Solutions revised

Author: bvanderlei

Vector_Spaces_Solutions.ipynb

@@ -1940,6 +1940,7 @@
     "#### Solution:\n",
     "\n",
     "The matrix $P$ which contains $W_1$, $W_2$ as its columns is as follows:\n",
+    "\n",
     "$$\n",
     "\\begin{equation}\n",
     "P = \\left[ \\begin{array}{rrrr} 1 & 2  \\\\ 2 & 3  \\\\  0 & 0 \\\\ 0 & 0 \\end{array}\\right] \n",
@@ -2002,6 +2003,7 @@
     "$$\n",
     "\n",
     "The row reduced form of $S$ looks like:\n",
+    "\n",
     "$$\n",
     "\\begin{equation}\n",
     "\\left[ \\begin{array}{rrrr} 1 & 0 & 0  \\\\ 0 & a - 4 & 0  \\\\ 0 & 0 & a - 1  \\end{array}\\right] \n",
@@ -2384,14 +2386,11 @@
   },
   {
    "cell_type": "markdown",
-   "id": "b3024d1b",
+   "id": "4e7a71ab",
    "metadata": {},
    "source": [
     "#### Solution:\n",
-    "Any set of four polynomials in $\\mathbb{P}_4$ would generate a $5 \\times 4$ coefficient matrix. For the polynomials to span $\\mathbb{P}_4$, there should be a pivot in each row of the coefficient matrix. Since the matrix is $5 \\times 4$, there can be atmost 4 pivots as there are only four columns. This means that one of the rows will always be without a pivot. \n",
-    "Since there will always be atleast one row without a pivot, any set of four polynomials in $\\mathbb{P}_4$ cannot span $\\mathbb{P}_4$.\n",
     "\n",
-    "This is demonstrated by the following example:\n",
     "Let us consider any four polynomials in $\\mathbb{P}_4$.\n",
     "\n",
     "$$\n",
@@ -2403,19 +2402,43 @@
     "\\end{eqnarray*}\n",
     "$$\n",
     "\n",
-    "The corresponding coefficient matrix $A$ is as follows:\n",
+    "If the four polynomials span $\\mathbb{P}_4$, then we should be able to express every polynomial in $\\mathbb{P}_4$ as a linear combination of $p_1$, $p_2$, $p_3$ and $p_4$.  An arbitrary polynomial in $\\mathbb{P}_4$ has the form $ax^4 + bx^3 + cx^2 + dx  + e $. We are therefore concerned with solving the following equation.\n",
     "\n",
     "$$\n",
     "\\begin{equation}\n",
-    "A = \\left[ \\begin{array}{rrr} 1 & 1 & 2 & 3 \\\\ 1 & 0 & 1 & 1 \\\\ 1 & 2 & 1 & 2 \\\\ 1 & 1 & 1 & 1 \\\\2 & 1 & 2 & 2 \\end{array}\\right]\n",
+    "ax^4 + bx^3 + cx^2 + dx  + e  = c_1(x^4 + x^3 + x^2 + x + 2 ) + c_2(x^4 + 2x^2 + x + 1) + c_3(2x^4 + x^3 + x^2 + x + 2) + c_4(3x^4 + x^3 + 2x^2 + x + 2)\n",
     "\\end{equation}\n",
-    "$$"
+    "$$\n",
+    "\n",
+    "This equation is true for all values of $x$ only if the left and right side are the same polynomial.  Gathering the like terms on the right side gives a system of equations.\n",
+    "\n",
+    "$$\n",
+    "\\begin{eqnarray*}\n",
+    "c_1 + c_2 + 2c_3 + 3c_4 & = & a \\\\\n",
+    "c_1 +\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c_3 \\,\\,+ \\,\\,\\,c_4 & = & b \\\\\n",
+    "c_1 + 2c_2 + c_3 + 2c_4 & = & c  \\\\\n",
+    "c_1 + \\,c_2 \\,+\\, c_3\\, +\\, c_4 & = & d \\\\\n",
+    "2c_1 + c_2 + 2c_3 + 2c_4 & = & e \\\\\n",
+    "\\end{eqnarray*}\n",
+    "$$\n",
+    "\n",
+    "We write the system as a matrix equation and observe that the coefficient matrix $A$ is $5\\times 4$.\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "AX = \\left[ \\begin{array}{rrr} 1 & 1 & 2 & 3 \\\\ 1 & 0 & 1 & 1 \\\\ 1 & 2 & 1 & 2 \\\\ 1 & 1 & 1 & 1 \\\\2 & 1 & 2 &2\\end{array}\\right]\n",
+    "\\left[ \\begin{array}{r} c_1 \\\\ c_2 \\\\ c_3 \\\\c_4\\end{array}\\right]=\n",
+    "\\left[ \\begin{array}{r} a \\\\ b \\\\ c \\\\d\\\\e \\end{array}\\right]= B\n",
+    "\\end{equation}\n",
+    "$$\n",
+    "\n",
+    "The system is consistent for any vector $B$ if there is a pivot in each row of $A$."
