Sum of Linear Subspaces












0












$begingroup$


Let $V$ be a vector space over $K$ and $S,U,W subset V$ linear subspaces of $V$. $U+W$ is defined as ${ u+w | u in U, w in W }$. Is it true, that



a) $U+W = { u-w | u in U, w in W }$



b) $S cap (U + W) = (S cap U) + (S cap W)$?



My thoughts for a) so far had been that due to $-1 in K$, $-w$ also has to be in $W$ if $w in W$.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $win W$ is $-win W$?
    $endgroup$
    – Doug M
    Jan 7 at 19:27












  • $begingroup$
    @DougM OP already mentioned that
    $endgroup$
    – Shubham Johri
    Jan 7 at 19:28










  • $begingroup$
    Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
    $endgroup$
    – bounceback
    Jan 7 at 19:31










  • $begingroup$
    @bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
    $endgroup$
    – Tim
    Jan 7 at 19:35










  • $begingroup$
    No, your first equality is false - see Bernard's answer below for the counterexample
    $endgroup$
    – bounceback
    Jan 7 at 19:38
















0












$begingroup$


Let $V$ be a vector space over $K$ and $S,U,W subset V$ linear subspaces of $V$. $U+W$ is defined as ${ u+w | u in U, w in W }$. Is it true, that



a) $U+W = { u-w | u in U, w in W }$



b) $S cap (U + W) = (S cap U) + (S cap W)$?



My thoughts for a) so far had been that due to $-1 in K$, $-w$ also has to be in $W$ if $w in W$.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $win W$ is $-win W$?
    $endgroup$
    – Doug M
    Jan 7 at 19:27












  • $begingroup$
    @DougM OP already mentioned that
    $endgroup$
    – Shubham Johri
    Jan 7 at 19:28










  • $begingroup$
    Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
    $endgroup$
    – bounceback
    Jan 7 at 19:31










  • $begingroup$
    @bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
    $endgroup$
    – Tim
    Jan 7 at 19:35










  • $begingroup$
    No, your first equality is false - see Bernard's answer below for the counterexample
    $endgroup$
    – bounceback
    Jan 7 at 19:38














0












0








0





$begingroup$


Let $V$ be a vector space over $K$ and $S,U,W subset V$ linear subspaces of $V$. $U+W$ is defined as ${ u+w | u in U, w in W }$. Is it true, that



a) $U+W = { u-w | u in U, w in W }$



b) $S cap (U + W) = (S cap U) + (S cap W)$?



My thoughts for a) so far had been that due to $-1 in K$, $-w$ also has to be in $W$ if $w in W$.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$




Let $V$ be a vector space over $K$ and $S,U,W subset V$ linear subspaces of $V$. $U+W$ is defined as ${ u+w | u in U, w in W }$. Is it true, that



a) $U+W = { u-w | u in U, w in W }$



b) $S cap (U + W) = (S cap U) + (S cap W)$?



My thoughts for a) so far had been that due to $-1 in K$, $-w$ also has to be in $W$ if $w in W$.






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 7 at 19:30







Tim

















asked Jan 7 at 19:22









TimTim

246




246












  • $begingroup$
    If $win W$ is $-win W$?
    $endgroup$
    – Doug M
    Jan 7 at 19:27












  • $begingroup$
    @DougM OP already mentioned that
    $endgroup$
    – Shubham Johri
    Jan 7 at 19:28










  • $begingroup$
    Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
    $endgroup$
    – bounceback
    Jan 7 at 19:31










  • $begingroup$
    @bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
    $endgroup$
    – Tim
    Jan 7 at 19:35










  • $begingroup$
    No, your first equality is false - see Bernard's answer below for the counterexample
    $endgroup$
    – bounceback
    Jan 7 at 19:38


















  • $begingroup$
    If $win W$ is $-win W$?
    $endgroup$
    – Doug M
    Jan 7 at 19:27












  • $begingroup$
    @DougM OP already mentioned that
    $endgroup$
    – Shubham Johri
    Jan 7 at 19:28










  • $begingroup$
    Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
    $endgroup$
    – bounceback
    Jan 7 at 19:31










  • $begingroup$
    @bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
    $endgroup$
    – Tim
    Jan 7 at 19:35










  • $begingroup$
    No, your first equality is false - see Bernard's answer below for the counterexample
    $endgroup$
    – bounceback
    Jan 7 at 19:38
















$begingroup$
If $win W$ is $-win W$?
$endgroup$
– Doug M
Jan 7 at 19:27






$begingroup$
If $win W$ is $-win W$?
$endgroup$
– Doug M
Jan 7 at 19:27














$begingroup$
@DougM OP already mentioned that
$endgroup$
– Shubham Johri
Jan 7 at 19:28




$begingroup$
@DougM OP already mentioned that
$endgroup$
– Shubham Johri
Jan 7 at 19:28












$begingroup$
Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
$endgroup$
– bounceback
Jan 7 at 19:31




$begingroup$
Well, that takes care of part (a): ${u - w: u in U, w in W} = {u + (-w): u in U, -w in W} = U + W$. As for part (b), what have you tried there, OP?
$endgroup$
– bounceback
Jan 7 at 19:31












$begingroup$
@bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
$endgroup$
– Tim
Jan 7 at 19:35




$begingroup$
@bounceback I was thinking about rewriting it like $S cap { u + w | u in U, w in W } = { u + w | u in (U cap S), w in (W cap S) } = (S cap U) + (S cap W)$ but I am not sure if this is a valid operation.
$endgroup$
– Tim
Jan 7 at 19:35












$begingroup$
No, your first equality is false - see Bernard's answer below for the counterexample
$endgroup$
– bounceback
Jan 7 at 19:38




$begingroup$
No, your first equality is false - see Bernard's answer below for the counterexample
$endgroup$
– bounceback
Jan 7 at 19:38










1 Answer
1






active

oldest

votes


















1












$begingroup$

For $a)$ it's fine, as $win Wiff -win W$.



For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.



Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.






share|cite|improve this answer









$endgroup$














    Your Answer








    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%2f3065387%2fsum-of-linear-subspaces%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









    1












    $begingroup$

    For $a)$ it's fine, as $win Wiff -win W$.



    For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.



    Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      For $a)$ it's fine, as $win Wiff -win W$.



      For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.



      Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        For $a)$ it's fine, as $win Wiff -win W$.



        For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.



        Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.






        share|cite|improve this answer









        $endgroup$



        For $a)$ it's fine, as $win Wiff -win W$.



        For $b)$, a counter-example: take $V=K^2$ ($K$ is the base field)$, $U={(x,0),;xin K}, $;W={(0,x),;xin K}$ and $;S={(x,x),;xin K}$.



        Then $U+W=V$, so $Scap(U+W)=S$, but $Scap U={0}=Scap W$, so $;Scap U+Scap W={0}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 19:34









        BernardBernard

        124k741117




        124k741117






























            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%2f3065387%2fsum-of-linear-subspaces%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

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more