Prove that there is a basis for the $n$th degree polynomial vector space consisting $only$ of $n$th degree polynomials.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I tried looking around for an answer to this question, but I couldn't find anything specific enough. I'm thinking I need to use either Cartesian products or subspace sums to produce such a vector space, however I'm still fuzzy on how to actually implement it.
Any help would be great!
linear-algebra polynomials vector-spaces
asked Sep 10 at 18:53
Silverwing171
82
add a comment |Â
up vote
1
down vote
favorite
I tried looking around for an answer to this question, but I couldn't find anything specific enough. I'm thinking I need to use either Cartesian products or subspace sums to produce such a vector space, however I'm still fuzzy on how to actually implement it.
Any help would be great!
linear-algebra polynomials vector-spaces
asked Sep 10 at 18:53
Silverwing171
82
What is the $n$-th degree polynomial vector space? Is that the vector space of polynomials (coefficients in some field) whose degree is at most $n$?
– Dave
Sep 10 at 18:56
Yes, that's my understanding. My text uses the notation ð”½[$x$;$n$], without any explanation as to how it should be read, unfortunately.
– Silverwing171
Sep 10 at 19:00
Hint: if $v_1,v_2,dots,v_n$ is a basis for a vector space, then so is $v_1+v_n,v_2,dots,v_n$.
– amd
Sep 10 at 22:58
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I tried looking around for an answer to this question, but I couldn't find anything specific enough. I'm thinking I need to use either Cartesian products or subspace sums to produce such a vector space, however I'm still fuzzy on how to actually implement it.
Any help would be great!
linear-algebra polynomials vector-spaces
asked Sep 10 at 18:53
Silverwing171
82
I tried looking around for an answer to this question, but I couldn't find anything specific enough. I'm thinking I need to use either Cartesian products or subspace sums to produce such a vector space, however I'm still fuzzy on how to actually implement it.
Any help would be great!
linear-algebra polynomials vector-spaces
linear-algebra polynomials vector-spaces
asked Sep 10 at 18:53
Silverwing171
82
asked Sep 10 at 18:53
Silverwing171
82
asked Sep 10 at 18:53
Silverwing171
82
asked Sep 10 at 18:53
Silverwing171
82
asked Sep 10 at 18:53
Silverwing171
82
82
What is the $n$-th degree polynomial vector space? Is that the vector space of polynomials (coefficients in some field) whose degree is at most $n$?
– Dave
Sep 10 at 18:56
Yes, that's my understanding. My text uses the notation ð”½[$x$;$n$], without any explanation as to how it should be read, unfortunately.
– Silverwing171
Sep 10 at 19:00
Hint: if $v_1,v_2,dots,v_n$ is a basis for a vector space, then so is $v_1+v_n,v_2,dots,v_n$.
– amd
Sep 10 at 22:58
add a comment |Â
What is the $n$-th degree polynomial vector space? Is that the vector space of polynomials (coefficients in some field) whose degree is at most $n$?
– Dave
Sep 10 at 18:56
Yes, that's my understanding. My text uses the notation ð”½[$x$;$n$], without any explanation as to how it should be read, unfortunately.
– Silverwing171
Sep 10 at 19:00
Hint: if $v_1,v_2,dots,v_n$ is a basis for a vector space, then so is $v_1+v_n,v_2,dots,v_n$.
– amd
Sep 10 at 22:58
What is the $n$-th degree polynomial vector space? Is that the vector space of polynomials (coefficients in some field) whose degree is at most $n$?
– Dave
Sep 10 at 18:56
What is the $n$-th degree polynomial vector space? Is that the vector space of polynomials (coefficients in some field) whose degree is at most $n$?
– Dave
Sep 10 at 18:56
Yes, that's my understanding. My text uses the notation ð”½[$x$;$n$], without any explanation as to how it should be read, unfortunately.
– Silverwing171
Sep 10 at 19:00
Yes, that's my understanding. My text uses the notation ð”½[$x$;$n$], without any explanation as to how it should be read, unfortunately.
– Silverwing171
Sep 10 at 19:00
Hint: if $v_1,v_2,dots,v_n$ is a basis for a vector space, then so is $v_1+v_n,v_2,dots,v_n$.
– amd
Sep 10 at 22:58
Hint: if $v_1,v_2,dots,v_n$ is a basis for a vector space, then so is $v_1+v_n,v_2,dots,v_n$.
