Mixing
Let $ $ be a subspace of $ $ and let $ v ∈ S $ such that $ v ≠ 0 $, then, for every $ λ|λ∈ℝ $...
$begingroup$
Let $ < S, +, · > $ be a subspace of $ < ℝ ^ 2, +, · > $ and let $ v ∈
S $ such that $ v ≠ 0 $, then, for every $ λ|λ∈ℝ $ it is true that $
λ·v ∈ S $
I have almost completed the proof, but I'm stuck in, what I suppose is, the final part.
From the hypothesis we know $ < S, +, · > $ is a subspace of $ < ℝ ^ 2, +, · > $, which is an $ℝ$-vector space, so we have $S ⊆ ℝ ^ 2$, $ S ≠ ∅ $ and $ < S, +, · > $ an $ℝ$-vector space.
Suppose $ v = ( a, b ) | a, b ≠ 0 $ and $ λ∈ℝ $, then, since $·$ is defined as
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
This is where doubts begin to arise; according to the definition of subspace, if S is a subspace of K, then S has the same operations as K, but I don't know if that includes the sets on which this operations are defined; I'll explain further: in this case $·$ is defined in $ℝ ^ 2$ as I just mentioned:
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
does that mean that it is defined exactly like that for S, or the sets over which it is defined change like this? :
$$ ·: S×S ^ 2 ↦ S ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
I assumed it stayed like
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
and followed the proof ( however, I need to confirm if this is correct ).
then $ λ·v = λ·(a,b) = (λa,λb) $ and, because ℝ is closed under multiplication, then $λa,λb ∈ ℝ $ and $(λa,λb) ∈ ℝ^2 $, which is the same as $ λ·(a,b) ∈ ℝ^2 $ or $ λ·v ∈ ℝ^2 $
Here's where I get completely stuck, I already proved $λ·v ∈ ℝ^2$ but I still have to prove $λ·v ∈ S$, and I haven't figured out how to do so.
Any recommendation?
Thanks in advance.
linear-algebra abstract-algebra vector-spaces
asked Jan 3 at 3:55
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
$endgroup$
add a comment |
$begingroup$
Let $ < S, +, · > $ be a subspace of $ < ℝ ^ 2, +, · > $ and let $ v ∈
S $ such that $ v ≠ 0 $, then, for every $ λ|λ∈ℝ $ it is true that $
λ·v ∈ S $
I have almost completed the proof, but I'm stuck in, what I suppose is, the final part.
From the hypothesis we know $ < S, +, · > $ is a subspace of $ < ℝ ^ 2, +, · > $, which is an $ℝ$-vector space, so we have $S ⊆ ℝ ^ 2$, $ S ≠ ∅ $ and $ < S, +, · > $ an $ℝ$-vector space.
Suppose $ v = ( a, b ) | a, b ≠ 0 $ and $ λ∈ℝ $, then, since $·$ is defined as
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
This is where doubts begin to arise; according to the definition of subspace, if S is a subspace of K, then S has the same operations as K, but I don't know if that includes the sets on which this operations are defined; I'll explain further: in this case $·$ is defined in $ℝ ^ 2$ as I just mentioned:
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
does that mean that it is defined exactly like that for S, or the sets over which it is defined change like this? :
$$ ·: S×S ^ 2 ↦ S ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
I assumed it stayed like
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
and followed the proof ( however, I need to confirm if this is correct ).
then $ λ·v = λ·(a,b) = (λa,λb) $ and, because ℝ is closed under multiplication, then $λa,λb ∈ ℝ $ and $(λa,λb) ∈ ℝ^2 $, which is the same as $ λ·(a,b) ∈ ℝ^2 $ or $ λ·v ∈ ℝ^2 $
Here's where I get completely stuck, I already proved $λ·v ∈ ℝ^2$ but I still have to prove $λ·v ∈ S$, and I haven't figured out how to do so.
Any recommendation?
Thanks in advance.
linear-algebra abstract-algebra vector-spaces
asked Jan 3 at 3:55
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
$endgroup$
4
$begingroup$
Isn't that just the definition of subspace? I don't understand the question
$endgroup$
– Randall
Jan 3 at 3:56
$begingroup$
I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:05
$begingroup$
It's just my book that is asking me to prove this. It also says that if those elements of the form $λ·v$ are all the members of S, then the vector space S is graphically a line that passes through (0, 0).
