Proving Polynomial is a subspace of a vector space
$begingroup$
$W={f(x)in P(mathbb R) colon f(x)=0 text{ or } f(x)text{ has degree }5}$, $V=P(mathbb R)$
I'm really stuck on proving this question. I know that the first axioms stating that $0$ must be an element of $W$ is held, however I'm not sure how to prove closure under addition or scalar.
I've tried using arbitrary polynomials $a$, $b$ and letting them be an element of $W$. I then have:
$$ax^5+bx^5=(a+b)x^5$$
And I'm not sure how to prove that this addition is an element of W. I know intuitively it makes sense that it's an element of W, but I'm just not sure how to proof it mathematically.
Any help would be greatly appreciated!
linear-algebra
edited Nov 25 '16 at 6:20
Martin Sleziak
44.8k9118272
asked Feb 5 '16 at 4:07
Nikitau LNikitau L
112
$endgroup$
add a comment |
$begingroup$
$W={f(x)in P(mathbb R) colon f(x)=0 text{ or } f(x)text{ has degree }5}$, $V=P(mathbb R)$
I'm really stuck on proving this question. I know that the first axioms stating that $0$ must be an element of $W$ is held, however I'm not sure how to prove closure under addition or scalar.
I've tried using arbitrary polynomials $a$, $b$ and letting them be an element of $W$. I then have:
$$ax^5+bx^5=(a+b)x^5$$
And I'm not sure how to prove that this addition is an element of W. I know intuitively it makes sense that it's an element of W, but I'm just not sure how to proof it mathematically.
Any help would be greatly appreciated!
linear-algebra
edited Nov 25 '16 at 6:20
Martin Sleziak
44.8k9118272
asked Feb 5 '16 at 4:07
Nikitau LNikitau L
112
$endgroup$
1
$begingroup$
$x^5$ is not a general polynomial of degree 5. A general polynomial of degree 5 will contain $x^i$ terms where $i le 4$ besides a mandatory $ax^5$ term where $a$ is not $0$
$endgroup$
– Shailesh
Feb 5 '16 at 4:13
$begingroup$
The phrase "proving this question" doesn't quite make sense. Questions should be answered. Theorems, claims, and the likes are to be proved. And this is not just being picky with words. While it's still an open question, are you sure you want to prove or to disprove the claim that this $W$ is a subspace? Proving something that's not even true would be such a waste of time and effort...
$endgroup$
– zipirovich
Nov 25 '16 at 6:04
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Martin Sleziak
Nov 25 '16 at 6:20
add a comment |
$begingroup$
$W={f(x)in P(mathbb R) colon f(x)=0 text{ or } f(x)text{ has degree }5}$, $V=P(mathbb R)$
I'm really stuck on proving this question. I know that the first axioms stating that $0$ must be an element of $W$ is held, however I'm not sure how to prove closure under addition or scalar.
I've tried using arbitrary polynomials $a$, $b$ and letting them be an element of $W$. I then have:
$$ax^5+bx^5=(a+b)x^5$$
And I'm not sure how to prove that this addition is an element of W. I know intuitively it makes sense that it's an element of W, but I'm just not sure how to proof it mathematically.
Any help would be greatly appreciated!
linear-algebra
edited Nov 25 '16 at 6:20
Martin Sleziak
44.8k9118272
asked Feb 5 '16 at 4:07
Nikitau LNikitau L
112
$endgroup$
$W={f(x)in P(mathbb R) colon f(x)=0 text{ or } f(x)text{ has degree }5}$, $V=P(mathbb R)$
I'm really stuck on proving this question. I know that the first axioms stating that $0$ must be an element of $W$ is held, however I'm not sure how to prove closure under addition or scalar.
I've tried using arbitrary polynomials $a$, $b$ and letting them be an element of $W$. I then have:
$$ax^5+bx^5=(a+b)x^5$$
And I'm not sure how to prove that this addition is an element of W. I know intuitively it makes sense that it's an element of W, but I'm just not sure how to proof it mathematically.
