find a recursive relation for the characteristic polynomial of the $k times k $ matrix?












1














find a recursive relation for the characteristic polynomial of the $k times k $ matrix $$begin{pmatrix} 0 & 1 \ 1 & 0 & 1 \ mbox{ } & 1 & . & . \ mbox{ } &mbox{ } & . & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & 0end{pmatrix}$$



and compute the polynomial for $kle 5$



My attempt : Let $M_k$ be the $ktimes k$ matrix and $P_k(x)=det(M_k-xI_k)$ be its characteristic polynomial. We have



$$P_k(x)=detbegin{pmatrix} -x & 1 \ 1 & -x & 1 \ mbox{ } & 1 & ddots & ddots \ mbox{ } &mbox{ } & ddots & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & -xend{pmatrix}$$



after that im not able proceed further



Any hints/solution will be appreciated






linear-algebra







share|cite|improve this question


















  • 2




    do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
    – Will Jagy
    yesterday












  • ya ,,,im trying @WillJagy.....haa
    – jasmine
    yesterday










  • what do you get for $k=2?$
    – Will Jagy
    yesterday






  • 2




    now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
    – Will Jagy
    yesterday






  • 1




    Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
    – Will Jagy
    yesterday
















1














find a recursive relation for the characteristic polynomial of the $k times k $ matrix $$begin{pmatrix} 0 & 1 \ 1 & 0 & 1 \ mbox{ } & 1 & . & . \ mbox{ } &mbox{ } & . & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & 0end{pmatrix}$$



and compute the polynomial for $kle 5$



My attempt : Let $M_k$ be the $ktimes k$ matrix and $P_k(x)=det(M_k-xI_k)$ be its characteristic polynomial. We have



$$P_k(x)=detbegin{pmatrix} -x & 1 \ 1 & -x & 1 \ mbox{ } & 1 & ddots & ddots \ mbox{ } &mbox{ } & ddots & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & -xend{pmatrix}$$



after that im not able proceed further



Any hints/solution will be appreciated






linear-algebra







share|cite|improve this question


















  • 2




    do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
    – Will Jagy
    yesterday












  • ya ,,,im trying @WillJagy.....haa
    – jasmine
    yesterday










  • what do you get for $k=2?$
    – Will Jagy
    yesterday






  • 2




    now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
    – Will Jagy
    yesterday






  • 1




    Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
    – Will Jagy
    yesterday














1












1








1







find a recursive relation for the characteristic polynomial of the $k times k $ matrix $$begin{pmatrix} 0 & 1 \ 1 & 0 & 1 \ mbox{ } & 1 & . & . \ mbox{ } &mbox{ } & . & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & 0end{pmatrix}$$



and compute the polynomial for $kle 5$



My attempt : Let $M_k$ be the $ktimes k$ matrix and $P_k(x)=det(M_k-xI_k)$ be its characteristic polynomial. We have



$$P_k(x)=detbegin{pmatrix} -x & 1 \ 1 & -x & 1 \ mbox{ } & 1 & ddots & ddots \ mbox{ } &mbox{ } & ddots & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & -xend{pmatrix}$$



after that im not able proceed further



Any hints/solution will be appreciated






linear-algebra







share|cite|improve this question













find a recursive relation for the characteristic polynomial of the $k times k $ matrix $$begin{pmatrix} 0 & 1 \ 1 & 0 & 1 \ mbox{ } & 1 & . & . \ mbox{ } &mbox{ } & . & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & 0end{pmatrix}$$



and compute the polynomial for $kle 5$



My attempt : Let $M_k$ be the $ktimes k$ matrix and $P_k(x)=det(M_k-xI_k)$ be its characteristic polynomial. We have



$$P_k(x)=detbegin{pmatrix} -x & 1 \ 1 & -x & 1 \ mbox{ } & 1 & ddots & ddots \ mbox{ } &mbox{ } & ddots & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & -xend{pmatrix}$$



after that im not able proceed further



Any hints/solution will be appreciated






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









jasmine

1,589416




1,589416








  • 2




    do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
    – Will Jagy
    yesterday












  • ya ,,,im trying @WillJagy.....haa
    – jasmine
    yesterday










  • what do you get for $k=2?$
    – Will Jagy
    yesterday






  • 2




    now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
    – Will Jagy
    yesterday






  • 1




    Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
    – Will Jagy
    yesterday














  • 2




    do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
    – Will Jagy
    yesterday












  • ya ,,,im trying @WillJagy.....haa
    – jasmine
    yesterday










  • what do you get for $k=2?$
    – Will Jagy
    yesterday






  • 2




    now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
    – Will Jagy
    yesterday






  • 1




    Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
    – Will Jagy
    yesterday








2




2




do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
– Will Jagy
yesterday






do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
– Will Jagy
yesterday














ya ,,,im trying @WillJagy.....haa
– jasmine
yesterday




ya ,,,im trying @WillJagy.....haa
– jasmine
yesterday












what do you get for $k=2?$
– Will Jagy
yesterday




what do you get for $k=2?$
– Will Jagy
yesterday




2




2




now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
– Will Jagy
yesterday




now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
– Will Jagy
yesterday




1




1




Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
– Will Jagy
yesterday




Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
– Will Jagy
yesterday










1 Answer
1






active

oldest

votes


















2














Hint: Look at the formula for the determinant of using the first row.
Then you get a recursive definition for the determinant.



Solution:




$P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$







share|cite|improve this answer










New contributor




A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















    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%2f3062175%2ffind-a-recursive-relation-for-the-characteristic-polynomial-of-the-k-times-k%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









    2














    Hint: Look at the formula for the determinant of using the first row.
    Then you get a recursive definition for the determinant.



    Solution:




    $P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$







    share|cite|improve this answer










    New contributor




    A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      2














      Hint: Look at the formula for the determinant of using the first row.
      Then you get a recursive definition for the determinant.



      Solution:




      $P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$







      share|cite|improve this answer










      New contributor




      A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





















        2












        2








        2






        Hint: Look at the formula for the determinant of using the first row.
        Then you get a recursive definition for the determinant.



        Solution:




        $P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$







        share|cite|improve this answer










        New contributor




        A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        Hint: Look at the formula for the determinant of using the first row.
        Then you get a recursive definition for the determinant.



        Solution:




        $P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$








        share|cite|improve this answer










        New contributor




        A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday









        J.G.

        23.1k22137




        23.1k22137






        New contributor




        A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered yesterday









        A. P

        1085




        1085




        New contributor




        A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3062175%2ffind-a-recursive-relation-for-the-characteristic-polynomial-of-the-k-times-k%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

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            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