Minimal polynomial of a matrix having only 1s on the counter diagonal












3












$begingroup$


Consider the matrix $A=a_{ij}$ where $$a_{ij}=begin{cases}1 text{if} i+j=n+1\0 text{otherwise}end{cases}$$. Then, what can be said about the minimal polynomial of the matrix $A$.



Note that one eigenvalue is easily found by taking the eigenvector $begin{pmatrix}1\1\1\ldots\ldots\ldots\1end{pmatrix}$. Any hints. Thanks beforehand.






linear-algebra matrices minimal-polynomials







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$endgroup$








  • 1




    $begingroup$
    When $n>1$, what are $A-I, A+I$ and $(A-I)(A+I)$?
    $endgroup$
    – user1551
    Jan 9 at 11:50










  • $begingroup$
    @user1551 how do we compute $(A-I)(A+I)$ for large matrices? Is there any inductive/recursive process?
    $endgroup$
    – vidyarthi
    Jan 9 at 11:53






  • 1




    $begingroup$
    Hint : We have $A^2=I$ , so the minimal polynomial must divide $x^2-1$
    $endgroup$
    – Peter
    Jan 9 at 12:04
















3












$begingroup$


Consider the matrix $A=a_{ij}$ where $$a_{ij}=begin{cases}1 text{if} i+j=n+1\0 text{otherwise}end{cases}$$. Then, what can be said about the minimal polynomial of the matrix $A$.



Note that one eigenvalue is easily found by taking the eigenvector $begin{pmatrix}1\1\1\ldots\ldots\ldots\1end{pmatrix}$. Any hints. Thanks beforehand.






linear-algebra matrices minimal-polynomials







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    When $n>1$, what are $A-I, A+I$ and $(A-I)(A+I)$?
    $endgroup$
    – user1551
    Jan 9 at 11:50










  • $begingroup$
    @user1551 how do we compute $(A-I)(A+I)$ for large matrices? Is there any inductive/recursive process?
    $endgroup$
    – vidyarthi
    Jan 9 at 11:53






  • 1




    $begingroup$
    Hint : We have $A^2=I$ , so the minimal polynomial must divide $x^2-1$
    $endgroup$
    – Peter
    Jan 9 at 12:04














3












3








3


1



$begingroup$


Consider the matrix $A=a_{ij}$ where $$a_{ij}=begin{cases}1 text{if} i+j=n+1\0 text{otherwise}end{cases}$$. Then, what can be said about the minimal polynomial of the matrix $A$.



Note that one eigenvalue is easily found by taking the eigenvector $begin{pmatrix}1\1\1\ldots\ldots\ldots\1end{pmatrix}$. Any hints. Thanks beforehand.






linear-algebra matrices minimal-polynomials







share|cite|improve this question









$endgroup$




Consider the matrix $A=a_{ij}$ where $$a_{ij}=begin{cases}1 text{if} i+j=n+1\0 text{otherwise}end{cases}$$. Then, what can be said about the minimal polynomial of the matrix $A$.



Note that one eigenvalue is easily found by taking the eigenvector $begin{pmatrix}1\1\1\ldots\ldots\ldots\1end{pmatrix}$. Any hints. Thanks beforehand.






linear-algebra matrices minimal-polynomials




linear-algebra matrices minimal-polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 11:45









vidyarthividyarthi

2,9311832




2,9311832








  • 1




    $begingroup$
    When $n>1$, what are $A-I, A+I$ and $(A-I)(A+I)$?
    $endgroup$
    – user1551
    Jan 9 at 11:50










  • $begingroup$
    @user1551 how do we compute $(A-I)(A+I)$ for large matrices? Is there any inductive/recursive process?
    $endgroup$
    – vidyarthi
    Jan 9 at 11:53






  • 1




    $begingroup$
    Hint : We have $A^2=I$ , so the minimal polynomial must divide $x^2-1$
    $endgroup$
    – Peter
    Jan 9 at 12:04














  • 1




    $begingroup$
    When $n>1$, what are $A-I, A+I$ and $(A-I)(A+I)$?
    $endgroup$
    – user1551
    Jan 9 at 11:50










  • $begingroup$
    @user1551 how do we compute $(A-I)(A+I)$ for large matrices? Is there any inductive/recursive process?
    $endgroup$
    – vidyarthi
    Jan 9 at 11:53






  • 1




    $begingroup$
    Hint : We have $A^2=I$ , so the minimal polynomial must divide $x^2-1$
    $endgroup$
    – Peter
    Jan 9 at 12:04








1




1




$begingroup$
When $n>1$, what are $A-I, A+I$ and $(A-I)(A+I)$?
$endgroup$
– user1551
Jan 9 at 11:50




$begingroup$
When $n>1$, what are $A-I, A+I$ and $(A-I)(A+I)$?
$endgroup$
– user1551
Jan 9 at 11:50












$begingroup$
@user1551 how do we compute $(A-I)(A+I)$ for large matrices? Is there any inductive/recursive process?
$endgroup$
– vidyarthi
Jan 9 at 11:53




