If $U_1,U_2subseteq U, U$ linear independent set then $( cap )= $












-1












$begingroup$


If I take an element $v$ in $(<U_1> cap <U_2>)$ why this element can be described as :



$v=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$



?



where $z_1,..,z_nin U_1cap U_2,x_{k+1},...,x_{k+n}in U_1backslash U_2$ and $y_{k+1},...,y_{k+m}in U_2backslash U_1$



I thought every element of $(<U_1> cap <U_2>)$ must be a linear combination of vectors that are bothin $U_1$ and $U_2$



My tought process so far:



Let $xin U_1cap U_2Rightarrow xin U_1$ and $xin U_2$



We have
$$<U_1>cap <U_2>:= {sum_{i=1}^{n}lambda_iu_i,u_iin U_1,ninmathbb{N},lambda_iinmathbb{F}}cap{sum_{j=1}^{k}mu_ju_j,u_jin U_2,kinmathbb{N},mu_jinmathbb{F}}$$



I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $ and then I have proved



$xin <U_1>cap <U_2>Longrightarrow U_1cap U_2subseteq <U_1>cap <U_2>Longrightarrow <U_1cap U_2>subseteq <U_1>cap <U_2>$



Because Intersections of subspaces are again subspaces. And $<A>$ is the smalleset subspace that contains $A$.



To prove the next part the original proof proceeds with the Claim in the head of the Body of the Question.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    "I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
    $endgroup$
    – Did
    Jan 19 at 19:22
















-1












$begingroup$


If I take an element $v$ in $(<U_1> cap <U_2>)$ why this element can be described as :



$v=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$



?



where $z_1,..,z_nin U_1cap U_2,x_{k+1},...,x_{k+n}in U_1backslash U_2$ and $y_{k+1},...,y_{k+m}in U_2backslash U_1$



I thought every element of $(<U_1> cap <U_2>)$ must be a linear combination of vectors that are bothin $U_1$ and $U_2$



My tought process so far:



Let $xin U_1cap U_2Rightarrow xin U_1$ and $xin U_2$



We have
$$<U_1>cap <U_2>:= {sum_{i=1}^{n}lambda_iu_i,u_iin U_1,ninmathbb{N},lambda_iinmathbb{F}}cap{sum_{j=1}^{k}mu_ju_j,u_jin U_2,kinmathbb{N},mu_jinmathbb{F}}$$



I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $ and then I have proved



$xin <U_1>cap <U_2>Longrightarrow U_1cap U_2subseteq <U_1>cap <U_2>Longrightarrow <U_1cap U_2>subseteq <U_1>cap <U_2>$



Because Intersections of subspaces are again subspaces. And $<A>$ is the smalleset subspace that contains $A$.



To prove the next part the original proof proceeds with the Claim in the head of the Body of the Question.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    "I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
    $endgroup$
    – Did
    Jan 19 at 19:22














-1












-1








-1





$begingroup$


If I take an element $v$ in $(<U_1> cap <U_2>)$ why this element can be described as :



$v=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$



?



where $z_1,..,z_nin U_1cap U_2,x_{k+1},...,x_{k+n}in U_1backslash U_2$ and $y_{k+1},...,y_{k+m}in U_2backslash U_1$



I thought every element of $(<U_1> cap <U_2>)$ must be a linear combination of vectors that are bothin $U_1$ and $U_2$



My tought process so far:



Let $xin U_1cap U_2Rightarrow xin U_1$ and $xin U_2$



We have
$$<U_1>cap <U_2>:= {sum_{i=1}^{n}lambda_iu_i,u_iin U_1,ninmathbb{N},lambda_iinmathbb{F}}cap{sum_{j=1}^{k}mu_ju_j,u_jin U_2,kinmathbb{N},mu_jinmathbb{F}}$$



I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $ and then I have proved



$xin <U_1>cap <U_2>Longrightarrow U_1cap U_2subseteq <U_1>cap <U_2>Longrightarrow <U_1cap U_2>subseteq <U_1>cap <U_2>$



Because Intersections of subspaces are again subspaces. And $<A>$ is the smalleset subspace that contains $A$.



To prove the next part the original proof proceeds with the Claim in the head of the Body of the Question.






linear-algebra







share|cite|improve this question











$endgroup$




If I take an element $v$ in $(<U_1> cap <U_2>)$ why this element can be described as :



$v=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$



?



where $z_1,..,z_nin U_1cap U_2,x_{k+1},...,x_{k+n}in U_1backslash U_2$ and $y_{k+1},...,y_{k+m}in U_2backslash U_1$



I thought every element of $(<U_1> cap <U_2>)$ must be a linear combination of vectors that are bothin $U_1$ and $U_2$



My tought process so far:



Let $xin U_1cap U_2Rightarrow xin U_1$ and $xin U_2$



We have
$$<U_1>cap <U_2>:= {sum_{i=1}^{n}lambda_iu_i,u_iin U_1,ninmathbb{N},lambda_iinmathbb{F}}cap{sum_{j=1}^{k}mu_ju_j,u_jin U_2,kinmathbb{N},mu_jinmathbb{F}}$$



I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $ and then I have proved



$xin <U_1>cap <U_2>Longrightarrow U_1cap U_2subseteq <U_1>cap <U_2>Longrightarrow <U_1cap U_2>subseteq <U_1>cap <U_2>$



Because Intersections of subspaces are again subspaces. And $<A>$ is the smalleset subspace that contains $A$.



To prove the next part the original proof proceeds with the Claim in the head of the Body of the Question.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 19:58







RM777

















asked Jan 18 at 18:38









RM777RM777

38012




38012












  • $begingroup$
    "I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
    $endgroup$
    – Did
    Jan 19 at 19:22


















  • $begingroup$
    "I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
    $endgroup$
    – Did
    Jan 19 at 19:22
















$begingroup$
"I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
$endgroup$
– Did
Jan 19 at 19:22




$begingroup$
"I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
$endgroup$
– Did
Jan 19 at 19:22










1 Answer
1






active

oldest

votes


















0












$begingroup$

I take a $xin <U_1>cap<U_2>:=W$



If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



$$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



$$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$






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%2f3078613%2fif-u-1-u-2-subseteq-u-u-linear-independent-set-then-u-1-cap-u-2-u%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









    0












    $begingroup$

    I take a $xin <U_1>cap<U_2>:=W$



    If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



    $$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



    I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



    The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



    $$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



    The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I take a $xin <U_1>cap<U_2>:=W$



      If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



      $$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



      I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



      The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



      $$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



      The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I take a $xin <U_1>cap<U_2>:=W$



        If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



        $$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



        I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



        The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



        $$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



        The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$






        share|cite|improve this answer









        $endgroup$



        I take a $xin <U_1>cap<U_2>:=W$



        If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



        $$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



        I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



        The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



        $$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



        The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 20:24









        RM777RM777

        38012




        38012






























            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%2f3078613%2fif-u-1-u-2-subseteq-u-u-linear-independent-set-then-u-1-cap-u-2-u%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

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
            Read more

            Antonio Litta Visconti Arese

            Image
            Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
            Read more