Some questions regarding functions and vector space
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I have three simple questions about whether a function form a vector space or not I hope someone can verify my intuition.
- Do functions that vanish at end points x = 0 and x = L form a vector space?
Yes, because any function between these points satisfies the additivity and multiplicity with a scalar property. Furthermore, we can construct a null vector which is function f(x) = 0 over 0 and L. Lastly, -f(x) is the inverse of f(x) over this space.
- Do periodic functions obeying f(0) = f(L) form a vector space?
Yes, same as above.
- Are vectors that obey f(0) = 4 form a vector space?
This one is a bit weird. Intuitively there is no inverse because we have no function that satisfies f(0) = -4, so all these functions do not satisfy the property of a vector space.
Can anyone check if I understand this correctly? Thank you!!
vector-spaces
asked Feb 14 '15 at 7:29
OlórinOlórin
1,95411839
$endgroup$
add a comment |
$begingroup$
I have three simple questions about whether a function form a vector space or not I hope someone can verify my intuition.
- Do functions that vanish at end points x = 0 and x = L form a vector space?
Yes, because any function between these points satisfies the additivity and multiplicity with a scalar property. Furthermore, we can construct a null vector which is function f(x) = 0 over 0 and L. Lastly, -f(x) is the inverse of f(x) over this space.
- Do periodic functions obeying f(0) = f(L) form a vector space?
Yes, same as above.
- Are vectors that obey f(0) = 4 form a vector space?
This one is a bit weird. Intuitively there is no inverse because we have no function that satisfies f(0) = -4, so all these functions do not satisfy the property of a vector space.
Can anyone check if I understand this correctly? Thank you!!
vector-spaces
asked Feb 14 '15 at 7:29
OlórinOlórin
1,95411839
$endgroup$
add a comment |
$begingroup$
I have three simple questions about whether a function form a vector space or not I hope someone can verify my intuition.
- Do functions that vanish at end points x = 0 and x = L form a vector space?
Yes, because any function between these points satisfies the additivity and multiplicity with a scalar property. Furthermore, we can construct a null vector which is function f(x) = 0 over 0 and L. Lastly, -f(x) is the inverse of f(x) over this space.
- Do periodic functions obeying f(0) = f(L) form a vector space?
Yes, same as above.
- Are vectors that obey f(0) = 4 form a vector space?
This one is a bit weird. Intuitively there is no inverse because we have no function that satisfies f(0) = -4, so all these functions do not satisfy the property of a vector space.
Can anyone check if I understand this correctly? Thank you!!
vector-spaces
asked Feb 14 '15 at 7:29
OlórinOlórin
1,95411839
$endgroup$
I have three simple questions about whether a function form a vector space or not I hope someone can verify my intuition.
- Do functions that vanish at end points x = 0 and x = L form a vector space?
Yes, because any function between these points satisfies the additivity and multiplicity with a scalar property. Furthermore, we can construct a null vector which is function f(x) = 0 over 0 and L. Lastly, -f(x) is the inverse of f(x) over this space.
- Do periodic functions obeying f(0) = f(L) form a vector space?
Yes, same as above.
- Are vectors that obey f(0) = 4 form a vector space?
This one is a bit weird. Intuitively there is no inverse because we have no function that satisfies f(0) = -4, so all these functions do not satisfy the property of a vector space.
Can anyone check if I understand this correctly? Thank you!!
vector-spaces
vector-spaces
asked Feb 14 '15 at 7:29
OlórinOlórin
1,95411839
asked Feb 14 '15 at 7:29
OlórinOlórin
1,95411839
asked Feb 14 '15 at 7:29
OlórinOlórin
1,95411839
asked Feb 14 '15 at 7:29
OlórinOlórin
1,95411839
asked Feb 14 '15 at 7:29
OlórinOlórin
1,95411839
1,95411839
add a comment |
add a comment |
2 Answers
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Your reasoning in part 1 is sound. Part 2 however is not correct. Observe that if $f(0) = f(L) = 1$ then the collection of functions with these conditions isn't closed under scalar multiplication. For instance $2f$ wouldn't be in your space since $2f(0) = 2f(L) = 2 neq f(0)$. Your argument in 3 is also good, but why is it weird? I don't see it weird in the least: some collections have an inherent algebraic (or vector space) structure and some don't. There's no promise that any set you can come up with has structure.
answered Feb 14 '15 at 7:51
MnifldzMnifldz
6,85011634
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$begingroup$
The second part also forms a linear vector space. There is a flaw in the argument above.
