Is the vector space finite-dimensional?
$begingroup$
The vector subspace $U$ of $mathbb{R}[X]$ (polynomials)
$U:=${$f∈mathbb{R}[X] | f(alpha+1)-f(alpha)=f(beta+1)-f(beta)
Ɐalpha,beta∈mathbb{R}$}
is finite-dimensional?
I know that $mathbb{R}[X]$ is finite-dimensional, because it has a basis {$1,x,x^2,...,x^n$} and its dimension is $dim(mathbb{R}[X])=n+1$. So theoretically a subspace of $mathbb{R}[X]$ must be finite too. Is that right? Otherways I don't know how to demonstrate that. Any help please
linear-algebra vector-spaces
asked Nov 27 '18 at 15:16
DadaDada
7510
$endgroup$
add a comment |
$begingroup$
The vector subspace $U$ of $mathbb{R}[X]$ (polynomials)
$U:=${$f∈mathbb{R}[X] | f(alpha+1)-f(alpha)=f(beta+1)-f(beta)
Ɐalpha,beta∈mathbb{R}$}
is finite-dimensional?
I know that $mathbb{R}[X]$ is finite-dimensional, because it has a basis {$1,x,x^2,...,x^n$} and its dimension is $dim(mathbb{R}[X])=n+1$. So theoretically a subspace of $mathbb{R}[X]$ must be finite too. Is that right? Otherways I don't know how to demonstrate that. Any help please
linear-algebra vector-spaces
asked Nov 27 '18 at 15:16
DadaDada
7510
$endgroup$
$begingroup$
$mathbb R[X]$ is not finite dimensional, because the powers of $x$ don't stop at $n$, they keep going forever! As for the question, essentially $U$ consists of all polynomials $f$ such that the polynomial $g(x) = f(x+1) - f(x)$ is a constant. If $g$ is a constant, can you prove that $f$ must be of degree at most one?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 27 '18 at 15:20
add a comment |
$begingroup$
The vector subspace $U$ of $mathbb{R}[X]$ (polynomials)
$U:=${$f∈mathbb{R}[X] | f(alpha+1)-f(alpha)=f(beta+1)-f(beta)
Ɐalpha,beta∈mathbb{R}$}
is finite-dimensional?
I know that $mathbb{R}[X]$ is finite-dimensional, because it has a basis {$1,x,x^2,...,x^n$} and its dimension is $dim(mathbb{R}[X])=n+1$. So theoretically a subspace of $mathbb{R}[X]$ must be finite too. Is that right? Otherways I don't know how to demonstrate that. Any help please
linear-algebra vector-spaces
asked Nov 27 '18 at 15:16
DadaDada
7510
$endgroup$
The vector subspace $U$ of $mathbb{R}[X]$ (polynomials)
$U:=${$f∈mathbb{R}[X] | f(alpha+1)-f(alpha)=f(beta+1)-f(beta)
Ɐalpha,beta∈mathbb{R}$}
is finite-dimensional?
I know that $mathbb{R}[X]$ is finite-dimensional, because it has a basis {$1,x,x^2,...,x^n$} and its dimension is $dim(mathbb{R}[X])=n+1$. So theoretically a subspace of $mathbb{R}[X]$ must be finite too. Is that right? Otherways I don't know how to demonstrate that. Any help please
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Nov 27 '18 at 15:16
DadaDada
7510
asked Nov 27 '18 at 15:16
DadaDada
7510
asked Nov 27 '18 at 15:16
DadaDada
7510
asked Nov 27 '18 at 15:16
DadaDada
7510
asked Nov 27 '18 at 15:16
DadaDada
7510
7510
$begingroup$
$mathbb R[X]$ is not finite dimensional, because the powers of $x$ don't stop at $n$, they keep going forever! As for the question, essentially $U$ consists of all polynomials $f$ such that the polynomial $g(x) = f(x+1) - f(x)$ is a constant. If $g$ is a constant, can you prove that $f$ must be of degree at most one?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 27 '18 at 15:20
add a comment |
$begingroup$
$mathbb R[X]$ is not finite dimensional, because the powers of $x$ don't stop at $n$, they keep going forever! As for the question, essentially $U$ consists of all polynomials $f$ such that the polynomial $g(x) = f(x+1) - f(x)$ is a constant. If $g$ is a constant, can you prove that $f$ must be of degree at most one?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 27 '18 at 15:20
$begingroup$
$mathbb R[X]$ is not finite dimensional, because the powers of $x$ don't stop at $n$, they keep going forever! As for the question, essentially $U$ consists of all polynomials $f$ such that the polynomial $g(x) = f(x+1) - f(x)$ is a constant. If $g$ is a constant, can you prove that $f$ must be of degree at most one?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 27 '18 at 15:20
$begingroup$
$mathbb R[X]$ is not finite dimensional, because the powers of $x$ don't stop at $n$, they keep going forever! As for the question, essentially $U$ consists of all polynomials $f$ such that the polynomial $g(x) = f(x+1) - f(x)$ is a constant. If $g$ is a constant, can you prove that $f$ must be of degree at most one?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 27 '18 at 15:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, $mathbb R[X]$ is infinite-dimensional. A basis is ${1, x, x^2, ldots}$ (there is no $n$ where this stops). But $U$ is indeed finite-dimensional.