$endgroup$
– Daniel Bonilla Jaramillo
Jan 3 at 4:08
$begingroup$
@DanielBonillaJaramillo I don't think your book is asking you to prove this. The definition of a subspace is that it is closed under scalar multiplication. You can't prove a definition. Please verify that you copied the question prompt exactly.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:09
add a comment |
$begingroup$
Let $ < S, +, · > $ be a subspace of $ < ℝ ^ 2, +, · > $ and let $ v ∈
S $ such that $ v ≠ 0 $, then, for every $ λ|λ∈ℝ $ it is true that $
λ·v ∈ S $
I have almost completed the proof, but I'm stuck in, what I suppose is, the final part.
From the hypothesis we know $ < S, +, · > $ is a subspace of $ < ℝ ^ 2, +, · > $, which is an $ℝ$-vector space, so we have $S ⊆ ℝ ^ 2$, $ S ≠ ∅ $ and $ < S, +, · > $ an $ℝ$-vector space.
Suppose $ v = ( a, b ) | a, b ≠ 0 $ and $ λ∈ℝ $, then, since $·$ is defined as
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
This is where doubts begin to arise; according to the definition of subspace, if S is a subspace of K, then S has the same operations as K, but I don't know if that includes the sets on which this operations are defined; I'll explain further: in this case $·$ is defined in $ℝ ^ 2$ as I just mentioned:
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
does that mean that it is defined exactly like that for S, or the sets over which it is defined change like this? :
$$ ·: S×S ^ 2 ↦ S ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
I assumed it stayed like
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
and followed the proof ( however, I need to confirm if this is correct ).
then $ λ·v = λ·(a,b) = (λa,λb) $ and, because ℝ is closed under multiplication, then $λa,λb ∈ ℝ $ and $(λa,λb) ∈ ℝ^2 $, which is the same as $ λ·(a,b) ∈ ℝ^2 $ or $ λ·v ∈ ℝ^2 $
Here's where I get completely stuck, I already proved $λ·v ∈ ℝ^2$ but I still have to prove $λ·v ∈ S$, and I haven't figured out how to do so.
Any recommendation?
Thanks in advance.
linear-algebra abstract-algebra vector-spaces
asked Jan 3 at 3:55
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
$endgroup$
Let $ < S, +, · > $ be a subspace of $ < ℝ ^ 2, +, · > $ and let $ v ∈
S $ such that $ v ≠ 0 $, then, for every $ λ|λ∈ℝ $ it is true that $
λ·v ∈ S $
I have almost completed the proof, but I'm stuck in, what I suppose is, the final part.
From the hypothesis we know $ < S, +, · > $ is a subspace of $ < ℝ ^ 2, +, · > $, which is an $ℝ$-vector space, so we have $S ⊆ ℝ ^ 2$, $ S ≠ ∅ $ and $ < S, +, · > $ an $ℝ$-vector space.
Suppose $ v = ( a, b ) | a, b ≠ 0 $ and $ λ∈ℝ $, then, since $·$ is defined as
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
This is where doubts begin to arise; according to the definition of subspace, if S is a subspace of K, then S has the same operations as K, but I don't know if that includes the sets on which this operations are defined; I'll explain further: in this case $·$ is defined in $ℝ ^ 2$ as I just mentioned:
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
does that mean that it is defined exactly like that for S, or the sets over which it is defined change like this? :
$$ ·: S×S ^ 2 ↦ S ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
I assumed it stayed like
$$ ·: ℝ×ℝ ^ 2 ↦ ℝ ^ 2 $$
$$ [λ, ( a, b )] ↦ λ·(a,b) = (λa, λb) $$
and followed the proof ( however, I need to confirm if this is correct ).
then $ λ·v = λ·(a,b) = (λa,λb) $ and, because ℝ is closed under multiplication, then $λa,λb ∈ ℝ $ and $(λa,λb) ∈ ℝ^2 $, which is the same as $ λ·(a,b) ∈ ℝ^2 $ or $ λ·v ∈ ℝ^2 $
Here's where I get completely stuck, I already proved $λ·v ∈ ℝ^2$ but I still have to prove $λ·v ∈ S$, and I haven't figured out how to do so.