The requirement is that $f(0) = f(L)$. Suppose $2f = g$, then we must have $g(0) = g(L)$, which is true. $g(0)$ doesn't have to be equivalent to $f(0)$; it would still be closed under scalar multiplication.
edited Sep 8 '16 at 5:09
Parcly Taxel
42.2k1372101
answered Sep 8 '16 at 4:05
user367154user367154
1
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2 Answers
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2 Answers
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$begingroup$
Your reasoning in part 1 is sound. Part 2 however is not correct. Observe that if $f(0) = f(L) = 1$ then the collection of functions with these conditions isn't closed under scalar multiplication. For instance $2f$ wouldn't be in your space since $2f(0) = 2f(L) = 2 neq f(0)$. Your argument in 3 is also good, but why is it weird? I don't see it weird in the least: some collections have an inherent algebraic (or vector space) structure and some don't. There's no promise that any set you can come up with has structure.
answered Feb 14 '15 at 7:51
MnifldzMnifldz
6,85011634
$endgroup$
add a comment |
$begingroup$
Your reasoning in part 1 is sound. Part 2 however is not correct. Observe that if $f(0) = f(L) = 1$ then the collection of functions with these conditions isn't closed under scalar multiplication. For instance $2f$ wouldn't be in your space since $2f(0) = 2f(L) = 2 neq f(0)$. Your argument in 3 is also good, but why is it weird? I don't see it weird in the least: some collections have an inherent algebraic (or vector space) structure and some don't. There's no promise that any set you can come up with has structure.
answered Feb 14 '15 at 7:51
MnifldzMnifldz
6,85011634
$endgroup$
add a comment |
$begingroup$
Your reasoning in part 1 is sound. Part 2 however is not correct. Observe that if $f(0) = f(L) = 1$ then the collection of functions with these conditions isn't closed under scalar multiplication. For instance $2f$ wouldn't be in your space since $2f(0) = 2f(L) = 2 neq f(0)$. Your argument in 3 is also good, but why is it weird? I don't see it weird in the least: some collections have an inherent algebraic (or vector space) structure and some don't. There's no promise that any set you can come up with has structure.
answered Feb 14 '15 at 7:51
MnifldzMnifldz
6,85011634
$endgroup$
Your reasoning in part 1 is sound. Part 2 however is not correct. Observe that if $f(0) = f(L) = 1$ then the collection of functions with these conditions isn't closed under scalar multiplication. For instance $2f$ wouldn't be in your space since $2f(0) = 2f(L) = 2 neq f(0)$. Your argument in 3 is also good, but why is it weird? I don't see it weird in the least: some collections have an inherent algebraic (or vector space) structure and some don't. There's no promise that any set you can come up with has structure.
answered Feb 14 '15 at 7:51
MnifldzMnifldz
6,85011634
answered Feb 14 '15 at 7:51
MnifldzMnifldz
6,85011634
answered Feb 14 '15 at 7:51
MnifldzMnifldz
6,85011634
answered Feb 14 '15 at 7:51
MnifldzMnifldz
6,85011634
6,85011634
add a comment |
add a comment |
$begingroup$
The second part also forms a linear vector space. There is a flaw in the argument above.
The requirement is that $f(0) = f(L)$. Suppose $2f = g$, then we must have $g(0) = g(L)$, which is true. $g(0)$ doesn't have to be equivalent to $f(0)$; it would still be closed under scalar multiplication.
edited Sep 8 '16 at 5:09
Parcly Taxel
42.2k1372101
answered Sep 8 '16 at 4:05
user367154user367154
1
$endgroup$
add a comment |
$begingroup$
The second part also forms a linear vector space. There is a flaw in the argument above.
The requirement is that $f(0) = f(L)$. Suppose $2f = g$, then we must have $g(0) = g(L)$, which is true. $g(0)$ doesn't have to be equivalent to $f(0)$; it would still be closed under scalar multiplication.
edited Sep 8 '16 at 5:09
Parcly Taxel
42.2k1372101
answered Sep 8 '16 at 4:05
user367154user367154
1
$endgroup$
add a comment |
$begingroup$
The second part also forms a linear vector space. There is a flaw in the argument above.
The requirement is that $f(0) = f(L)$. Suppose $2f = g$, then we must have $g(0) = g(L)$, which is true. $g(0)$ doesn't have to be equivalent to $f(0)$; it would still be closed under scalar multiplication.
edited Sep 8 '16 at 5:09
Parcly Taxel
42.2k1372101
answered Sep 8 '16 at 4:05
user367154user367154
1
$endgroup$
The second part also forms a linear vector space. There is a flaw in the argument above.
The requirement is that $f(0) = f(L)$. Suppose $2f = g$, then we must have $g(0) = g(L)$, which is true. $g(0)$ doesn't have to be equivalent to $f(0)$; it would still be closed under scalar multiplication.
edited Sep 8 '16 at 5:09
Parcly Taxel
42.2k1372101
answered Sep 8 '16 at 4:05
user367154user367154
1
edited Sep 8 '16 at 5:09
Parcly Taxel
42.2k1372101
edited Sep 8 '16 at 5:09
Parcly Taxel
42.2k1372101
edited Sep 8 '16 at 5:09
Parcly Taxel
42.2k1372101
42.2k1372101
answered Sep 8 '16 at 4:05
user367154user367154
1
answered Sep 8 '16 at 4:05
user367154user367154
1
answered Sep 8 '16 at 4:05
user367154user367154
1
1
add a comment |
add a comment |
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