Mixing
Positive definite quadratic form over positive integer vectors
$begingroup$
Let $mathcal{M}$ denote the set of $d times d$ positive definite matrices with real entries for some positive integer $d$. Now consider the set
$$
mathcal{N}={A in M_n(mathbb{R}): z^TAz > 0, z in (mathbb{N} cup{0})^d - {mathbb{0}}}.
$$
Certainly we have $mathcal{M} subset mathcal{N}$, however these sets are not equivalent - for example, when $d=2$, the quadratic form generating $x^2 + y^2 + 3xy in mathcal{N}$ but not in $mathcal{M}$. What additional conditions need to be imposed on quadratic forms to guarantee that they're in the set $mathcal{N}$?
linear-algebra matrices quadratic-forms
asked Jan 24 at 17:16
ChrisChris
341112
$endgroup$
add a comment |
$begingroup$
Let $mathcal{M}$ denote the set of $d times d$ positive definite matrices with real entries for some positive integer $d$. Now consider the set
$$
mathcal{N}={A in M_n(mathbb{R}): z^TAz > 0, z in (mathbb{N} cup{0})^d - {mathbb{0}}}.
$$
Certainly we have $mathcal{M} subset mathcal{N}$, however these sets are not equivalent - for example, when $d=2$, the quadratic form generating $x^2 + y^2 + 3xy in mathcal{N}$ but not in $mathcal{M}$. What additional conditions need to be imposed on quadratic forms to guarantee that they're in the set $mathcal{N}$?
linear-algebra matrices quadratic-forms
asked Jan 24 at 17:16
ChrisChris
341112
$endgroup$
add a comment |
$begingroup$
Let $mathcal{M}$ denote the set of $d times d$ positive definite matrices with real entries for some positive integer $d$. Now consider the set
$$
mathcal{N}={A in M_n(mathbb{R}): z^TAz > 0, z in (mathbb{N} cup{0})^d - {mathbb{0}}}.
$$
Certainly we have $mathcal{M} subset mathcal{N}$, however these sets are not equivalent - for example, when $d=2$, the quadratic form generating $x^2 + y^2 + 3xy in mathcal{N}$ but not in $mathcal{M}$. What additional conditions need to be imposed on quadratic forms to guarantee that they're in the set $mathcal{N}$?
linear-algebra matrices quadratic-forms
asked Jan 24 at 17:16
ChrisChris
341112
$endgroup$
Let $mathcal{M}$ denote the set of $d times d$ positive definite matrices with real entries for some positive integer $d$. Now consider the set
$$
mathcal{N}={A in M_n(mathbb{R}): z^TAz > 0, z in (mathbb{N} cup{0})^d - {mathbb{0}}}.
$$
Certainly we have $mathcal{M} subset mathcal{N}$, however these sets are not equivalent - for example, when $d=2$, the quadratic form generating $x^2 + y^2 + 3xy in mathcal{N}$ but not in $mathcal{M}$. What additional conditions need to be imposed on quadratic forms to guarantee that they're in the set $mathcal{N}$?
linear-algebra matrices quadratic-forms
linear-algebra matrices quadratic-forms
asked Jan 24 at 17:16
ChrisChris
341112
asked Jan 24 at 17:16
ChrisChris
341112
asked Jan 24 at 17:16
ChrisChris
341112
asked Jan 24 at 17:16
ChrisChris
341112
asked Jan 24 at 17:16
ChrisChris
341112
341112
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Of course, $mathcal{N}$ contains all matrices with $>0$ entries.
On the other hand, changing $A$ into $A+A^T$, the problem reduces to symmetric matrices $A$.
Consider the case $n=2$ (in fact, it was your business) and $A=begin{pmatrix}a&b\b&cend{pmatrix}$. The condition is $ax^2+cy^2+2bxy>0$ when $(x,y)in mathbb{N}^2setminus {0,0}$; necessarily $a>0,c>0$; moreover, $at^2+2bt+c>0$ for every $tinmathbb{Q}^+$.
That is, $at^2+2bt+c>0$ for every $tgeq 0$ OR
$(*)$ $at^2+2bt+c$ admits a double root in $t_0>0$ a non-rational number.
