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hw2 complete #26

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6452c70
hw2 complete
burak0206 Feb 25, 2017
3004c67
fixed
burak0206 Feb 25, 2017
7c86350
done
burak0206 Feb 25, 2017
7b473df
errors are fixed
burak0206 Feb 25, 2017
18c45eb
fixing go on
burak0206 Feb 25, 2017
cc8c353
continue to fix
burak0206 Feb 25, 2017
4d0181c
merged
burak0206 Feb 25, 2017
3764de1
parse error
burak0206 Feb 26, 2017
19cfc54
sqrt script is cleared
burak0206 Feb 26, 2017
ab1080b
sort function can not be rendered
burak0206 Feb 26, 2017
f1d186c
try to solve render problem
burak0206 Feb 26, 2017
79d2074
attemp to problem
burak0206 Feb 26, 2017
d245a75
split a and b
burak0206 Feb 26, 2017
1d2de13
render problems are fixed
burak0206 Feb 26, 2017
ddf4524
render problem can not be solved
burak0206 Feb 26, 2017
8c25ad5
\sum\limit is changed by \sum
burak0206 Feb 26, 2017
477850f
all \sum_\limits are changed by \sum function
burak0206 Feb 26, 2017
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170 changes: 170 additions & 0 deletions .ipynb_checkpoints/hw2-checkpoint.ipynb
View file
Open in desktop
Original file line number Diff line number Diff line change
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Homework 2\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exercise 3.3"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### (a) $\\|x\\|_\\infty \\leq \\|x\\|_{2}$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ \\|x\\|_\\infty = \\max\\limits_{i} |x_{i}| = \\sqrt{(\\max\\limits_{i} |x_{i}|)^2} \\leq \\sqrt{ \\sum_{i=1}^{m} |x_{i}|^2} = \\|x\\|_{2} $"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ x_{m*1} = [2,1,\\dots,0]^T \\hspace{5mm}, then \\hspace{3mm} \\|x\\|_\\infty = 2 \\hspace{3mm} and \\hspace{3mm} \\|x\\|_{2} = \\sqrt{5}$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### (b) $\\|x\\|_{2} \\leq \\sqrt{m}\\|x\\|_\\infty$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ \\|x\\|_{2} = \\sqrt {\\sum_{i=1}^{m} |x_{i}|^2} \\leq \\sqrt{m.(\\max\\limits_{i} |x_{i}|)^2} = \\sqrt{m}.\\|x\\|_\\infty $"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ x_{m*1} = [1,1,1,1,0]^T \\hspace{5mm}, then \\hspace{3mm} m = 5 \\hspace{3mm} \\|x\\|_\\infty = 1 \\hspace{3mm} and \\hspace{3mm} \\|x\\|_{2} = \\sqrt{4}=2$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### (c) $\\|A\\|_\\infty \\leq \\sqrt{n}\\|A\\|_{2}$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ \\|A\\|_{\\infty}= \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|_{\\infty}} {\\|x\\|_{\\infty}} \\leq \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|_{2}} {\\|x\\|_{2}} \\leq \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|_{2}} {\\|x\\|_{2}/ \\sqrt{n}} = \\sqrt{n}\\|A\\|_{2}$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ A_{mxn} = \\begin{bmatrix}\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 0 & 0 & 0 & 0 \\\\\\\\\n",
" 0 & 0 & 0 & 0\n",
" \\end{bmatrix}_{mxn} A ^T = \\begin{bmatrix}\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" \\end{bmatrix}_{nxm} A ^T A = \\begin{bmatrix}\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 1 & 1 & 1 & 1 \n",
" \\end{bmatrix}_{nxn}$\n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$\\|A\\|_\\infty = 4 = n$\n",
"\n",
"$\\|A\\|_{2} =\\sqrt{tr(A^TA)}= \\sqrt{4} = \\sqrt{n} = 2$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### (d) $\\|A\\|_{2} \\leq \\sqrt{m}\\|A\\|_\\infty$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ \\|A\\|_{{2}}= \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|_{{2}}} {\\|x\\|_{{2}}} \\leq \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|\\infty} {\\|x\\|\\infty} \\leq \\sup\\limits_{x \\neq 0}\\frac{\\sqrt{m} \\|Ax\\|\\infty} {\\|x\\|\\infty} = \\sqrt{m}\\|A\\|\\infty $"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ A_{mxn} = \\begin{bmatrix}\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \n",