Hint: if $f$ is a polynomial of degree $n$, what is the degree of $f(X+1)-f(X)$?
answered Nov 27 '18 at 15:19
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Ok, then $f(X+1)-f(X)$ would have a degree $0$. So $U$ is finite-dimensional, but what would be a basis of $U$?
$endgroup$
– Dada
Nov 27 '18 at 15:26
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
No, $mathbb R[X]$ is infinite-dimensional. A basis is ${1, x, x^2, ldots}$ (there is no $n$ where this stops). But $U$ is indeed finite-dimensional.
Hint: if $f$ is a polynomial of degree $n$, what is the degree of $f(X+1)-f(X)$?
answered Nov 27 '18 at 15:19
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Ok, then $f(X+1)-f(X)$ would have a degree $0$. So $U$ is finite-dimensional, but what would be a basis of $U$?
$endgroup$
– Dada
Nov 27 '18 at 15:26
add a comment |
$begingroup$
No, $mathbb R[X]$ is infinite-dimensional. A basis is ${1, x, x^2, ldots}$ (there is no $n$ where this stops). But $U$ is indeed finite-dimensional.
Hint: if $f$ is a polynomial of degree $n$, what is the degree of $f(X+1)-f(X)$?
answered Nov 27 '18 at 15:19
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Ok, then $f(X+1)-f(X)$ would have a degree $0$. So $U$ is finite-dimensional, but what would be a basis of $U$?
$endgroup$
– Dada
Nov 27 '18 at 15:26
add a comment |
$begingroup$
No, $mathbb R[X]$ is infinite-dimensional. A basis is ${1, x, x^2, ldots}$ (there is no $n$ where this stops). But $U$ is indeed finite-dimensional.
Hint: if $f$ is a polynomial of degree $n$, what is the degree of $f(X+1)-f(X)$?
answered Nov 27 '18 at 15:19
Robert IsraelRobert Israel
321k23210462
$endgroup$
No, $mathbb R[X]$ is infinite-dimensional. A basis is ${1, x, x^2, ldots}$ (there is no $n$ where this stops). But $U$ is indeed finite-dimensional.
Hint: if $f$ is a polynomial of degree $n$, what is the degree of $f(X+1)-f(X)$?
answered Nov 27 '18 at 15:19
Robert IsraelRobert Israel
321k23210462
answered Nov 27 '18 at 15:19
Robert IsraelRobert Israel
321k23210462
answered Nov 27 '18 at 15:19
Robert IsraelRobert Israel
321k23210462
answered Nov 27 '18 at 15:19
Robert IsraelRobert Israel
321k23210462
321k23210462
$begingroup$
Ok, then $f(X+1)-f(X)$ would have a degree $0$. So $U$ is finite-dimensional, but what would be a basis of $U$?
$endgroup$
– Dada
Nov 27 '18 at 15:26
add a comment |
$begingroup$
Ok, then $f(X+1)-f(X)$ would have a degree $0$. So $U$ is finite-dimensional, but what would be a basis of $U$?
$endgroup$
– Dada
Nov 27 '18 at 15:26
$begingroup$
Ok, then $f(X+1)-f(X)$ would have a degree $0$. So $U$ is finite-dimensional, but what would be a basis of $U$?
$endgroup$
– Dada
Nov 27 '18 at 15:26
$begingroup$
Ok, then $f(X+1)-f(X)$ would have a degree $0$. So $U$ is finite-dimensional, but what would be a basis of $U$?
$endgroup$
– Dada
Nov 27 '18 at 15:26
add a comment |
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$begingroup$
$mathbb R[X]$ is not finite dimensional, because the powers of $x$ don't stop at $n$, they keep going forever! As for the question, essentially $U$ consists of all polynomials $f$ such that the polynomial $g(x) = f(x+1) - f(x)$ is a constant. If $g$ is a constant, can you prove that $f$ must be of degree at most one?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 27 '18 at 15:20