
     "\n",
-    "print(\"X_cap: \\n\", X_cap, '\\n')"
+    "print(\"A_aug_red: \\n\", A_aug_red, '\\n')"
    ]
   },
   {
    "cell_type": "markdown",
-   "id": "de71d3e3",
+   "id": "b9155d0c",
    "metadata": {},
    "source": [
-    "There are an infinite number of least square solutions for the given system. One such solution is \n",
+    "Let\n",
     "\n",
     "$$\n",
     "\\begin{equation}\n",
-    "\\hat{X} = \\left[\\begin{array}{r} 0.16666667 \\\\0.66666667 \\\\ 0\\end{array}\\right]  \n",
+    "\\hat{X} = \\left[\\begin{array}{r} a \\\\b \\\\ c\\end{array}\\right]  \n",
     "\\end{equation}\n",
-    "$$\n"
+    "$$\n",
+    "\n",
+    "\n",
+    "We can see that $c$ is a free variable. So, let $c=t$ where $t$ is a scalar. Then, $b = 0.67$, $a = 0.167 - 2t$.\n",
+    "\n",
+    "\n",
+    "So, \n",
+    "\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "\\hat{X} = \\left[\\begin{array}{r} 0.167-2t \\\\ 0.67 \\\\ t\\end{array}\\right]    \n",
+    "= \\left[\\begin{array}{r} 0.167 \\\\ 0.67 \\\\ 0\\end{array}\\right]  +t\\left[\\begin{array}{r} -2 \\\\ 0 \\\\ 1\\end{array}\\right]\n",
+    "\\end{equation}\n",
+    "$$\n",
+    "\n",
+    "Hence, we can conclude that there are infintely many solutions for the system $ A\\hat{X} = \\hat{B}$. One possible solution is:\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "\\hat{X} = \\left[\\begin{array}{r} 0.167 \\\\ 0.67 \\\\ 0\\end{array}\\right]    \n",
+    "\\end{equation}\n",
+    "$$\n",
+    "\n",
+    "when $t=0$."
    ]
   },
   {