Real n-by-n Matrices…












5












$begingroup$


Let $M_n(mathbb{R})$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbb{R})$.



Part (a) of this question says: Suppose $B in M_n(mathbb{R})$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbb{R})$ such that $CA = 0$, then prove $C = 0$.



I have already proven part (a).



Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbb{R}$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbb{R})$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).



My idea is to use induction on $m$. That is, suppose



$$t_0I = 0.$$



But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as



begin{align*}
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
end{align*}



But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".



I am studying for my linear algebra comp in a few weeks so any help is appreciated!






linear-algebra matrices vector-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
    $endgroup$
    – A.Γ.
    Jan 1 at 20:44
















5












$begingroup$


Let $M_n(mathbb{R})$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbb{R})$.



Part (a) of this question says: Suppose $B in M_n(mathbb{R})$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbb{R})$ such that $CA = 0$, then prove $C = 0$.



I have already proven part (a).



Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbb{R}$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbb{R})$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).



My idea is to use induction on $m$. That is, suppose



$$t_0I = 0.$$



But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as



begin{align*}
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
end{align*}



But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".



I am studying for my linear algebra comp in a few weeks so any help is appreciated!






linear-algebra matrices vector-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
    $endgroup$
    – A.Γ.
    Jan 1 at 20:44














5












5








5





$begingroup$


Let $M_n(mathbb{R})$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbb{R})$.



Part (a) of this question says: Suppose $B in M_n(mathbb{R})$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbb{R})$ such that $CA = 0$, then prove $C = 0$.



I have already proven part (a).



Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbb{R}$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbb{R})$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).



My idea is to use induction on $m$. That is, suppose



$$t_0I = 0.$$



But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as



begin{align*}
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
end{align*}



But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".



I am studying for my linear algebra comp in a few weeks so any help is appreciated!






linear-algebra matrices vector-spaces







share|cite|improve this question









$endgroup$




Let $M_n(mathbb{R})$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbb{R})$.



Part (a) of this question says: Suppose $B in M_n(mathbb{R})$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbb{R})$ such that $CA = 0$, then prove $C = 0$.



I have already proven part (a).



Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbb{R}$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbb{R})$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).



My idea is to use induction on $m$. That is, suppose



$$t_0I = 0.$$



But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as



begin{align*}
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
end{align*}



But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".



I am studying for my linear algebra comp in a few weeks so any help is appreciated!






linear-algebra matrices vector-spaces




linear-algebra matrices vector-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 1 at 20:25









Taylor McMillanTaylor McMillan

695




695












  • $begingroup$
    Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
    $endgroup$
    – A.Γ.
    Jan 1 at 20:44


















  • $begingroup$
    Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
    $endgroup$
    – A.Γ.
    Jan 1 at 20:44
















$begingroup$
Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
$endgroup$
– A.Γ.
Jan 1 at 20:44




$begingroup$
Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
$endgroup$
– A.Γ.
Jan 1 at 20:44










3 Answers
3






active

oldest

votes


















7












$begingroup$

Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
$$
t_0 I + t_1 A + cdots + t_m A^m = 0
$$



If $t_0=0$ you have,
$$
t_1 A + cdots + t_m A^m = 0
$$



Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
$$
t_0'I + t_1'A + cdots t_n ' A^n = 0
$$
where $n<m$.



Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
    $endgroup$
    – Taylor McMillan
    Jan 1 at 21:05










  • $begingroup$
    That is what I am doing. My university has several past exams posted I am doing all of those problems.
    $endgroup$
    – Taylor McMillan
    Jan 1 at 21:07



















4












$begingroup$

Hint: assume $t_0=0$ then
$$
t_1A+ldots+t_mA^m=0.
$$
Multiply by $B$ and get the contradiction with the minimality of $m$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Here is another hint/solution:




    • assume it to equal $0$,

    • multiply your equation by $A$, and factor by $A$,

    • using part (a), get a contradiction.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
      $endgroup$
      – timtfj
      Jan 1 at 22:18











    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%2f3058840%2freal-n-by-n-matrices%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
    $$
    t_0 I + t_1 A + cdots + t_m A^m = 0
    $$



    If $t_0=0$ you have,
    $$
    t_1 A + cdots + t_m A^m = 0
    $$



    Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
    $$
    t_0'I + t_1'A + cdots t_n ' A^n = 0
    $$
    where $n<m$.



    Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



    I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
      $endgroup$
      – Taylor McMillan
      Jan 1 at 21:05










    • $begingroup$
      That is what I am doing. My university has several past exams posted I am doing all of those problems.
      $endgroup$
      – Taylor McMillan
      Jan 1 at 21:07
















    7












    $begingroup$

    Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
    $$
    t_0 I + t_1 A + cdots + t_m A^m = 0
    $$



    If $t_0=0$ you have,
    $$
    t_1 A + cdots + t_m A^m = 0
    $$



    Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
    $$
    t_0'I + t_1'A + cdots t_n ' A^n = 0
    $$
    where $n<m$.



    Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



    I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
      $endgroup$
      – Taylor McMillan
      Jan 1 at 21:05










    • $begingroup$
      That is what I am doing. My university has several past exams posted I am doing all of those problems.
      $endgroup$
      – Taylor McMillan
      Jan 1 at 21:07














    7












    7








    7





    $begingroup$

    Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
    $$
    t_0 I + t_1 A + cdots + t_m A^m = 0
    $$



    If $t_0=0$ you have,
    $$
    t_1 A + cdots + t_m A^m = 0
    $$



    Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
    $$
    t_0'I + t_1'A + cdots t_n ' A^n = 0
    $$
    where $n<m$.



    Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



    I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.






    share|cite|improve this answer









    $endgroup$



    Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
    $$
    t_0 I + t_1 A + cdots + t_m A^m = 0
    $$



    If $t_0=0$ you have,
    $$
    t_1 A + cdots + t_m A^m = 0
    $$



    Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
    $$
    t_0'I + t_1'A + cdots t_n ' A^n = 0
    $$
    where $n<m$.



    Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



    I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 1 at 20:46









    tchtch

    803310




    803310












    • $begingroup$
      Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
      $endgroup$
      – Taylor McMillan
      Jan 1 at 21:05










    • $begingroup$
      That is what I am doing. My university has several past exams posted I am doing all of those problems.
      $endgroup$
      – Taylor McMillan
      Jan 1 at 21:07


















    • $begingroup$
      Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
      $endgroup$
      – Taylor McMillan
      Jan 1 at 21:05










    • $begingroup$
      That is what I am doing. My university has several past exams posted I am doing all of those problems.
      $endgroup$
      – Taylor McMillan
      Jan 1 at 21:07
















    $begingroup$
    Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
    $endgroup$
    – Taylor McMillan
    Jan 1 at 21:05




    $begingroup$
    Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
    $endgroup$
    – Taylor McMillan
    Jan 1 at 21:05












    $begingroup$
    That is what I am doing. My university has several past exams posted I am doing all of those problems.
    $endgroup$
    – Taylor McMillan
    Jan 1 at 21:07




    $begingroup$
    That is what I am doing. My university has several past exams posted I am doing all of those problems.
    $endgroup$
    – Taylor McMillan
    Jan 1 at 21:07











    4












    $begingroup$

    Hint: assume $t_0=0$ then
    $$
    t_1A+ldots+t_mA^m=0.
    $$
    Multiply by $B$ and get the contradiction with the minimality of $m$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Hint: assume $t_0=0$ then
      $$
      t_1A+ldots+t_mA^m=0.
      $$
      Multiply by $B$ and get the contradiction with the minimality of $m$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Hint: assume $t_0=0$ then
        $$
        t_1A+ldots+t_mA^m=0.
        $$
        Multiply by $B$ and get the contradiction with the minimality of $m$.






        share|cite|improve this answer









        $endgroup$



        Hint: assume $t_0=0$ then
        $$
        t_1A+ldots+t_mA^m=0.
        $$
        Multiply by $B$ and get the contradiction with the minimality of $m$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 20:46









        A.Γ.A.Γ.

