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
    Jan 4 at 22:15












  • ya ,,,im trying @WillJagy.....haa
    – jasmine
    Jan 4 at 22:17










  • what do you get for $k=2?$
    – Will Jagy
    Jan 4 at 22:19






  • 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
    Jan 4 at 22:24






  • 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
    Jan 4 at 22:35
















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
    Jan 4 at 22:15












  • ya ,,,im trying @WillJagy.....haa
    – jasmine
    Jan 4 at 22:17










  • what do you get for $k=2?$
    – Will Jagy
    Jan 4 at 22:19






  • 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
    Jan 4 at 22:24






  • 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
    Jan 4 at 22:35














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 Jan 4 at 22:13









jasminejasmine

1,640416




1,640416








  • 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
    Jan 4 at 22:15












  • ya ,,,im trying @WillJagy.....haa
    – jasmine
    Jan 4 at 22:17










  • what do you get for $k=2?$
    – Will Jagy
    Jan 4 at 22:19






  • 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
    Jan 4 at 22:24






  • 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
    Jan 4 at 22:35














  • 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
    Jan 4 at 22:15












  • ya ,,,im trying @WillJagy.....haa
    – jasmine
    Jan 4 at 22:17










  • what do you get for $k=2?$
    – Will Jagy
    Jan 4 at 22:19






  • 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
    Jan 4 at 22:24






  • 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
    Jan 4 at 22:35








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
Jan 4 at 22:15






do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
– Will Jagy
Jan 4 at 22:15














ya ,,,im trying @WillJagy.....haa
– jasmine
Jan 4 at 22:17




ya ,,,im trying @WillJagy.....haa
– jasmine
Jan 4 at 22:17












what do you get for $k=2?$
– Will Jagy
Jan 4 at 22:19




what do you get for $k=2?$
– Will Jagy
Jan 4 at 22:19




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
Jan 4 at 22:24




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
Jan 4 at 22:24




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
Jan 4 at 22:35




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
Jan 4 at 22:35










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 Jan 4 at 23:05









        J.G.

        23.5k22237




        23.5k22237






        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 Jan 4 at 22:34









        A. PA. 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.




            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

            An IMO inspired problem

            Image
            14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
            Read more

            Management

            Image
            Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
            Read more

            Has there ever been an instance of an active nuclear power plant within or near a war zone?

            Image
            36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
            Read more