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 11,
-   "id": "f779ac11",
+   "execution_count": 6,
+   "id": "9824ac29",
    "metadata": {},
    "outputs": [
     {
@@ -2450,10 +2473,10 @@
   },
   {
    "cell_type": "markdown",
-   "id": "e6d3adbc",
+   "id": "73872fae",
    "metadata": {},
    "source": [
-    "We can see from the computaions done in the above code cell that there is no pivot in the fifth row. Therefore, any set of four polynomials cannot span $\\mathbb{P}_4$."
+    "The computations show that there is no pivot in the fifth row, which implies that the set $\\{ p_1, p_2, p_3, p_4 \\}$ does not span $\\mathbb{P}_4$.  In general, a $5\\times 4$ matrix can have at most $4$ pivots and can thus never have a pivot in each row.  We can therefore conclude that *any* set of four polynomials does not span $\\mathbb{P}_4$."
    ]
   },
   {
@@ -2626,6 +2649,80 @@
     "($c$)  Determine the dimension of $\\mathbb{D}_{2\\times 2}$."
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "9fdb0982",
+   "metadata": {},
+   "source": [
+    "#### Solution:\n",
+    "\n",
+    "$(a)$ \n",
+    "\n",
+    "Let us consider two arbitrary matrices $X$ and $Y$ in $\\mathbb{D}_{2\\times 2}$:\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "A = \\left[ \\begin{array}{ccc} a & 0  \\\\ 0 & b  \\end{array}\\right] \\hspace{1cm}\n",
+    "B = \\left[ \\begin{array}{ccc} c & 0 \\\\ 0 & d  \\end{array}\\right] \\hspace{1cm}\n",
+    "\\end{equation}\n",
+    "$$\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "A + B = \\left[ \\begin{array}{ccc} a + c & 0  \\\\ 0 & b + d \\end{array}\\right] \n",
+    "\\end{equation}\n",
+    "$$ \n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "kA  = \\left[ \\begin{array}{ccc} ka & 0  \\\\ 0 & kb  \\end{array}\\right] \n",
+    "\\end{equation}\n",
+    "$$ \n",
+    "\n",
+    "For any two arbitrary matrices $X$ and $Y$ in $\\mathbb{D}_{2\\times 2}$, $X+Y$ and $kA$ are also in $\\mathbb{D}_{2\\times 2}$. Therefore, $\\mathbb{D}_{2\\times 2}$ is a subspace of $\\mathbb{M}_{2\\times 2}$.\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "$(b)$ \n",
+    "\n",
+    "Any matrix in $\\mathbb{D}_{2\\times 2}$ has the form:\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "\\left[ \\begin{array}{ccc} a & 0  \\\\ 0 & b  \\end{array}\\right] \\hspace{1cm}\n",
+    "\\end{equation}\n",
+    "$$ where $a$ and $b$ are some scalars.\n",
+    "\n",
+    "This matrix can also be written as:\n",
+    "\n",
+    "$$\n",
+    "\\begin{equation}\n",
+    "\\left[ \\begin{array}{ccc}a & 0  \\\\0 & b  \\end{array}\\right] =\n",
+    "a\\left[ \\begin{array}{ccc} 1& 0\\\\ 0 &0  \\end{array}\\right] +\n",
+    "b\\left[ \\begin{array}{ccc} 0 & 0 \\\\ 0 & 1 \\end{array}\\right]  \n",
+    "\\end{equation}\n",
+    "$$\n",
+    "\n",
+    "Let $$\n",
+    "\\begin{equation}\n",
+    "P = \\left[ \\begin{array}{ccc} 1 & 0  \\\\ 0 & 0  \\end{array}\\right] \\hspace{1cm}\n",
+    "Q = \\left[ \\begin{array}{ccc} 0 & 0  \\\\ 0 & 1 \\end{array}\\right] \\hspace{1cm}\n",
+    "\\end{equation}\n",
+    "$$\n",
+    "\n",
+    "Since every matrix in $\\mathbb{D}_{2\\times 2}$ can be written as a unique linear combination of $P$ and $Q$, $\\{P,Q\\}$ is a basis for $\\mathbb{D}_{2\\times 2}$.\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n",
+    "$(c)$ \n",
+    "\n",
+    "Since there are two matrices in the basis for $\\mathbb{D}_{2\\times 2}$, the dimension of $\\mathbb{D}_{2\\times 2}$ is 2. "
+   ]
+  },
   {
    "cell_type": "markdown",
    "id": "c48349a9",