Proving this is a linear subspace
up vote
5
down vote
favorite
I need to prove the following is a subspace:
Let $V$ be a set of vectors over $F=mathbb{R}$, $V=operatorname{Functions}(mathbb{R} ,mathbb{R})$ and $W$ is a subgroup of $V$ such that $$W={fin V| , f(x)=f(-x)}$$
I'm not sure about the "close under vector addition" part.
My solution: Let be $w_1 , w_2 in W. w_1={fin V| , f(x_1)=f(-x_1)},,,,w_2={fin V| , f(x_2)=f(-x_2)}$. Therefore:
$$w_1+w_2=f(x_1+x_2)=f(x_1)+f(x_2)=f(-x_1)+f(-x_2)=fbig(-(x_1+x_2)big)\
$$
Please tell me if something is wrong here (I feel like something is wrong...). Many thanks for you time and effort.
linear-algebra
edited Apr 17 '13 at 18:20
amWhy
191k28224439
asked Apr 17 '13 at 18:03
yuvalz
357112
add a comment |
up vote
5
down vote
favorite
I need to prove the following is a subspace:
Let $V$ be a set of vectors over $F=mathbb{R}$, $V=operatorname{Functions}(mathbb{R} ,mathbb{R})$ and $W$ is a subgroup of $V$ such that $$W={fin V| , f(x)=f(-x)}$$
I'm not sure about the "close under vector addition" part.
My solution: Let be $w_1 , w_2 in W. w_1={fin V| , f(x_1)=f(-x_1)},,,,w_2={fin V| , f(x_2)=f(-x_2)}$. Therefore:
$$w_1+w_2=f(x_1+x_2)=f(x_1)+f(x_2)=f(-x_1)+f(-x_2)=fbig(-(x_1+x_2)big)\
$$
Please tell me if something is wrong here (I feel like something is wrong...). Many thanks for you time and effort.
linear-algebra
edited Apr 17 '13 at 18:20
amWhy
191k28224439
asked Apr 17 '13 at 18:03
yuvalz
357112
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I need to prove the following is a subspace:
Let $V$ be a set of vectors over $F=mathbb{R}$, $V=operatorname{Functions}(mathbb{R} ,mathbb{R})$ and $W$ is a subgroup of $V$ such that $$W={fin V| , f(x)=f(-x)}$$
I'm not sure about the "close under vector addition" part.
My solution: Let be $w_1 , w_2 in W. w_1={fin V| , f(x_1)=f(-x_1)},,,,w_2={fin V| , f(x_2)=f(-x_2)}$. Therefore:
$$w_1+w_2=f(x_1+x_2)=f(x_1)+f(x_2)=f(-x_1)+f(-x_2)=fbig(-(x_1+x_2)big)\
$$
Please tell me if something is wrong here (I feel like something is wrong...). Many thanks for you time and effort.
linear-algebra
edited Apr 17 '13 at 18:20
amWhy
191k28224439
asked Apr 17 '13 at 18:03
yuvalz
357112
I need to prove the following is a subspace:
Let $V$ be a set of vectors over $F=mathbb{R}$, $V=operatorname{Functions}(mathbb{R} ,mathbb{R})$ and $W$ is a subgroup of $V$ such that $$W={fin V| , f(x)=f(-x)}$$
I'm not sure about the "close under vector addition" part.
My solution: Let be $w_1 , w_2 in W. w_1={fin V| , f(x_1)=f(-x_1)},,,,w_2={fin V| , f(x_2)=f(-x_2)}$. Therefore:
$$w_1+w_2=f(x_1+x_2)=f(x_1)+f(x_2)=f(-x_1)+f(-x_2)=fbig(-(x_1+x_2)big)\
$$
Please tell me if something is wrong here (I feel like something is wrong...). Many thanks for you time and effort.
linear-algebra
linear-algebra
edited Apr 17 '13 at 18:20
amWhy
191k28224439
asked Apr 17 '13 at 18:03
yuvalz
357112
edited Apr 17 '13 at 18:20
amWhy
191k28224439
asked Apr 17 '13 at 18:03
yuvalz
357112
edited Apr 17 '13 at 18:20
amWhy
191k28224439
edited Apr 17 '13 at 18:20
amWhy
191k28224439
edited Apr 17 '13 at 18:20
amWhy
191k28224439
191k28224439
asked Apr 17 '13 at 18:03
yuvalz
357112
asked Apr 17 '13 at 18:03
yuvalz
357112
asked Apr 17 '13 at 18:03
yuvalz
357112
357112
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Your thinking about $W$ is muddled, as evidenced when you refer to $w_1$ and $w_2$ as sets. You may as well call them $f_1$ and $f_2$ because they are functions, and they are the elements of $W$ (vectors) that you are working with.
