Let $ $ be a subspace of $ $ and let $ v ∈ S $ such that $ v ≠ 0 $, then, for every $ λ|λ∈ℝ $...












1















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







share|cite|improve this question


















  • 4




    Isn't that just the definition of subspace? I don't understand the question
    – Randall
    yesterday










  • I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
    – ItsJustASeriesBro
    yesterday










  • 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).
    – Daniel Bonilla Jaramillo
    yesterday












  • @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.
    – ItsJustASeriesBro
    yesterday


















1















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







share|cite|improve this question


















  • 4




    Isn't that just the definition of subspace? I don't understand the question
    – Randall
    yesterday










  • I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
    – ItsJustASeriesBro
    yesterday










  • 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).
    – Daniel Bonilla Jaramillo
    yesterday












  • @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.
    – ItsJustASeriesBro
    yesterday
















1












1








1








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







share|cite|improve this question














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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









Daniel Bonilla Jaramillo

464310




464310








  • 4




    Isn't that just the definition of subspace? I don't understand the question
    – Randall
    yesterday










  • I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
    – ItsJustASeriesBro
    yesterday










  • 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).
    – Daniel Bonilla Jaramillo
    yesterday












  • @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.
    – ItsJustASeriesBro
    yesterday
















  • 4




    Isn't that just the definition of subspace? I don't understand the question
    – Randall
    yesterday










  • I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
    – ItsJustASeriesBro
    yesterday










  • 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).
    – Daniel Bonilla Jaramillo
    yesterday












  • @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.
    – ItsJustASeriesBro
    yesterday










4




4




Isn't that just the definition of subspace? I don't understand the question
– Randall
yesterday




Isn't that just the definition of subspace? I don't understand the question
– Randall
yesterday












I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
– ItsJustASeriesBro
yesterday




I am also unclear as to why we wanted $vneq 0$. It is certainly true in that case as well.
– ItsJustASeriesBro
yesterday












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).
– Daniel Bonilla Jaramillo
yesterday






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).
– Daniel Bonilla Jaramillo
yesterday














@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.
– ItsJustASeriesBro
yesterday






@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.
– ItsJustASeriesBro
yesterday












1 Answer
1






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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 $0cdot v+lambdacdot v=lambdacdot vin S$.






share|cite|improve this answer























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    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 $0cdot v+lambdacdot v=lambdacdot vin S$.






    share|cite|improve this answer




























      3














      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 $0cdot v+lambdacdot v=lambdacdot vin S$.






      share|cite|improve this answer


























        3












        3








        3






        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 $0cdot v+lambdacdot v=lambdacdot vin S$.






        share|cite|improve this answer














        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 $0cdot v+lambdacdot v=lambdacdot vin S$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 15 hours ago

























        answered yesterday









        John Wayland Bales

        13.9k21237




        13.9k21237






























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