" \\end{bmatrix}_{mxn} A ^T = \\begin{bmatrix}\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 0 & 0 & 0 & 0 \\\\\\\\\n",
" 0 & 0 & 0 & 0\n",
" \\end{bmatrix}_{nxm} A ^T A = \\begin{bmatrix}\n",
" 4 & 0 & 0 \\\\\\\\\n",
" 0 & 0 & 0 \\\\\\\\\n",
" 0 & 0 & 0 \n",
" \\end{bmatrix}_{nxn}$\n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$\\|A\\|_\\infty = 1$\n",
"\n",
"$\\|A\\|_{2} =\\sqrt{tr(A^TA)}= \\sqrt{4} = 2$"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.13"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
170 changes: 170 additions & 0 deletions hw2.ipynb
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,170 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Homework 2\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exercise 3.3"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### (a) $\\|x\\|_\\infty \\leq \\|x\\|_{2}$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ \\|x\\|_\\infty = \\max\\limits_{i} |x_{i}| = \\sqrt{(\\max\\limits_{i} |x_{i}|)^2} \\leq \\sqrt{ \\sum_{i=1}^{m} |x_{i}|^2} = \\|x\\|_{2} $"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ x_{m*1} = [2,1,\\dots,0]^T \\hspace{5mm}, then \\hspace{3mm} \\|x\\|_\\infty = 2 \\hspace{3mm} and \\hspace{3mm} \\|x\\|_{2} = \\sqrt{5}$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### (b) $\\|x\\|_{2} \\leq \\sqrt{m}\\|x\\|_\\infty$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ \\|x\\|_{2} = \\sqrt {\\sum_{i=1}^{m} |x_{i}|^2} \\leq \\sqrt{m.(\\max\\limits_{i} |x_{i}|)^2} = \\sqrt{m}.\\|x\\|_\\infty $"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ x_{m*1} = [1,1,1,1,0]^T \\hspace{5mm}, then \\hspace{3mm} m = 5 \\hspace{3mm} \\|x\\|_\\infty = 1 \\hspace{3mm} and \\hspace{3mm} \\|x\\|_{2} = \\sqrt{4}=2$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### (c) $\\|A\\|_\\infty \\leq \\sqrt{n}\\|A\\|_{2}$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ \\|A\\|_{\\infty}= \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|_{\\infty}} {\\|x\\|_{\\infty}} \\leq \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|_{2}} {\\|x\\|_{2}} \\leq \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|_{2}} {\\|x\\|_{2}/ \\sqrt{n}} = \\sqrt{n}\\|A\\|_{2}$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ A_{mxn} = \\begin{bmatrix}\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 0 & 0 & 0 & 0 \\\\\\\\\n",
" 0 & 0 & 0 & 0\n",
" \\end{bmatrix}_{mxn} A ^T = \\begin{bmatrix}\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" \\end{bmatrix}_{nxm} A ^T A = \\begin{bmatrix}\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 1 & 1 & 1 & 1 \n",
" \\end{bmatrix}_{nxn}$\n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$\\|A\\|_\\infty = 4 = n$\n",
"\n",
"$\\|A\\|_{2} =\\sqrt{tr(A^TA)}= \\sqrt{4} = \\sqrt{n} = 2$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### (d) $\\|A\\|_{2} \\leq \\sqrt{m}\\|A\\|_\\infty$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ \\|A\\|_{{2}}= \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|_{{2}}} {\\|x\\|_{{2}}} \\leq \\sup\\limits_{x \\neq 0}\\frac{ \\|Ax\\|\\infty} {\\|x\\|\\infty} \\leq \\sup\\limits_{x \\neq 0}\\frac{\\sqrt{m} \\|Ax\\|\\infty} {\\|x\\|\\infty} = \\sqrt{m}\\|A\\|\\infty $"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$ A_{mxn} = \\begin{bmatrix}\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \\\\\\\\\n",
" 1 & 0 & 0 \n",
" \\end{bmatrix}_{mxn} A ^T = \\begin{bmatrix}\n",
" 1 & 1 & 1 & 1 \\\\\\\\\n",
" 0 & 0 & 0 & 0 \\\\\\\\\n",
" 0 & 0 & 0 & 0\n",
" \\end{bmatrix}_{nxm} A ^T A = \\begin{bmatrix}\n",
" 4 & 0 & 0 \\\\\\\\\n",
" 0 & 0 & 0 \\\\\\\\\n",
" 0 & 0 & 0 \n",
" \\end{bmatrix}_{nxn}$\n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$\\|A\\|_\\infty = 1$\n",
"\n",
"$\\|A\\|_{2} =\\sqrt{tr(A^TA)}= \\sqrt{4} = 2$"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.13"
}
},
"nbformat": 4,
"nbformat_minor": 2
}