Any help would be greatly appreciated!
linear-algebra
linear-algebra
edited Nov 25 '16 at 6:20
Martin Sleziak
44.8k9118272
asked Feb 5 '16 at 4:07
Nikitau LNikitau L
112
edited Nov 25 '16 at 6:20
Martin Sleziak
44.8k9118272
asked Feb 5 '16 at 4:07
Nikitau LNikitau L
112
edited Nov 25 '16 at 6:20
Martin Sleziak
44.8k9118272
edited Nov 25 '16 at 6:20
Martin Sleziak
44.8k9118272
edited Nov 25 '16 at 6:20
Martin Sleziak
44.8k9118272
44.8k9118272
asked Feb 5 '16 at 4:07
Nikitau LNikitau L
112
asked Feb 5 '16 at 4:07
Nikitau LNikitau L
112
asked Feb 5 '16 at 4:07
Nikitau LNikitau L
112
112
1
$begingroup$
$x^5$ is not a general polynomial of degree 5. A general polynomial of degree 5 will contain $x^i$ terms where $i le 4$ besides a mandatory $ax^5$ term where $a$ is not $0$
$endgroup$
– Shailesh
Feb 5 '16 at 4:13
$begingroup$
The phrase "proving this question" doesn't quite make sense. Questions should be answered. Theorems, claims, and the likes are to be proved. And this is not just being picky with words. While it's still an open question, are you sure you want to prove or to disprove the claim that this $W$ is a subspace? Proving something that's not even true would be such a waste of time and effort...
$endgroup$
– zipirovich
Nov 25 '16 at 6:04
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Martin Sleziak
Nov 25 '16 at 6:20
add a comment |
1
$begingroup$
$x^5$ is not a general polynomial of degree 5. A general polynomial of degree 5 will contain $x^i$ terms where $i le 4$ besides a mandatory $ax^5$ term where $a$ is not $0$
$endgroup$
– Shailesh
Feb 5 '16 at 4:13
$begingroup$
The phrase "proving this question" doesn't quite make sense. Questions should be answered. Theorems, claims, and the likes are to be proved. And this is not just being picky with words. While it's still an open question, are you sure you want to prove or to disprove the claim that this $W$ is a subspace? Proving something that's not even true would be such a waste of time and effort...
$endgroup$
– zipirovich
Nov 25 '16 at 6:04
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Martin Sleziak
Nov 25 '16 at 6:20
1
1
$begingroup$
$x^5$ is not a general polynomial of degree 5. A general polynomial of degree 5 will contain $x^i$ terms where $i le 4$ besides a mandatory $ax^5$ term where $a$ is not $0$
$endgroup$
– Shailesh
Feb 5 '16 at 4:13
$begingroup$
$x^5$ is not a general polynomial of degree 5. A general polynomial of degree 5 will contain $x^i$ terms where $i le 4$ besides a mandatory $ax^5$ term where $a$ is not $0$
$endgroup$
– Shailesh
Feb 5 '16 at 4:13
$begingroup$
The phrase "proving this question" doesn't quite make sense. Questions should be answered. Theorems, claims, and the likes are to be proved. And this is not just being picky with words. While it's still an open question, are you sure you want to prove or to disprove the claim that this $W$ is a subspace? Proving something that's not even true would be such a waste of time and effort...
$endgroup$
– zipirovich
Nov 25 '16 at 6:04
$begingroup$
The phrase "proving this question" doesn't quite make sense. Questions should be answered. Theorems, claims, and the likes are to be proved. And this is not just being picky with words. While it's still an open question, are you sure you want to prove or to disprove the claim that this $W$ is a subspace? Proving something that's not even true would be such a waste of time and effort...
$endgroup$
– zipirovich
Nov 25 '16 at 6:04
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Martin Sleziak
Nov 25 '16 at 6:20
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Martin Sleziak
Nov 25 '16 at 6:20
add a comment |
1 Answer
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$begingroup$
To be a vector (sub)space $V$, you have to check that $f,g in V$ implies that $af + bg in V$ for all $a,b in Bbb R$. Notice that all polynomials of degree five or fewer take the form $a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$, where $a_i in Bbb R$ and that addition of polynomials is done componentwise.
Edit: Notice that you've shown that IF you add two polynomials off degree less five, what you get back is a polynomial of degree less or equal to five. Since you've already noted that $0$ is in your space, all you have to do is show that multiplying by a real number gives a polynomial of degree less than or equal to five. By showing this for any two fixed polynomials, you show this for any polynomials.
Second edit: Don't forget your constant terms; they are important. Remember your goal when proving something is a vector space is to say "IF I add two elements that I know are in the space together, they give me a third element also in the space."
answered Feb 5 '16 at 4:20
RandomWalkerRandomWalker
1549
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add a comment |
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1 Answer
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$begingroup$
To be a vector (sub)space $V$, you have to check that $f,g in V$ implies that $af + bg in V$ for all $a,b in Bbb R$. Notice that all polynomials of degree five or fewer take the form $a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$, where $a_i in Bbb R$ and that addition of polynomials is done componentwise.