There is no reason to index the $x$: it is just an arbitrary element of $Bbb R$. I suspect you have fallen into a common misunderstanding among beginners. The fact is that the functions are the elements of the vector space. (Whereas beginners sometimes are overly attached to the $x$ being an element of a vector space.)
By virtue of being in $W$, both of them have the property that
$f_i(x)=f_i(-x)$. Then
$$
(f_1+f_2)(x)=dots=(f_1+f_2)(-x)
$$
proving that $f_1+f_2$ is also a member of $W$. (I omitted the middle computation so you could puzzle it out :) )
See if you can do the closure under scalars now: the goal is to show that if $lambdain Bbb R$, if $fin W$, then $lambda fin W$. Finally, show that $0in W$, "0" here denoting the function that is constantly 0 on $Bbb R$.
In words, what you are doing is showing that the even functions on $Bbb R$ form a subspace. As a followup exercise, you could additionally show that the odd functions also form a subspace of $V$.
answered Apr 17 '13 at 18:09
rschwieb
104k1299241
Is that the missing part?: $$(f_1 + f_2)(x)=f_1(x)+f_2(x)=f_1(-x)+f_2(-x)=(f_1+f_2)(-x)$$ Also, for the scalar ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
– yuvalz
Apr 17 '13 at 18:32
@yuvalz You are perfectly right on the first part... but the second part has a mistake. There is no reason to believe $alpha f(x)=f(alpha x)$. It should look like this: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
– rschwieb
Apr 17 '13 at 18:48
add a comment |
up vote
0
down vote
Your understanding is a little off. Here, a vector in $V$ is a function $f:mathbb{R}tomathbb{R}$, and those in $W$ are exactly the functions satisfying $f(x)=f(-x)$. Such functions are called "even." Now suppose that $f_1$ and $f_2$ are two even functions. See if you can show that the function $f_1+f_2$ defined by $(f_1+f_2)(x)=f_1(x)+f_2(x)$ is also even.
answered Apr 17 '13 at 18:12
Jared
24.1k104599
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f364703%2fproving-this-is-a-linear-subspace%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your thinking about $W$ is muddled, as evidenced when you refer to $w_1$ and $w_2$ as sets. You may as well call them $f_1$ and $f_2$ because they are functions, and they are the elements of $W$ (vectors) that you are working with.
There is no reason to index the $x$: it is just an arbitrary element of $Bbb R$. I suspect you have fallen into a common misunderstanding among beginners. The fact is that the functions are the elements of the vector space. (Whereas beginners sometimes are overly attached to the $x$ being an element of a vector space.)
By virtue of being in $W$, both of them have the property that
$f_i(x)=f_i(-x)$. Then
$$
(f_1+f_2)(x)=dots=(f_1+f_2)(-x)
$$
proving that $f_1+f_2$ is also a member of $W$. (I omitted the middle computation so you could puzzle it out :) )
See if you can do the closure under scalars now: the goal is to show that if $lambdain Bbb R$, if $fin W$, then $lambda fin W$. Finally, show that $0in W$, "0" here denoting the function that is constantly 0 on $Bbb R$.
In words, what you are doing is showing that the even functions on $Bbb R$ form a subspace. As a followup exercise, you could additionally show that the odd functions also form a subspace of $V$.
answered Apr 17 '13 at 18:09
rschwieb
104k1299241
Is that the missing part?: $$(f_1 + f_2)(x)=f_1(x)+f_2(x)=f_1(-x)+f_2(-x)=(f_1+f_2)(-x)$$ Also, for the scalar ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
– yuvalz
Apr 17 '13 at 18:32
@yuvalz You are perfectly right on the first part... but the second part has a mistake. There is no reason to believe $alpha f(x)=f(alpha x)$. It should look like this: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
– rschwieb
Apr 17 '13 at 18:48
add a comment |
up vote
2
down vote
accepted
Your thinking about $W$ is muddled, as evidenced when you refer to $w_1$ and $w_2$ as sets. You may as well call them $f_1$ and $f_2$ because they are functions, and they are the elements of $W$ (vectors) that you are working with.
