Prove a set of functions is a sub space
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Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The set of functions $f:mathbb{R}→mathbb{R}$ so that $f(-1) + f(1) = 0$ are an example of subspace?
I know I have to prove 0 is in the set, the sum of two vectors (functions) is in the set and the scalar product is in the set.
I have 9 different groups to check whether they are sub spaces or not, but really don't know the method to proceed with the question, so I would like this as an example.
Thanks!
linear-algebra functions
asked Nov 22 at 20:21
Tegernako
596
add a comment |
up vote
0
down vote
favorite
Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The set of functions $f:mathbb{R}→mathbb{R}$ so that $f(-1) + f(1) = 0$ are an example of subspace?
I know I have to prove 0 is in the set, the sum of two vectors (functions) is in the set and the scalar product is in the set.
I have 9 different groups to check whether they are sub spaces or not, but really don't know the method to proceed with the question, so I would like this as an example.
Thanks!
linear-algebra functions
asked Nov 22 at 20:21
Tegernako
596
If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
– JMoravitz
Nov 22 at 20:28
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The set of functions $f:mathbb{R}→mathbb{R}$ so that $f(-1) + f(1) = 0$ are an example of subspace?
I know I have to prove 0 is in the set, the sum of two vectors (functions) is in the set and the scalar product is in the set.
I have 9 different groups to check whether they are sub spaces or not, but really don't know the method to proceed with the question, so I would like this as an example.
Thanks!
linear-algebra functions
asked Nov 22 at 20:21
Tegernako
596
Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The set of functions $f:mathbb{R}→mathbb{R}$ so that $f(-1) + f(1) = 0$ are an example of subspace?
I know I have to prove 0 is in the set, the sum of two vectors (functions) is in the set and the scalar product is in the set.
I have 9 different groups to check whether they are sub spaces or not, but really don't know the method to proceed with the question, so I would like this as an example.
Thanks!
linear-algebra functions
linear-algebra functions
asked Nov 22 at 20:21
Tegernako
596
asked Nov 22 at 20:21
Tegernako
596
asked Nov 22 at 20:21
Tegernako
596
asked Nov 22 at 20:21
Tegernako
596
asked Nov 22 at 20:21
Tegernako
596
596
If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
– JMoravitz
Nov 22 at 20:28
add a comment |
If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
– JMoravitz
Nov 22 at 20:28
If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
– JMoravitz
Nov 22 at 20:28
If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
– JMoravitz
Nov 22 at 20:28
add a comment |
2 Answers
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Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace
answered Nov 22 at 21:29
Mostafa Ayaz
13.6k3836
I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
– Tegernako
Nov 23 at 6:28
That's right. Furthermore $0notin S$ which is another argument ....
– Mostafa Ayaz
Nov 23 at 7:26
add a comment |
up vote
0
down vote
$f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.
answered Nov 22 at 20:27
John_Wick
1,199111
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
0
down vote
accepted
Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace
answered Nov 22 at 21:29
Mostafa Ayaz
13.6k3836
I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
– Tegernako
Nov 23 at 6:28
That's right. Furthermore $0notin S$ which is another argument ....
– Mostafa Ayaz
Nov 23 at 7:26
add a comment |
up vote
0
down vote
accepted
Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace
answered Nov 22 at 21:29
Mostafa Ayaz
13.6k3836
I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
– Tegernako
Nov 23 at 6:28
That's right. Furthermore $0notin S$ which is another argument ....
– Mostafa Ayaz
Nov 23 at 7:26
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace
answered Nov 22 at 21:29
Mostafa Ayaz
13.6k3836
Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace
answered Nov 22 at 21:29
Mostafa Ayaz
13.6k3836
answered Nov 22 at 21:29
Mostafa Ayaz
13.6k3836
answered Nov 22 at 21:29
Mostafa Ayaz
13.6k3836
answered Nov 22 at 21:29
Mostafa Ayaz
13.6k3836
13.6k3836
I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
– Tegernako
Nov 23 at 6:28
That's right. Furthermore $0notin S$ which is another argument ....
– Mostafa Ayaz
Nov 23 at 7:26
add a comment |
I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
– Tegernako
Nov 23 at 6:28
That's right. Furthermore $0notin S$ which is another argument ....
– Mostafa Ayaz
Nov 23 at 7:26
I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
– Tegernako
Nov 23 at 6:28
I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
– Tegernako
Nov 23 at 6:28
That's right. Furthermore $0notin S$ which is another argument ....
– Mostafa Ayaz
Nov 23 at 7:26
That's right. Furthermore $0notin S$ which is another argument ....
– Mostafa Ayaz
Nov 23 at 7:26
add a comment |
up vote
0
down vote
$f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.
answered Nov 22 at 20:27
John_Wick
1,199111
add a comment |
up vote
0
down vote
$f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.
answered Nov 22 at 20:27
John_Wick
1,199111
add a comment |
up vote
0
down vote
up vote
0
down vote
$f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.
answered Nov 22 at 20:27
John_Wick
1,199111
$f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.
answered Nov 22 at 20:27
John_Wick
1,199111
answered Nov 22 at 20:27
John_Wick
1,199111
answered Nov 22 at 20:27
John_Wick
1,199111
answered Nov 22 at 20:27
John_Wick
1,199111
1,199111
add a comment |
add a comment |
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If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
– JMoravitz
Nov 22 at 20:28