Edit: Notice that you've shown that IF you add two polynomials off degree less five, what you get back is a polynomial of degree less or equal to five. Since you've already noted that $0$ is in your space, all you have to do is show that multiplying by a real number gives a polynomial of degree less than or equal to five. By showing this for any two fixed polynomials, you show this for any polynomials.
Second edit: Don't forget your constant terms; they are important. Remember your goal when proving something is a vector space is to say "IF I add two elements that I know are in the space together, they give me a third element also in the space."
answered Feb 5 '16 at 4:20
RandomWalkerRandomWalker
1549
$endgroup$
add a comment |
$begingroup$
To be a vector (sub)space $V$, you have to check that $f,g in V$ implies that $af + bg in V$ for all $a,b in Bbb R$. Notice that all polynomials of degree five or fewer take the form $a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$, where $a_i in Bbb R$ and that addition of polynomials is done componentwise.
Edit: Notice that you've shown that IF you add two polynomials off degree less five, what you get back is a polynomial of degree less or equal to five. Since you've already noted that $0$ is in your space, all you have to do is show that multiplying by a real number gives a polynomial of degree less than or equal to five. By showing this for any two fixed polynomials, you show this for any polynomials.
Second edit: Don't forget your constant terms; they are important. Remember your goal when proving something is a vector space is to say "IF I add two elements that I know are in the space together, they give me a third element also in the space."
answered Feb 5 '16 at 4:20
RandomWalkerRandomWalker
1549
$endgroup$
add a comment |
$begingroup$
To be a vector (sub)space $V$, you have to check that $f,g in V$ implies that $af + bg in V$ for all $a,b in Bbb R$. Notice that all polynomials of degree five or fewer take the form $a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$, where $a_i in Bbb R$ and that addition of polynomials is done componentwise.
Edit: Notice that you've shown that IF you add two polynomials off degree less five, what you get back is a polynomial of degree less or equal to five. Since you've already noted that $0$ is in your space, all you have to do is show that multiplying by a real number gives a polynomial of degree less than or equal to five. By showing this for any two fixed polynomials, you show this for any polynomials.
Second edit: Don't forget your constant terms; they are important. Remember your goal when proving something is a vector space is to say "IF I add two elements that I know are in the space together, they give me a third element also in the space."
answered Feb 5 '16 at 4:20
RandomWalkerRandomWalker
1549
$endgroup$
To be a vector (sub)space $V$, you have to check that $f,g in V$ implies that $af + bg in V$ for all $a,b in Bbb R$. Notice that all polynomials of degree five or fewer take the form $a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$, where $a_i in Bbb R$ and that addition of polynomials is done componentwise.
Edit: Notice that you've shown that IF you add two polynomials off degree less five, what you get back is a polynomial of degree less or equal to five. Since you've already noted that $0$ is in your space, all you have to do is show that multiplying by a real number gives a polynomial of degree less than or equal to five. By showing this for any two fixed polynomials, you show this for any polynomials.
Second edit: Don't forget your constant terms; they are important. Remember your goal when proving something is a vector space is to say "IF I add two elements that I know are in the space together, they give me a third element also in the space."
answered Feb 5 '16 at 4:20
RandomWalkerRandomWalker
1549
edited Feb 5 '16 at 4:25
answered Feb 5 '16 at 4:20
RandomWalkerRandomWalker
1549
answered Feb 5 '16 at 4:20
RandomWalkerRandomWalker
1549
answered Feb 5 '16 at 4:20
RandomWalkerRandomWalker
1549
1549
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$begingroup$
$x^5$ is not a general polynomial of degree 5. A general polynomial of degree 5 will contain $x^i$ terms where $i le 4$ besides a mandatory $ax^5$ term where $a$ is not $0$
$endgroup$
– Shailesh
Feb 5 '16 at 4:13
$begingroup$
The phrase "proving this question" doesn't quite make sense. Questions should be answered. Theorems, claims, and the likes are to be proved. And this is not just being picky with words. While it's still an open question, are you sure you want to prove or to disprove the claim that this $W$ is a subspace? Proving something that's not even true would be such a waste of time and effort...
$endgroup$
– zipirovich
Nov 25 '16 at 6:04
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here.
$endgroup$
– Martin Sleziak
Nov 25 '16 at 6:20