There is no reason to index the $x$: it is just an arbitrary element of $Bbb R$. I suspect you have fallen into a common misunderstanding among beginners. The fact is that the functions are the elements of the vector space. (Whereas beginners sometimes are overly attached to the $x$ being an element of a vector space.)
By virtue of being in $W$, both of them have the property that
$f_i(x)=f_i(-x)$. Then
$$
(f_1+f_2)(x)=dots=(f_1+f_2)(-x)
$$
proving that $f_1+f_2$ is also a member of $W$. (I omitted the middle computation so you could puzzle it out :) )
See if you can do the closure under scalars now: the goal is to show that if $lambdain Bbb R$, if $fin W$, then $lambda fin W$. Finally, show that $0in W$, "0" here denoting the function that is constantly 0 on $Bbb R$.
In words, what you are doing is showing that the even functions on $Bbb R$ form a subspace. As a followup exercise, you could additionally show that the odd functions also form a subspace of $V$.
answered Apr 17 '13 at 18:09
rschwieb
104k1299241
Is that the missing part?: $$(f_1 + f_2)(x)=f_1(x)+f_2(x)=f_1(-x)+f_2(-x)=(f_1+f_2)(-x)$$ Also, for the scalar ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
– yuvalz
Apr 17 '13 at 18:32
@yuvalz You are perfectly right on the first part... but the second part has a mistake. There is no reason to believe $alpha f(x)=f(alpha x)$. It should look like this: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
– rschwieb
Apr 17 '13 at 18:48
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your thinking about $W$ is muddled, as evidenced when you refer to $w_1$ and $w_2$ as sets. You may as well call them $f_1$ and $f_2$ because they are functions, and they are the elements of $W$ (vectors) that you are working with.
There is no reason to index the $x$: it is just an arbitrary element of $Bbb R$. I suspect you have fallen into a common misunderstanding among beginners. The fact is that the functions are the elements of the vector space. (Whereas beginners sometimes are overly attached to the $x$ being an element of a vector space.)
By virtue of being in $W$, both of them have the property that
$f_i(x)=f_i(-x)$. Then
$$
(f_1+f_2)(x)=dots=(f_1+f_2)(-x)
$$
proving that $f_1+f_2$ is also a member of $W$. (I omitted the middle computation so you could puzzle it out :) )
See if you can do the closure under scalars now: the goal is to show that if $lambdain Bbb R$, if $fin W$, then $lambda fin W$. Finally, show that $0in W$, "0" here denoting the function that is constantly 0 on $Bbb R$.
In words, what you are doing is showing that the even functions on $Bbb R$ form a subspace. As a followup exercise, you could additionally show that the odd functions also form a subspace of $V$.
answered Apr 17 '13 at 18:09
rschwieb
104k1299241
Your thinking about $W$ is muddled, as evidenced when you refer to $w_1$ and $w_2$ as sets. You may as well call them $f_1$ and $f_2$ because they are functions, and they are the elements of $W$ (vectors) that you are working with.
There is no reason to index the $x$: it is just an arbitrary element of $Bbb R$. I suspect you have fallen into a common misunderstanding among beginners. The fact is that the functions are the elements of the vector space. (Whereas beginners sometimes are overly attached to the $x$ being an element of a vector space.)
By virtue of being in $W$, both of them have the property that
$f_i(x)=f_i(-x)$. Then
$$
(f_1+f_2)(x)=dots=(f_1+f_2)(-x)
$$
proving that $f_1+f_2$ is also a member of $W$. (I omitted the middle computation so you could puzzle it out :) )
See if you can do the closure under scalars now: the goal is to show that if $lambdain Bbb R$, if $fin W$, then $lambda fin W$. Finally, show that $0in W$, "0" here denoting the function that is constantly 0 on $Bbb R$.
