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






active

oldest

votes


















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























    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',
    autoActivateHeartbeat: false,
    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060236%2flet-s-be-a-subspace-of-%25e2%2584%259d-2-and-let-v-%25e2%2588%2588-s-such%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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














      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






























            draft saved

            draft discarded




















































            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060236%2flet-s-be-a-subspace-of-%25e2%2584%259d-2-and-let-v-%25e2%2588%2588-s-such%23new-answer', 'question_page');
            }
            );

            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







            -

            Popular posts from this blog

            An IMO inspired problem

            Image
            14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
            Read more

            Management

            Image
            Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
            Read more

            Has there ever been an instance of an active nuclear power plant within or near a war zone?

            Image
            36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
            Read more