Any recommendation?
Thanks in advance.
linear-algebra abstract-algebra vector-spaces
linear-algebra abstract-algebra vector-spaces
asked Jan 3 at 3:55
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
asked Jan 3 at 3:55
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
asked Jan 3 at 3:55
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
asked Jan 3 at 3:55
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
asked Jan 3 at 3:55
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
464310
4
$begingroup$
Isn't that just the definition of subspace? I don't understand the question
$endgroup$
– Randall
Jan 3 at 3:56
$begingroup$
I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:05
$begingroup$
It's just my book that is asking me to prove this. It also says that if those elements of the form $λ·v$ are all the members of S, then the vector space S is graphically a line that passes through (0, 0).
$endgroup$
– Daniel Bonilla Jaramillo
Jan 3 at 4:08
$begingroup$
@DanielBonillaJaramillo I don't think your book is asking you to prove this. The definition of a subspace is that it is closed under scalar multiplication. You can't prove a definition. Please verify that you copied the question prompt exactly.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:09
add a comment |
4
$begingroup$
Isn't that just the definition of subspace? I don't understand the question
$endgroup$
– Randall
Jan 3 at 3:56
$begingroup$
I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:05
$begingroup$
It's just my book that is asking me to prove this. It also says that if those elements of the form $λ·v$ are all the members of S, then the vector space S is graphically a line that passes through (0, 0).
$endgroup$
– Daniel Bonilla Jaramillo
Jan 3 at 4:08
$begingroup$
@DanielBonillaJaramillo I don't think your book is asking you to prove this. The definition of a subspace is that it is closed under scalar multiplication. You can't prove a definition. Please verify that you copied the question prompt exactly.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:09
4
4
$begingroup$
Isn't that just the definition of subspace? I don't understand the question
$endgroup$
– Randall
Jan 3 at 3:56
$begingroup$
Isn't that just the definition of subspace? I don't understand the question
$endgroup$
– Randall
Jan 3 at 3:56
$begingroup$
I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:05
$begingroup$
I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:05
$begingroup$
It's just my book that is asking me to prove this. It also says that if those elements of the form $λ·v$ are all the members of S, then the vector space S is graphically a line that passes through (0, 0).
$endgroup$
– Daniel Bonilla Jaramillo
Jan 3 at 4:08
$begingroup$
It's just my book that is asking me to prove this. It also says that if those elements of the form $λ·v$ are all the members of S, then the vector space S is graphically a line that passes through (0, 0).
$endgroup$
– Daniel Bonilla Jaramillo
Jan 3 at 4:08
$begingroup$
@DanielBonillaJaramillo I don't think your book is asking you to prove this. The definition of a subspace is that it is closed under scalar multiplication. You can't prove a definition. Please verify that you copied the question prompt exactly.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:09
$begingroup$
@DanielBonillaJaramillo I don't think your book is asking you to prove this. The definition of a subspace is that it is closed under scalar multiplication. You can't prove a definition. Please verify that you copied the question prompt exactly.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The statement that $S$ is a subspace of $mathbb{R}$ means that if $x_1,x_2in S$ and $lambda_1,lambda_2inmathbb{R}$ then $lambda_1cdot x_1+lambda_2cdot x_2in S$.
Thus if $lambdainmathbb{R}$ and $vin S$ it follows that $(-lambda)cdot v+lambdacdot v=0in S$ and therefore $lambdacdot 0+lambdacdot v=lambdacdot vin S$
Note: In this context, of course, $0$ is $(0,0)$.
ADDENDUM: My answer above and OP's approach are both examples of "overthinking a problem."
Since $0,lambdainmathbb{R}$ and $vin S$ it follows that $0cdot v+lambdacdot v=lambdacdot vin S$.
answered Jan 3 at 5:08
John Wayland BalesJohn Wayland Bales
13.9k21238
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The statement that $S$ is a subspace of $mathbb{R}$ means that if $x_1,x_2in S$ and $lambda_1,lambda_2inmathbb{R}$ then $lambda_1cdot x_1+lambda_2cdot x_2in S$.
