Mixing
Use linear algebra to find an $n$th-degree polynomial with a zero at $x = sqrt{3} + sqrt{5} + sqrt{3 +...
$begingroup$
It was easy enough (though tedious) to remove the radicals in the given zero algebraically. I got (and double-checked with software): $$x^8 - 44x^6 - 160x^5 + 254x^4 + 1600x^3 + 4740x^2 + 8800x - 5975$$But, I'm wondering how I might solve it with matrix methods. I can see a system of equations. Letting $a=sqrt{3}, b=sqrt{5}$, and $c =sqrt{3 + 4sqrt{5}}$, we get $$x-a-b-c = 0$$ $$ a^2 - 3 = 0$$ $$b^2 -5 = 0$$ $$c^2-4b-3 = 0$$But I'm stuck here. How can I set this up (and solve it) as a linear algebra problem?
linear-algebra extension-field
edited Jan 24 at 6:24
Jyrki Lahtonen
110k13171385
asked Jan 24 at 6:16
user636164user636164
234
$endgroup$
add a comment |
$begingroup$
It was easy enough (though tedious) to remove the radicals in the given zero algebraically. I got (and double-checked with software): $$x^8 - 44x^6 - 160x^5 + 254x^4 + 1600x^3 + 4740x^2 + 8800x - 5975$$But, I'm wondering how I might solve it with matrix methods. I can see a system of equations. Letting $a=sqrt{3}, b=sqrt{5}$, and $c =sqrt{3 + 4sqrt{5}}$, we get $$x-a-b-c = 0$$ $$ a^2 - 3 = 0$$ $$b^2 -5 = 0$$ $$c^2-4b-3 = 0$$But I'm stuck here. How can I set this up (and solve it) as a linear algebra problem?
linear-algebra extension-field
edited Jan 24 at 6:24
Jyrki Lahtonen
110k13171385
asked Jan 24 at 6:16
user636164user636164
234
$endgroup$
$begingroup$
Groebner bases?
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:20
2
$begingroup$
Calling this linear algebra feels bizarre to me. May be you are expected to use a natural basis of the field extension $Bbb{Q}(sqrt3,sqrt5,sqrt{3+4sqrt5})/Bbb{Q}$, write the matrix corresponding to multiplication by that number, and then work out the characteristic polynomial of that $8times8$ matrix? Bizarre!
$endgroup$
– Jyrki Lahtonen
Jan 24 at 6:28
add a comment |
$begingroup$
It was easy enough (though tedious) to remove the radicals in the given zero algebraically. I got (and double-checked with software): $$x^8 - 44x^6 - 160x^5 + 254x^4 + 1600x^3 + 4740x^2 + 8800x - 5975$$But, I'm wondering how I might solve it with matrix methods. I can see a system of equations. Letting $a=sqrt{3}, b=sqrt{5}$, and $c =sqrt{3 + 4sqrt{5}}$, we get $$x-a-b-c = 0$$ $$ a^2 - 3 = 0$$ $$b^2 -5 = 0$$ $$c^2-4b-3 = 0$$But I'm stuck here. How can I set this up (and solve it) as a linear algebra problem?
linear-algebra extension-field
edited Jan 24 at 6:24
Jyrki Lahtonen
110k13171385
asked Jan 24 at 6:16
user636164user636164
234
$endgroup$
It was easy enough (though tedious) to remove the radicals in the given zero algebraically. I got (and double-checked with software): $$x^8 - 44x^6 - 160x^5 + 254x^4 + 1600x^3 + 4740x^2 + 8800x - 5975$$But, I'm wondering how I might solve it with matrix methods. I can see a system of equations. Letting $a=sqrt{3}, b=sqrt{5}$, and $c =sqrt{3 + 4sqrt{5}}$, we get $$x-a-b-c = 0$$ $$ a^2 - 3 = 0$$ $$b^2 -5 = 0$$ $$c^2-4b-3 = 0$$But I'm stuck here. How can I set this up (and solve it) as a linear algebra problem?
linear-algebra extension-field
linear-algebra extension-field
edited Jan 24 at 6:24
Jyrki Lahtonen
110k13171385
asked Jan 24 at 6:16
user636164user636164
234
edited Jan 24 at 6:24
Jyrki Lahtonen
110k13171385
asked Jan 24 at 6:16
user636164user636164
234
edited Jan 24 at 6:24
Jyrki Lahtonen
110k13171385
edited Jan 24 at 6:24
Jyrki Lahtonen
110k13171385
edited Jan 24 at 6:24
Jyrki Lahtonen
110k13171385
110k13171385
asked Jan 24 at 6:16
user636164user636164
234
asked Jan 24 at 6:16
user636164user636164
234
asked Jan 24 at 6:16
user636164user636164
234
234
$begingroup$
Groebner bases?
