Proving this is a linear subspace
$begingroup$
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
192k28225439
asked Apr 17 '13 at 18:03
yuvalzyuvalz
357112
$endgroup$
add a comment |
$begingroup$
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
192k28225439
asked Apr 17 '13 at 18:03
yuvalzyuvalz
357112
$endgroup$
add a comment |
$begingroup$
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
192k28225439
asked Apr 17 '13 at 18:03
yuvalzyuvalz
357112
$endgroup$
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
192k28225439
asked Apr 17 '13 at 18:03
yuvalzyuvalz
357112
edited Apr 17 '13 at 18:20
amWhy
192k28225439
asked Apr 17 '13 at 18:03
yuvalzyuvalz
357112
edited Apr 17 '13 at 18:20
amWhy
192k28225439
edited Apr 17 '13 at 18:20
amWhy
192k28225439
edited Apr 17 '13 at 18:20
amWhy
192k28225439
192k28225439
asked Apr 17 '13 at 18:03
yuvalzyuvalz
357112
asked Apr 17 '13 at 18:03
yuvalzyuvalz
357112
asked Apr 17 '13 at 18:03
yuvalzyuvalz
357112
357112
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
rschwiebrschwieb
105k12100245
$endgroup$
$begingroup$
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 ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
$endgroup$
– yuvalz
Apr 17 '13 at 18:32
$begingroup$
@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: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
$endgroup$
– rschwieb
Apr 17 '13 at 18:48
add a comment |
$begingroup$
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
JaredJared
24.1k104599
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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
rschwiebrschwieb
105k12100245
$endgroup$
$begingroup$
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 ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
$endgroup$
– yuvalz
Apr 17 '13 at 18:32
$begingroup$
@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: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
$endgroup$
– rschwieb
Apr 17 '13 at 18:48
add a comment |
$begingroup$
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
rschwiebrschwieb
105k12100245
$endgroup$
$begingroup$
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 ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
$endgroup$
– yuvalz
Apr 17 '13 at 18:32
$begingroup$
@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: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
$endgroup$
– rschwieb
Apr 17 '13 at 18:48
add a comment |
$begingroup$
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
rschwiebrschwieb
105k12100245
$endgroup$
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
rschwiebrschwieb
105k12100245
edited Apr 17 '13 at 18:18
answered Apr 17 '13 at 18:09
rschwiebrschwieb
105k12100245
answered Apr 17 '13 at 18:09
rschwiebrschwieb
105k12100245
answered Apr 17 '13 at 18:09
rschwiebrschwieb
105k12100245
105k12100245
$begingroup$
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 ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
$endgroup$
– yuvalz
Apr 17 '13 at 18:32
$begingroup$
@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: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
$endgroup$
– rschwieb
Apr 17 '13 at 18:48
add a comment |
$begingroup$
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 ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
$endgroup$
– yuvalz
Apr 17 '13 at 18:32
$begingroup$
@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: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
$endgroup$
– rschwieb
Apr 17 '13 at 18:48
$begingroup$
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 ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
$endgroup$
– yuvalz
Apr 17 '13 at 18:32
$begingroup$
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 ($alpha in mathbb{R}$): $$alpha f(x)=f(alpha x) = f(-alpha x)=alpha f(-x)$$
$endgroup$
– yuvalz
Apr 17 '13 at 18:32
$begingroup$
@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: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
$endgroup$
– rschwieb
Apr 17 '13 at 18:48
$begingroup$
@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: $(alpha f)(x):=alpha cdot f(x)=dots=(alpha f)(-x)$.
$endgroup$
– rschwieb
Apr 17 '13 at 18:48
add a comment |
$begingroup$
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
JaredJared
24.1k104599
$endgroup$
add a comment |
$begingroup$
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
JaredJared
24.1k104599
$endgroup$
add a comment |
$begingroup$
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
JaredJared
24.1k104599
$endgroup$
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
JaredJared
24.1k104599
edited Apr 19 '13 at 0:12
answered Apr 17 '13 at 18:12
JaredJared
24.1k104599
answered Apr 17 '13 at 18:12
JaredJared
24.1k104599
answered Apr 17 '13 at 18:12
JaredJared
24.1k104599
24.1k104599
add a comment |
add a comment |
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