        22.8k32656




        22.8k32656























            1












            $begingroup$

            Here is another hint/solution:




            • assume it to equal $0$,

            • multiply your equation by $A$, and factor by $A$,

            • using part (a), get a contradiction.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
              $endgroup$
              – timtfj
              Jan 1 at 22:18
















            1












            $begingroup$

            Here is another hint/solution:




            • assume it to equal $0$,

            • multiply your equation by $A$, and factor by $A$,

            • using part (a), get a contradiction.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
              $endgroup$
              – timtfj
              Jan 1 at 22:18














            1












            1








            1





            $begingroup$

            Here is another hint/solution:




            • assume it to equal $0$,

            • multiply your equation by $A$, and factor by $A$,

            • using part (a), get a contradiction.






            share|cite|improve this answer











            $endgroup$



            Here is another hint/solution:




            • assume it to equal $0$,

            • multiply your equation by $A$, and factor by $A$,

            • using part (a), get a contradiction.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 1 at 22:14









            amWhy

            1




            1










            answered Jan 1 at 21:25









            Cauchy is my masterCauchy is my master

            413




            413












            • $begingroup$
              Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
              $endgroup$
              – timtfj
              Jan 1 at 22:18


















            • $begingroup$
              Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
              $endgroup$
              – timtfj
              Jan 1 at 22:18
















            $begingroup$
            Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
            $endgroup$
            – timtfj
            Jan 1 at 22:18




            $begingroup$
            Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
            $endgroup$
            – timtfj
            Jan 1 at 22:18


















            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%2f3058840%2freal-n-by-n-matrices%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

            Angular Downloading a file using contenturl with Basic Authentication

            Image
            0 I have a contentUrl when user clicks on that,file should be downloaded. If I try to browse the contentUrl in the browser it asks for Userid and password.I believe this is Basic Authentication.I dont want user to enter UserId and Password so I am passing it Authorization(Please see below code). Url looks some thing like this: http://XXX.XXX.XX.XXX:8080/flowable-rest/service/runtime/tasks/90875/attachments/90555/content let username : string = 'admin'; let pwd : string = 'test'; let headers = new Headers(); headers.set('Accept', 'text/json'); headers.append('Content-Type', 'application/json'); headers.set('Access-Control-Allow-Origin', '*'); headers.set('Access-Control-Allow-Methods','POST, GET, OPTIONS, PUT, DELETE'); head
            Read more

            Olmecas

            Image
            Monumento 1, uma das quatro cabeças colossais encontradas em La Venta, com altura aproximada de 3 metros Olmecas é a designação do povo e da civilização que estiveram na origem da antiga cultura pré-colombiana da Mesoamérica e que se desenvolveram nas regiões tropicais do centro-sul do atual México durante o pré-clássico, próximo de onde hoje estão localizados os estados mexicanos de Veracruz e Tabasco, no Istmo de Tehuantepec, numa zona designada área nuclear olmeca. A cultura olmeca floresceu nesta região aproximadamente entre 1500 e 400 a.C., [ 1 ] e crê-se que tenha sido a civilização-mãe de todas as civilizações mesoamericanas que se desenvolveram posteriormente. [ 2 ] No entanto, desconhece-se a sua exacta filiação étnica, ainda que existam numerosas hipóteses colocadas para tentar resolver esta questão. O etnónimo olmeca foi cunhado pelos arqueólogos do século XX, e não devem confundir-se com os muito posteriores olmecas-xicalancas que ocuparam vários locais do México
            Read more

            Can't read property showImagePicker of undefined in react native iOS

            Image
            1 I have tired with this issue npm install react-native-image-picker@latest --save react-native link react-native-image-picker import ImagePicker from 'react-native-image-picker'; const options = { quality: 1.0, maxWidth: 500, maxHeight: 500, storageOptions: { skipBackup: true } }; ImagePicker.showImagePicker(options, (response) => { console.log('Response = ', response); if (response.didCancel) { console.log('User cancelled image picker'); } else if (response.error) { console.log('ImagePicker Error: ', response.error); } else if (response.customButton) { console.log('User tapped custom button: ', response.customButton); } else {
            Read more