Not able to solve a matrix problem












1












$begingroup$


This is my first question of SO so sorry if I made a mistake somewhere.



I have this matrices problem which I'm not able to solve. I asked many of my friends but they just don't get the question.
This is the question:



$text{Let n} geq 3. text{Let A} = ((a_{ij}))_{1 leq i, j leq n} text{be an n} times text{n matrix such that a}_{ij} in {1, -1} text{for all 0} leq i, j leq n.
text{Suppose a}_{k1} = text{1 for all 1} leq k leq text{n. } \ text{And} sum_{k = 1}^{n} a_{ki}cdot a_{kj}text{= 0. For all i} neq j.\ text{Show that n is a multiple of 4.}$



(sorry I can't add an image and I'm not so good at latex too)



I think I've understood the summation part: when k = 1, the argument of summation becomes $a_{1i} cdot a_{1j}$. Which is the sum of product of elements of 1st row taken 2 at a time. (that's what I interpreted $a_{1i} cdot a_{1j}$ as. Is this right?)
I'm not able to utilize the summation to get an equation in n.



Thanx in advance!






matrices summation







share|cite|improve this question









$endgroup$












  • $begingroup$
    You are looking at Hadamard matrices, so now you know what to look for.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:54










  • $begingroup$
    See, e.g., math.stackexchange.com/questions/753867/…
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:59
















1












$begingroup$


This is my first question of SO so sorry if I made a mistake somewhere.



I have this matrices problem which I'm not able to solve. I asked many of my friends but they just don't get the question.
This is the question:



$text{Let n} geq 3. text{Let A} = ((a_{ij}))_{1 leq i, j leq n} text{be an n} times text{n matrix such that a}_{ij} in {1, -1} text{for all 0} leq i, j leq n.
text{Suppose a}_{k1} = text{1 for all 1} leq k leq text{n. } \ text{And} sum_{k = 1}^{n} a_{ki}cdot a_{kj}text{= 0. For all i} neq j.\ text{Show that n is a multiple of 4.}$



(sorry I can't add an image and I'm not so good at latex too)



I think I've understood the summation part: when k = 1, the argument of summation becomes $a_{1i} cdot a_{1j}$. Which is the sum of product of elements of 1st row taken 2 at a time. (that's what I interpreted $a_{1i} cdot a_{1j}$ as. Is this right?)
I'm not able to utilize the summation to get an equation in n.



Thanx in advance!






matrices summation







share|cite|improve this question









$endgroup$












  • $begingroup$
    You are looking at Hadamard matrices, so now you know what to look for.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:54










  • $begingroup$
    See, e.g., math.stackexchange.com/questions/753867/…
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:59














1












1








1


1



$begingroup$


This is my first question of SO so sorry if I made a mistake somewhere.



I have this matrices problem which I'm not able to solve. I asked many of my friends but they just don't get the question.
This is the question:



$text{Let n} geq 3. text{Let A} = ((a_{ij}))_{1 leq i, j leq n} text{be an n} times text{n matrix such that a}_{ij} in {1, -1} text{for all 0} leq i, j leq n.
text{Suppose a}_{k1} = text{1 for all 1} leq k leq text{n. } \ text{And} sum_{k = 1}^{n} a_{ki}cdot a_{kj}text{= 0. For all i} neq j.\ text{Show that n is a multiple of 4.}$



(sorry I can't add an image and I'm not so good at latex too)



I think I've understood the summation part: when k = 1, the argument of summation becomes $a_{1i} cdot a_{1j}$. Which is the sum of product of elements of 1st row taken 2 at a time. (that's what I interpreted $a_{1i} cdot a_{1j}$ as. Is this right?)
I'm not able to utilize the summation to get an equation in n.



Thanx in advance!






matrices summation







share|cite|improve this question









$endgroup$




This is my first question of SO so sorry if I made a mistake somewhere.



I have this matrices problem which I'm not able to solve. I asked many of my friends but they just don't get the question.
This is the question:



$text{Let n} geq 3. text{Let A} = ((a_{ij}))_{1 leq i, j leq n} text{be an n} times text{n matrix such that a}_{ij} in {1, -1} text{for all 0} leq i, j leq n.
text{Suppose a}_{k1} = text{1 for all 1} leq k leq text{n. } \ text{And} sum_{k = 1}^{n} a_{ki}cdot a_{kj}text{= 0. For all i} neq j.\ text{Show that n is a multiple of 4.}$



(sorry I can't add an image and I'm not so good at latex too)



I think I've understood the summation part: when k = 1, the argument of summation becomes $a_{1i} cdot a_{1j}$. Which is the sum of product of elements of 1st row taken 2 at a time. (that's what I interpreted $a_{1i} cdot a_{1j}$ as. Is this right?)
I'm not able to utilize the summation to get an equation in n.



Thanx in advance!






matrices summation




matrices summation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 10:58









YashKandalkarYashKandalkar

82




82












  • $begingroup$
    You are looking at Hadamard matrices, so now you know what to look for.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:54










  • $begingroup$
    See, e.g., math.stackexchange.com/questions/753867/…
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:59


















  • $begingroup$
    You are looking at Hadamard matrices, so now you know what to look for.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:54










  • $begingroup$
    See, e.g., math.stackexchange.com/questions/753867/…
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:59
















$begingroup$
You are looking at Hadamard matrices, so now you know what to look for.
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:54




$begingroup$
You are looking at Hadamard matrices, so now you know what to look for.
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:54












$begingroup$
See, e.g., math.stackexchange.com/questions/753867/…
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:59




$begingroup$
See, e.g., math.stackexchange.com/questions/753867/…
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:59










1 Answer
1






active

oldest

votes


















1












$begingroup$

Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.






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%2f3019953%2fnot-able-to-solve-a-matrix-problem%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












    $begingroup$

    Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



    Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



    So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



      Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



      So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



        Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



        So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.






        share|cite|improve this answer









        $endgroup$



        Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



        Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



        So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 12:40









        Michael HoppeMichael Hoppe

        10.8k31834




        10.8k31834






























            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%2f3019953%2fnot-able-to-solve-a-matrix-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

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more