– amd
Sep 10 at 22:58
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Hint:
We know that $1, x, x^2, ldots, x^n$ is a basis for your space. But then $$1+x^n, x+x^n, x^2+x^n, ldots, 2x^n$$
is also a basis for your space.
answered Sep 10 at 19:44
mechanodroid
24.7k62245
add a comment |Â
up vote
0
down vote
Pick any basis $f_1, ldots, f_n, f_n+1$. At least one of these polynomials must have degree exactly $n$, otherwise $x^n$ would not be in the generated subspace.
WLOG $deg f_1 =n$.
If some of the other basis vectors have smaller degree, add $f_1$ to them (and only to those with degree strictly less than $n$).
That yields a solution you were looking for.
answered Sep 10 at 19:09
A. Pongrácz
4,465725
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Hint:
We know that $1, x, x^2, ldots, x^n$ is a basis for your space. But then $$1+x^n, x+x^n, x^2+x^n, ldots, 2x^n$$
is also a basis for your space.
answered Sep 10 at 19:44
mechanodroid
24.7k62245
add a comment |Â
up vote
0
down vote
accepted
Hint:
We know that $1, x, x^2, ldots, x^n$ is a basis for your space. But then $$1+x^n, x+x^n, x^2+x^n, ldots, 2x^n$$
is also a basis for your space.
answered Sep 10 at 19:44
mechanodroid
24.7k62245
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint:
We know that $1, x, x^2, ldots, x^n$ is a basis for your space. But then $$1+x^n, x+x^n, x^2+x^n, ldots, 2x^n$$
is also a basis for your space.
answered Sep 10 at 19:44
mechanodroid
24.7k62245
Hint:
We know that $1, x, x^2, ldots, x^n$ is a basis for your space. But then $$1+x^n, x+x^n, x^2+x^n, ldots, 2x^n$$
is also a basis for your space.
answered Sep 10 at 19:44
mechanodroid
24.7k62245
answered Sep 10 at 19:44
mechanodroid
24.7k62245
answered Sep 10 at 19:44
mechanodroid
24.7k62245
answered Sep 10 at 19:44
mechanodroid
24.7k62245
24.7k62245
add a comment |Â
add a comment |Â
up vote
0
down vote
Pick any basis $f_1, ldots, f_n, f_n+1$. At least one of these polynomials must have degree exactly $n$, otherwise $x^n$ would not be in the generated subspace.
WLOG $deg f_1 =n$.
If some of the other basis vectors have smaller degree, add $f_1$ to them (and only to those with degree strictly less than $n$).
That yields a solution you were looking for.
answered Sep 10 at 19:09
A. Pongrácz
4,465725
add a comment |Â
up vote
0
down vote
Pick any basis $f_1, ldots, f_n, f_n+1$. At least one of these polynomials must have degree exactly $n$, otherwise $x^n$ would not be in the generated subspace.
WLOG $deg f_1 =n$.
If some of the other basis vectors have smaller degree, add $f_1$ to them (and only to those with degree strictly less than $n$).
That yields a solution you were looking for.
answered Sep 10 at 19:09
A. Pongrácz
4,465725
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Pick any basis $f_1, ldots, f_n, f_n+1$. At least one of these polynomials must have degree exactly $n$, otherwise $x^n$ would not be in the generated subspace.
WLOG $deg f_1 =n$.
If some of the other basis vectors have smaller degree, add $f_1$ to them (and only to those with degree strictly less than $n$).
That yields a solution you were looking for.
answered Sep 10 at 19:09
A. Pongrácz
4,465725
Pick any basis $f_1, ldots, f_n, f_n+1$. At least one of these polynomials must have degree exactly $n$, otherwise $x^n$ would not be in the generated subspace.
WLOG $deg f_1 =n$.
If some of the other basis vectors have smaller degree, add $f_1$ to them (and only to those with degree strictly less than $n$).
That yields a solution you were looking for.
answered Sep 10 at 19:09
A. Pongrácz
4,465725
answered Sep 10 at 19:09
A. Pongrácz
4,465725
answered Sep 10 at 19:09
A. Pongrácz
4,465725
answered Sep 10 at 19:09
A. Pongrácz
4,465725
4,465725
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2912227%2fprove-that-there-is-a-basis-for-the-nth-degree-polynomial-vector-space-consist%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
What is the $n$-th degree polynomial vector space? Is that the vector space of polynomials (coefficients in some field) whose degree is at most $n$?
– Dave
Sep 10 at 18:56
Yes, that's my understanding. My text uses the notation ð”½[$x$;$n$], without any explanation as to how it should be read, unfortunately.
– Silverwing171
Sep 10 at 19:00
Hint: if $v_1,v_2,dots,v_n$ is a basis for a vector space, then so is $v_1+v_n,v_2,dots,v_n$.
– amd
Sep 10 at 22:58