In words, what you are doing is showing that the even functions on $Bbb R$ form a subspace. As a followup exercise, you could additionally show that the odd functions also form a subspace of $V$.
answered Apr 17 '13 at 18:09
rschwieb
104k1299241
edited Apr 17 '13 at 18:18
answered Apr 17 '13 at 18:09
rschwieb
104k1299241
answered Apr 17 '13 at 18:09
rschwieb
104k1299241
answered Apr 17 '13 at 18:09
rschwieb
104k1299241
104k1299241
Is that the missing part?: $$(f_1 + f_2)(x)=f_1(x)+f_2(x)=f_1(-x)+f_2(-x)=(f_1+f_2)(-x)$$ Also, for the scalar ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
– yuvalz
Apr 17 '13 at 18:32
@yuvalz You are perfectly right on the first part... but the second part has a mistake. There is no reason to believe $alpha f(x)=f(alpha x)$. It should look like this: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
– rschwieb
Apr 17 '13 at 18:48
add a comment |
Is that the missing part?: $$(f_1 + f_2)(x)=f_1(x)+f_2(x)=f_1(-x)+f_2(-x)=(f_1+f_2)(-x)$$ Also, for the scalar ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
– yuvalz
Apr 17 '13 at 18:32
@yuvalz You are perfectly right on the first part... but the second part has a mistake. There is no reason to believe $alpha f(x)=f(alpha x)$. It should look like this: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
– rschwieb
Apr 17 '13 at 18:48
Is that the missing part?: $$(f_1 + f_2)(x)=f_1(x)+f_2(x)=f_1(-x)+f_2(-x)=(f_1+f_2)(-x)$$ Also, for the scalar ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
– yuvalz
Apr 17 '13 at 18:32
Is that the missing part?: $$(f_1 + f_2)(x)=f_1(x)+f_2(x)=f_1(-x)+f_2(-x)=(f_1+f_2)(-x)$$ Also, for the scalar ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
– yuvalz
Apr 17 '13 at 18:32
@yuvalz You are perfectly right on the first part... but the second part has a mistake. There is no reason to believe $alpha f(x)=f(alpha x)$. It should look like this: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
– rschwieb
Apr 17 '13 at 18:48
@yuvalz You are perfectly right on the first part... but the second part has a mistake. There is no reason to believe $alpha f(x)=f(alpha x)$. It should look like this: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
– rschwieb
Apr 17 '13 at 18:48
add a comment |
up vote
0
down vote
Your understanding is a little off. Here, a vector in $V$ is a function $f:mathbb{R}tomathbb{R}$, and those in $W$ are exactly the functions satisfying $f(x)=f(-x)$. Such functions are called "even." Now suppose that $f_1$ and $f_2$ are two even functions. See if you can show that the function $f_1+f_2$ defined by $(f_1+f_2)(x)=f_1(x)+f_2(x)$ is also even.
answered Apr 17 '13 at 18:12
Jared
24.1k104599
add a comment |
up vote
0
down vote
Your understanding is a little off. Here, a vector in $V$ is a function $f:mathbb{R}tomathbb{R}$, and those in $W$ are exactly the functions satisfying $f(x)=f(-x)$. Such functions are called "even." Now suppose that $f_1$ and $f_2$ are two even functions. See if you can show that the function $f_1+f_2$ defined by $(f_1+f_2)(x)=f_1(x)+f_2(x)$ is also even.
answered Apr 17 '13 at 18:12
Jared
24.1k104599
add a comment |
up vote
0
down vote
up vote
0
down vote
Your understanding is a little off. Here, a vector in $V$ is a function $f:mathbb{R}tomathbb{R}$, and those in $W$ are exactly the functions satisfying $f(x)=f(-x)$. Such functions are called "even." Now suppose that $f_1$ and $f_2$ are two even functions. See if you can show that the function $f_1+f_2$ defined by $(f_1+f_2)(x)=f_1(x)+f_2(x)$ is also even.
answered Apr 17 '13 at 18:12
Jared
24.1k104599
Your understanding is a little off. Here, a vector in $V$ is a function $f:mathbb{R}tomathbb{R}$, and those in $W$ are exactly the functions satisfying $f(x)=f(-x)$. Such functions are called "even." Now suppose that $f_1$ and $f_2$ are two even functions. See if you can show that the function $f_1+f_2$ defined by $(f_1+f_2)(x)=f_1(x)+f_2(x)$ is also even.
answered Apr 17 '13 at 18:12
Jared
24.1k104599
edited Apr 19 '13 at 0:12
answered Apr 17 '13 at 18:12
Jared
24.1k104599
answered Apr 17 '13 at 18:12
Jared
24.1k104599
answered Apr 17 '13 at 18:12
Jared
24.1k104599
24.1k104599
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f364703%2fproving-this-is-a-linear-subspace%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