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:20
2
$begingroup$
Calling this linear algebra feels bizarre to me. May be you are expected to use a natural basis of the field extension $Bbb{Q}(sqrt3,sqrt5,sqrt{3+4sqrt5})/Bbb{Q}$, write the matrix corresponding to multiplication by that number, and then work out the characteristic polynomial of that $8times8$ matrix? Bizarre!
$endgroup$
– Jyrki Lahtonen
Jan 24 at 6:28
add a comment |
$begingroup$
Groebner bases?
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:20
2
$begingroup$
Calling this linear algebra feels bizarre to me. May be you are expected to use a natural basis of the field extension $Bbb{Q}(sqrt3,sqrt5,sqrt{3+4sqrt5})/Bbb{Q}$, write the matrix corresponding to multiplication by that number, and then work out the characteristic polynomial of that $8times8$ matrix? Bizarre!
$endgroup$
– Jyrki Lahtonen
Jan 24 at 6:28
$begingroup$
Groebner bases?
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:20
$begingroup$
Groebner bases?
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:20
2
2
$begingroup$
Calling this linear algebra feels bizarre to me. May be you are expected to use a natural basis of the field extension $Bbb{Q}(sqrt3,sqrt5,sqrt{3+4sqrt5})/Bbb{Q}$, write the matrix corresponding to multiplication by that number, and then work out the characteristic polynomial of that $8times8$ matrix? Bizarre!
$endgroup$
– Jyrki Lahtonen
Jan 24 at 6:28
$begingroup$
Calling this linear algebra feels bizarre to me. May be you are expected to use a natural basis of the field extension $Bbb{Q}(sqrt3,sqrt5,sqrt{3+4sqrt5})/Bbb{Q}$, write the matrix corresponding to multiplication by that number, and then work out the characteristic polynomial of that $8times8$ matrix? Bizarre!
$endgroup$
– Jyrki Lahtonen
Jan 24 at 6:28
add a comment |
1 Answer
1
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votes
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Here's a way to solve it using matrix methods, but it's very tedious.
The field $K$ generated over the rationals by $sqrt3$, $sqrt5$, and $sqrt{3+4sqrt5}$ is an 8-dimensional vector space over the rationals, with basis
$$
{1,sqrt3,sqrt5,sqrt{15},sqrt{3+4sqrt5},sqrt{9+12sqrt5},sqrt{15+20sqrt5},sqrt{45+60sqrt5}}
$$
Express each of the nine numbers $1,x,x^2,dots,x^8$ as a linear combination of these basis elements. Then $a_0+a_1x+a_2x^2+cdots+a_7x^7+x^8=0$ leads to eight linear equations in the eight unknowns $a_0,dots,a_7$.
answered Jan 24 at 6:32
Gerry MyersonGerry Myerson
147k8150302
$endgroup$
$begingroup$
Thank you. I started down something like this path, but hesitated, thinking I wouldn't be asked to do something so messy, and was missing some trick...
$endgroup$
– user636164
Jan 24 at 6:39
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's a way to solve it using matrix methods, but it's very tedious.
The field $K$ generated over the rationals by $sqrt3$, $sqrt5$, and $sqrt{3+4sqrt5}$ is an 8-dimensional vector space over the rationals, with basis
$$
{1,sqrt3,sqrt5,sqrt{15},sqrt{3+4sqrt5},sqrt{9+12sqrt5},sqrt{15+20sqrt5},sqrt{45+60sqrt5}}
$$
Express each of the nine numbers $1,x,x^2,dots,x^8$ as a linear combination of these basis elements. Then $a_0+a_1x+a_2x^2+cdots+a_7x^7+x^8=0$ leads to eight linear equations in the eight unknowns $a_0,dots,a_7$.
answered Jan 24 at 6:32
Gerry MyersonGerry Myerson
147k8150302
$endgroup$
$begingroup$
Thank you. I started down something like this path, but hesitated, thinking I wouldn't be asked to do something so messy, and was missing some trick...