Thus if $lambdainmathbb{R}$ and $vin S$ it follows that $(-lambda)cdot v+lambdacdot v=0in S$ and therefore $lambdacdot 0+lambdacdot v=lambdacdot vin S$
Note: In this context, of course, $0$ is $(0,0)$.
ADDENDUM: My answer above and OP's approach are both examples of "overthinking a problem."
Since $0,lambdainmathbb{R}$ and $vin S$ it follows that $0cdot v+lambdacdot v=lambdacdot vin S$.
answered Jan 3 at 5:08
John Wayland BalesJohn Wayland Bales
13.9k21238
$endgroup$
add a comment |
$begingroup$
The statement that $S$ is a subspace of $mathbb{R}$ means that if $x_1,x_2in S$ and $lambda_1,lambda_2inmathbb{R}$ then $lambda_1cdot x_1+lambda_2cdot x_2in S$.
Thus if $lambdainmathbb{R}$ and $vin S$ it follows that $(-lambda)cdot v+lambdacdot v=0in S$ and therefore $lambdacdot 0+lambdacdot v=lambdacdot vin S$
Note: In this context, of course, $0$ is $(0,0)$.
ADDENDUM: My answer above and OP's approach are both examples of "overthinking a problem."
Since $0,lambdainmathbb{R}$ and $vin S$ it follows that $0cdot v+lambdacdot v=lambdacdot vin S$.
answered Jan 3 at 5:08
John Wayland BalesJohn Wayland Bales
13.9k21238
$endgroup$
add a comment |
$begingroup$
The statement that $S$ is a subspace of $mathbb{R}$ means that if $x_1,x_2in S$ and $lambda_1,lambda_2inmathbb{R}$ then $lambda_1cdot x_1+lambda_2cdot x_2in S$.
Thus if $lambdainmathbb{R}$ and $vin S$ it follows that $(-lambda)cdot v+lambdacdot v=0in S$ and therefore $lambdacdot 0+lambdacdot v=lambdacdot vin S$
Note: In this context, of course, $0$ is $(0,0)$.
ADDENDUM: My answer above and OP's approach are both examples of "overthinking a problem."
Since $0,lambdainmathbb{R}$ and $vin S$ it follows that $0cdot v+lambdacdot v=lambdacdot vin S$.
answered Jan 3 at 5:08
John Wayland BalesJohn Wayland Bales
13.9k21238
$endgroup$
The statement that $S$ is a subspace of $mathbb{R}$ means that if $x_1,x_2in S$ and $lambda_1,lambda_2inmathbb{R}$ then $lambda_1cdot x_1+lambda_2cdot x_2in S$.
Thus if $lambdainmathbb{R}$ and $vin S$ it follows that $(-lambda)cdot v+lambdacdot v=0in S$ and therefore $lambdacdot 0+lambdacdot v=lambdacdot vin S$
Note: In this context, of course, $0$ is $(0,0)$.
ADDENDUM: My answer above and OP's approach are both examples of "overthinking a problem."
Since $0,lambdainmathbb{R}$ and $vin S$ it follows that $0cdot v+lambdacdot v=lambdacdot vin S$.
answered Jan 3 at 5:08
John Wayland BalesJohn Wayland Bales
13.9k21238
edited Jan 3 at 15:23
answered Jan 3 at 5:08
John Wayland BalesJohn Wayland Bales
13.9k21238
answered Jan 3 at 5:08
John Wayland BalesJohn Wayland Bales
13.9k21238
answered Jan 3 at 5:08
John Wayland BalesJohn Wayland Bales
13.9k21238
13.9k21238
add a comment |
add a comment |
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4
$begingroup$
Isn't that just the definition of subspace? I don't understand the question
$endgroup$
– Randall
Jan 3 at 3:56
$begingroup$
I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:05
$begingroup$
It's just my book that is asking me to prove this. It also says that if those elements of the form $λ·v$ are all the members of S, then the vector space S is graphically a line that passes through (0, 0).
$endgroup$
– Daniel Bonilla Jaramillo
Jan 3 at 4:08
$begingroup$
@DanielBonillaJaramillo I don't think your book is asking you to prove this. The definition of a subspace is that it is closed under scalar multiplication. You can't prove a definition. Please verify that you copied the question prompt exactly.
$endgroup$
– ItsJustASeriesBro
Jan 3 at 4:09