That implies that, in general, the roots of our polynomial are $<0$ or non-real.
In the first case, the entries of $A$ are $>0$ and, in the second one, $A$ is symmetric $>0$.
An example of the exceptional case $(*)$: $A=begin{pmatrix}1&-pi\-pi&pi^2end{pmatrix}$. Note that $A$ is only symmetric $geq 0$.
Now, have a look at a generalization, for $n>2$, of the previous result.
answered Jan 25 at 12:30
loup blancloup blanc
23.7k21851
$endgroup$
$begingroup$
This is a very clever way of approaching the problem - however, I am having troubles reducing a general quadratic form to a single variable polynomial for n>2.
$endgroup$
– Chris
Jan 25 at 17:22
$begingroup$
Actually, never mind - I think I know how to solve the problem now. Thanks!
$endgroup$
– Chris
Jan 25 at 17:31
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
Of course, $mathcal{N}$ contains all matrices with $>0$ entries.
On the other hand, changing $A$ into $A+A^T$, the problem reduces to symmetric matrices $A$.
Consider the case $n=2$ (in fact, it was your business) and $A=begin{pmatrix}a&b\b&cend{pmatrix}$. The condition is $ax^2+cy^2+2bxy>0$ when $(x,y)in mathbb{N}^2setminus {0,0}$; necessarily $a>0,c>0$; moreover, $at^2+2bt+c>0$ for every $tinmathbb{Q}^+$.
That is, $at^2+2bt+c>0$ for every $tgeq 0$ OR
$(*)$ $at^2+2bt+c$ admits a double root in $t_0>0$ a non-rational number.
That implies that, in general, the roots of our polynomial are $<0$ or non-real.
In the first case, the entries of $A$ are $>0$ and, in the second one, $A$ is symmetric $>0$.
An example of the exceptional case $(*)$: $A=begin{pmatrix}1&-pi\-pi&pi^2end{pmatrix}$. Note that $A$ is only symmetric $geq 0$.
Now, have a look at a generalization, for $n>2$, of the previous result.
answered Jan 25 at 12:30
loup blancloup blanc
23.7k21851
$endgroup$
$begingroup$
This is a very clever way of approaching the problem - however, I am having troubles reducing a general quadratic form to a single variable polynomial for n>2.
$endgroup$
– Chris
Jan 25 at 17:22
$begingroup$
Actually, never mind - I think I know how to solve the problem now. Thanks!
$endgroup$
– Chris
Jan 25 at 17:31
add a comment |
$begingroup$
Of course, $mathcal{N}$ contains all matrices with $>0$ entries.
On the other hand, changing $A$ into $A+A^T$, the problem reduces to symmetric matrices $A$.
Consider the case $n=2$ (in fact, it was your business) and $A=begin{pmatrix}a&b\b&cend{pmatrix}$. The condition is $ax^2+cy^2+2bxy>0$ when $(x,y)in mathbb{N}^2setminus {0,0}$; necessarily $a>0,c>0$; moreover, $at^2+2bt+c>0$ for every $tinmathbb{Q}^+$.
That is, $at^2+2bt+c>0$ for every $tgeq 0$ OR
$(*)$ $at^2+2bt+c$ admits a double root in $t_0>0$ a non-rational number.
That implies that, in general, the roots of our polynomial are $<0$ or non-real.
In the first case, the entries of $A$ are $>0$ and, in the second one, $A$ is symmetric $>0$.
An example of the exceptional case $(*)$: $A=begin{pmatrix}1&-pi\-pi&pi^2end{pmatrix}$. Note that $A$ is only symmetric $geq 0$.
Now, have a look at a generalization, for $n>2$, of the previous result.
answered Jan 25 at 12:30
loup blancloup blanc
23.7k21851
$endgroup$
$begingroup$
This is a very clever way of approaching the problem - however, I am having troubles reducing a general quadratic form to a single variable polynomial for n>2.
$endgroup$
– Chris
Jan 25 at 17:22
$begingroup$
Actually, never mind - I think I know how to solve the problem now. Thanks!
$endgroup$
– Chris
Jan 25 at 17:31
add a comment |
$begingroup$
Of course, $mathcal{N}$ contains all matrices with $>0$ entries.
