Subspace of a vector space problem












2












$begingroup$


So the task says:



for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?



I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:



$Ax + Cy ∈ U$,



I could not find the answer so i checked the answer.



enter image description here



I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.



Thank you!






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Well if $B$ is not zero then 0 is not in the subspace.
    $endgroup$
    – Calvin Khor
    Jan 20 at 19:34










  • $begingroup$
    Of course, but the OP wanted to understand that specific proof.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 20:13










  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    Jan 20 at 20:35
















2












$begingroup$


So the task says:



for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?



I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:



$Ax + Cy ∈ U$,



I could not find the answer so i checked the answer.



enter image description here



I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.



Thank you!






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Well if $B$ is not zero then 0 is not in the subspace.
    $endgroup$
    – Calvin Khor
    Jan 20 at 19:34










  • $begingroup$
    Of course, but the OP wanted to understand that specific proof.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 20:13










  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    Jan 20 at 20:35














2












2








2





$begingroup$


So the task says:



for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?



I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:



$Ax + Cy ∈ U$,



I could not find the answer so i checked the answer.



enter image description here



I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.



Thank you!






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$




So the task says:



for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?



I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:



$Ax + Cy ∈ U$,



I could not find the answer so i checked the answer.



enter image description here



I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.



Thank you!






linear-algebra vector-spaces




linear-algebra vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 15:02









José Carlos Santos

156k22126227




156k22126227










asked Jan 20 at 19:30









PetarPetar

203




203








  • 2




    $begingroup$
    Well if $B$ is not zero then 0 is not in the subspace.
    $endgroup$
    – Calvin Khor
    Jan 20 at 19:34










  • $begingroup$
    Of course, but the OP wanted to understand that specific proof.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 20:13










  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    Jan 20 at 20:35














  • 2




    $begingroup$
    Well if $B$ is not zero then 0 is not in the subspace.
    $endgroup$
    – Calvin Khor
    Jan 20 at 19:34










  • $begingroup$
    Of course, but the OP wanted to understand that specific proof.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 20:13










  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    Jan 20 at 20:35








2




2




$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
Jan 20 at 19:34




$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
Jan 20 at 19:34












$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
Jan 20 at 20:13




$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
Jan 20 at 20:13












$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:35




$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:35










2 Answers
2






active

oldest

votes


















4












$begingroup$

Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    Jan 20 at 20:05










  • $begingroup$
    If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 21:34












  • $begingroup$
    Ok thank you! I understand it now.
    $endgroup$
    – Petar
    Jan 20 at 21:51










  • $begingroup$
    If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 21:53










  • $begingroup$
    yes i did it now ,tnx
    $endgroup$
    – Petar
    Jan 20 at 22:09



















2












$begingroup$

Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.






share|cite|improve this answer









$endgroup$













    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%2f3081040%2fsubspace-of-a-vector-space-problem%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









    4












    $begingroup$

    Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
      $endgroup$
      – Petar
      Jan 20 at 20:05










    • $begingroup$
      If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
      $endgroup$
      – José Carlos Santos
      Jan 20 at 21:34












    • $begingroup$
      Ok thank you! I understand it now.
      $endgroup$
      – Petar
      Jan 20 at 21:51










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      Jan 20 at 21:53










    • $begingroup$
      yes i did it now ,tnx
      $endgroup$
      – Petar
      Jan 20 at 22:09
















    4












    $begingroup$

    Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
      $endgroup$
      – Petar
      Jan 20 at 20:05










    • $begingroup$
      If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
      $endgroup$
      – José Carlos Santos
      Jan 20 at 21:34












    • $begingroup$
      Ok thank you! I understand it now.
      $endgroup$
      – Petar
      Jan 20 at 21:51










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      Jan 20 at 21:53










    • $begingroup$
      yes i did it now ,tnx
      $endgroup$
      – Petar
      Jan 20 at 22:09














    4












    4








    4





    $begingroup$

    Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.






    share|cite|improve this answer









    $endgroup$



    Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 20 at 19:33









    José Carlos SantosJosé Carlos Santos

    156k22126227




    156k22126227












    • $begingroup$
      Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
      $endgroup$
      – Petar
      Jan 20 at 20:05










    • $begingroup$
      If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
      $endgroup$
      – José Carlos Santos
      Jan 20 at 21:34












    • $begingroup$
      Ok thank you! I understand it now.
      $endgroup$
      – Petar
      Jan 20 at 21:51










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      Jan 20 at 21:53










    • $begingroup$
      yes i did it now ,tnx
      $endgroup$
      – Petar
      Jan 20 at 22:09


















    • $begingroup$
      Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
      $endgroup$
      – Petar
      Jan 20 at 20:05










    • $begingroup$
      If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
      $endgroup$
      – José Carlos Santos
      Jan 20 at 21:34












    • $begingroup$
      Ok thank you! I understand it now.
      $endgroup$
      – Petar
      Jan 20 at 21:51










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      Jan 20 at 21:53










    • $begingroup$
      yes i did it now ,tnx
      $endgroup$
      – Petar
      Jan 20 at 22:09
















    $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    Jan 20 at 20:05




    $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    Jan 20 at 20:05












    $begingroup$
    If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 21:34






    $begingroup$
    If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 21:34














    $begingroup$
    Ok thank you! I understand it now.
    $endgroup$
    – Petar
    Jan 20 at 21:51




    $begingroup$
    Ok thank you! I understand it now.
    $endgroup$
    – Petar
    Jan 20 at 21:51












    $begingroup$
    If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 21:53




    $begingroup$
    If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    Jan 20 at 21:53












    $begingroup$
    yes i did it now ,tnx
    $endgroup$
    – Petar
    Jan 20 at 22:09




    $begingroup$
    yes i did it now ,tnx
    $endgroup$
    – Petar
    Jan 20 at 22:09











    2












    $begingroup$

    Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.






        share|cite|improve this answer









        $endgroup$



        Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 20 at 19:41









        Shubham JohriShubham Johri

        4,992717




        4,992717






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081040%2fsubspace-of-a-vector-space-problem%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

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            Display a custom attribute below product name in the front-end Magento 1.9.3.8

            Image
            0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
            Read more