$endgroup$
– user636164
Jan 24 at 6:39
add a comment |
$begingroup$
Here's a way to solve it using matrix methods, but it's very tedious.
The field $K$ generated over the rationals by $sqrt3$, $sqrt5$, and $sqrt{3+4sqrt5}$ is an 8-dimensional vector space over the rationals, with basis
$$
{1,sqrt3,sqrt5,sqrt{15},sqrt{3+4sqrt5},sqrt{9+12sqrt5},sqrt{15+20sqrt5},sqrt{45+60sqrt5}}
$$
Express each of the nine numbers $1,x,x^2,dots,x^8$ as a linear combination of these basis elements. Then $a_0+a_1x+a_2x^2+cdots+a_7x^7+x^8=0$ leads to eight linear equations in the eight unknowns $a_0,dots,a_7$.
answered Jan 24 at 6:32
Gerry MyersonGerry Myerson
147k8150302
$endgroup$
$begingroup$
Thank you. I started down something like this path, but hesitated, thinking I wouldn't be asked to do something so messy, and was missing some trick...
$endgroup$
– user636164
Jan 24 at 6:39
add a comment |
$begingroup$
Here's a way to solve it using matrix methods, but it's very tedious.
The field $K$ generated over the rationals by $sqrt3$, $sqrt5$, and $sqrt{3+4sqrt5}$ is an 8-dimensional vector space over the rationals, with basis
$$
{1,sqrt3,sqrt5,sqrt{15},sqrt{3+4sqrt5},sqrt{9+12sqrt5},sqrt{15+20sqrt5},sqrt{45+60sqrt5}}
$$
Express each of the nine numbers $1,x,x^2,dots,x^8$ as a linear combination of these basis elements. Then $a_0+a_1x+a_2x^2+cdots+a_7x^7+x^8=0$ leads to eight linear equations in the eight unknowns $a_0,dots,a_7$.
answered Jan 24 at 6:32
Gerry MyersonGerry Myerson
147k8150302
$endgroup$
Here's a way to solve it using matrix methods, but it's very tedious.
The field $K$ generated over the rationals by $sqrt3$, $sqrt5$, and $sqrt{3+4sqrt5}$ is an 8-dimensional vector space over the rationals, with basis
$$
{1,sqrt3,sqrt5,sqrt{15},sqrt{3+4sqrt5},sqrt{9+12sqrt5},sqrt{15+20sqrt5},sqrt{45+60sqrt5}}
$$
Express each of the nine numbers $1,x,x^2,dots,x^8$ as a linear combination of these basis elements. Then $a_0+a_1x+a_2x^2+cdots+a_7x^7+x^8=0$ leads to eight linear equations in the eight unknowns $a_0,dots,a_7$.
answered Jan 24 at 6:32
Gerry MyersonGerry Myerson
147k8150302
answered Jan 24 at 6:32
Gerry MyersonGerry Myerson
147k8150302
answered Jan 24 at 6:32
Gerry MyersonGerry Myerson
147k8150302
answered Jan 24 at 6:32
Gerry MyersonGerry Myerson
147k8150302
147k8150302
$begingroup$
Thank you. I started down something like this path, but hesitated, thinking I wouldn't be asked to do something so messy, and was missing some trick...
$endgroup$
– user636164
Jan 24 at 6:39
add a comment |
$begingroup$
Thank you. I started down something like this path, but hesitated, thinking I wouldn't be asked to do something so messy, and was missing some trick...
$endgroup$
– user636164
Jan 24 at 6:39
$begingroup$
Thank you. I started down something like this path, but hesitated, thinking I wouldn't be asked to do something so messy, and was missing some trick...
$endgroup$
– user636164
Jan 24 at 6:39
$begingroup$
Thank you. I started down something like this path, but hesitated, thinking I wouldn't be asked to do something so messy, and was missing some trick...
$endgroup$
– user636164
Jan 24 at 6:39
add a comment |
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$begingroup$
Groebner bases?
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:20
2
$begingroup$
Calling this linear algebra feels bizarre to me. May be you are expected to use a natural basis of the field extension $Bbb{Q}(sqrt3,sqrt5,sqrt{3+4sqrt5})/Bbb{Q}$, write the matrix corresponding to multiplication by that number, and then work out the characteristic polynomial of that $8times8$ matrix? Bizarre!
$endgroup$
– Jyrki Lahtonen
Jan 24 at 6:28