On the other hand, changing $A$ into $A+A^T$, the problem reduces to symmetric matrices $A$.
Consider the case $n=2$ (in fact, it was your business) and $A=begin{pmatrix}a&b\b&cend{pmatrix}$. The condition is $ax^2+cy^2+2bxy>0$ when $(x,y)in mathbb{N}^2setminus {0,0}$; necessarily $a>0,c>0$; moreover, $at^2+2bt+c>0$ for every $tinmathbb{Q}^+$.
That is, $at^2+2bt+c>0$ for every $tgeq 0$ OR
$(*)$ $at^2+2bt+c$ admits a double root in $t_0>0$ a non-rational number.
That implies that, in general, the roots of our polynomial are $<0$ or non-real.
In the first case, the entries of $A$ are $>0$ and, in the second one, $A$ is symmetric $>0$.
An example of the exceptional case $(*)$: $A=begin{pmatrix}1&-pi\-pi&pi^2end{pmatrix}$. Note that $A$ is only symmetric $geq 0$.
Now, have a look at a generalization, for $n>2$, of the previous result.
answered Jan 25 at 12:30
loup blancloup blanc
23.7k21851
$endgroup$
Of course, $mathcal{N}$ contains all matrices with $>0$ entries.
On the other hand, changing $A$ into $A+A^T$, the problem reduces to symmetric matrices $A$.
Consider the case $n=2$ (in fact, it was your business) and $A=begin{pmatrix}a&b\b&cend{pmatrix}$. The condition is $ax^2+cy^2+2bxy>0$ when $(x,y)in mathbb{N}^2setminus {0,0}$; necessarily $a>0,c>0$; moreover, $at^2+2bt+c>0$ for every $tinmathbb{Q}^+$.
That is, $at^2+2bt+c>0$ for every $tgeq 0$ OR
$(*)$ $at^2+2bt+c$ admits a double root in $t_0>0$ a non-rational number.
That implies that, in general, the roots of our polynomial are $<0$ or non-real.
In the first case, the entries of $A$ are $>0$ and, in the second one, $A$ is symmetric $>0$.
An example of the exceptional case $(*)$: $A=begin{pmatrix}1&-pi\-pi&pi^2end{pmatrix}$. Note that $A$ is only symmetric $geq 0$.
Now, have a look at a generalization, for $n>2$, of the previous result.
answered Jan 25 at 12:30
loup blancloup blanc
23.7k21851
answered Jan 25 at 12:30
loup blancloup blanc
23.7k21851
answered Jan 25 at 12:30
loup blancloup blanc
23.7k21851
answered Jan 25 at 12:30
loup blancloup blanc
23.7k21851
23.7k21851
$begingroup$
This is a very clever way of approaching the problem - however, I am having troubles reducing a general quadratic form to a single variable polynomial for n>2.
$endgroup$
– Chris
Jan 25 at 17:22
$begingroup$
Actually, never mind - I think I know how to solve the problem now. Thanks!
$endgroup$
– Chris
Jan 25 at 17:31
add a comment |
$begingroup$
This is a very clever way of approaching the problem - however, I am having troubles reducing a general quadratic form to a single variable polynomial for n>2.
$endgroup$
– Chris
Jan 25 at 17:22
$begingroup$
Actually, never mind - I think I know how to solve the problem now. Thanks!
$endgroup$
– Chris
Jan 25 at 17:31
$begingroup$
This is a very clever way of approaching the problem - however, I am having troubles reducing a general quadratic form to a single variable polynomial for n>2.
$endgroup$
– Chris
Jan 25 at 17:22
$begingroup$
This is a very clever way of approaching the problem - however, I am having troubles reducing a general quadratic form to a single variable polynomial for n>2.
$endgroup$
– Chris
Jan 25 at 17:22
$begingroup$
Actually, never mind - I think I know how to solve the problem now. Thanks!
$endgroup$
– Chris
Jan 25 at 17:31
$begingroup$
Actually, never mind - I think I know how to solve the problem now. Thanks!
$endgroup$
– Chris
Jan 25 at 17:31
add a comment |
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