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












1












$begingroup$


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









$endgroup$








  • 2




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












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










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






  • 2




    $begingroup$
    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
    $endgroup$
    – Will Jagy
    Jan 4 at 22:24






  • 1




    $begingroup$
    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
    $endgroup$
    – Will Jagy
    Jan 4 at 22:35
















1












$begingroup$


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









$endgroup$








  • 2




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












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










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






  • 2




    $begingroup$
    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
    $endgroup$
    – Will Jagy
    Jan 4 at 22:24






  • 1




    $begingroup$
    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
    $endgroup$
    – Will Jagy
    Jan 4 at 22:35














1












1








1





$begingroup$


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









$endgroup$




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

1,884418




1,884418








  • 2




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












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










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






  • 2




    $begingroup$
    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
    $endgroup$
    – Will Jagy
    Jan 4 at 22:24






  • 1




    $begingroup$
    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
    $endgroup$
    – Will Jagy
    Jan 4 at 22:35














  • 2




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












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










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






  • 2




    $begingroup$
    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
    $endgroup$
    – Will Jagy
    Jan 4 at 22:24






  • 1




    $begingroup$
    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
    $endgroup$
    – Will Jagy
    Jan 4 at 22:35








2




2




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






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














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




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












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




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




2




2




$begingroup$
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
$endgroup$
– Will Jagy
Jan 4 at 22:24




$begingroup$
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
$endgroup$
– Will Jagy
Jan 4 at 22:24




1




1




$begingroup$
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
$endgroup$
– Will Jagy
Jan 4 at 22:35




$begingroup$
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
$endgroup$
– Will Jagy
Jan 4 at 22:35










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$begingroup$

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











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    2












    $begingroup$

    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











    $endgroup$


















      2












      $begingroup$

      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











      $endgroup$
















        2












        2








        2





        $begingroup$

        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











        $endgroup$



        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














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 23:05









        J.G.

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        30.3k23148










        answered Jan 4 at 22:34









        A. PA. P

        1086




        1086






























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