$begingroup$
@user1551 how do we compute $(A-I)(A+I)$ for large matrices? Is there any inductive/recursive process?
$endgroup$
– vidyarthi
Jan 9 at 11:53




1




1




$begingroup$
Hint : We have $A^2=I$ , so the minimal polynomial must divide $x^2-1$
$endgroup$
– Peter
Jan 9 at 12:04




$begingroup$
Hint : We have $A^2=I$ , so the minimal polynomial must divide $x^2-1$
$endgroup$
– Peter
Jan 9 at 12:04










3 Answers
3






active

oldest

votes


















2












$begingroup$

Hint: Assuming $A$ is supposed to be an $n times n$ matrix, $A^2$ is very easy to calculate, and this gives you your answer almost immediately. The entry in row $i$ and column $j$ of $A^2$ is the product of the $i^mbox{th}$ row and $j^mbox{th}$ colmn of $A$ , which you know.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Thanks to @user1551, the answer is deceptively simple in this case. We have, $$A-I=begin{pmatrix}-1&0&ldots&0&1\0&-1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&-1end{pmatrix}$$. Similarly, $$A+I=begin{pmatrix}1&0&ldots&0&1\0&1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&1end{pmatrix}$$. Now multiplying would naturally give us the zero matrix as the $1$s and $-1$s cancel out in pairs at each non zero product entry .






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You'd need to distinguish odd/even order of the matrix (for the form of the sum and difference).
      $endgroup$
      – user376343
      Jan 9 at 19:25





















    1












    $begingroup$

    First by some computation, we can quickly find that its characteristic polynomial is



    $ch_A(x) = (x-1)^{big{lceil}frac{n}{2}big{rceil}}(x+1)^{big{lfloor}frac{n}{2}big{rfloor}}$,



    which means that $1$ and $-1$ are the only eigenvalues of $A$.



    It turns out that geometric multiplicity of both eigenvalues are only one (i.e. $m(x) = x^2-1$), since $A^2 = I$.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Hint: Assuming $A$ is supposed to be an $n times n$ matrix, $A^2$ is very easy to calculate, and this gives you your answer almost immediately. The entry in row $i$ and column $j$ of $A^2$ is the product of the $i^mbox{th}$ row and $j^mbox{th}$ colmn of $A$ , which you know.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hint: Assuming $A$ is supposed to be an $n times n$ matrix, $A^2$ is very easy to calculate, and this gives you your answer almost immediately. The entry in row $i$ and column $j$ of $A^2$ is the product of the $i^mbox{th}$ row and $j^mbox{th}$ colmn of $A$ , which you know.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint: Assuming $A$ is supposed to be an $n times n$ matrix, $A^2$ is very easy to calculate, and this gives you your answer almost immediately. The entry in row $i$ and column $j$ of $A^2$ is the product of the $i^mbox{th}$ row and $j^mbox{th}$ colmn of $A$ , which you know.






          share|cite|improve this answer









          $endgroup$



          Hint: Assuming $A$ is supposed to be an $n times n$ matrix, $A^2$ is very easy to calculate, and this gives you your answer almost immediately. The entry in row $i$ and column $j$ of $A^2$ is the product of the $i^mbox{th}$ row and $j^mbox{th}$ colmn of $A$ , which you know.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 12:07









          lonza leggieralonza leggiera

          3715




          3715























              2












              $begingroup$

              Thanks to @user1551, the answer is deceptively simple in this case. We have, $$A-I=begin{pmatrix}-1&0&ldots&0&1\0&-1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&-1end{pmatrix}$$. Similarly, $$A+I=begin{pmatrix}1&0&ldots&0&1\0&1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&1end{pmatrix}$$. Now multiplying would naturally give us the zero matrix as the $1$s and $-1$s cancel out in pairs at each non zero product entry .






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                You'd need to distinguish odd/even order of the matrix (for the form of the sum and difference).
                $endgroup$
                – user376343
                Jan 9 at 19:25


















              2












              $begingroup$

              Thanks to @user1551, the answer is deceptively simple in this case. We have, $$A-I=begin{pmatrix}-1&0&ldots&0&1\0&-1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&-1end{pmatrix}$$. Similarly, $$A+I=begin{pmatrix}1&0&ldots&0&1\0&1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&1end{pmatrix}$$. Now multiplying would naturally give us the zero matrix as the $1$s and $-1$s cancel out in pairs at each non zero product entry .






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                You'd need to distinguish odd/even order of the matrix (for the form of the sum and difference).
                $endgroup$
                – user376343
                Jan 9 at 19:25
















              2












              2








              2





              $begingroup$

              Thanks to @user1551, the answer is deceptively simple in this case. We have, $$A-I=begin{pmatrix}-1&0&ldots&0&1\0&-1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&-1end{pmatrix}$$. Similarly, $$A+I=begin{pmatrix}1&0&ldots&0&1\0&1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&1end{pmatrix}$$. Now multiplying would naturally give us the zero matrix as the $1$s and $-1$s cancel out in pairs at each non zero product entry .






              share|cite|improve this answer









              $endgroup$



              Thanks to @user1551, the answer is deceptively simple in this case. We have, $$A-I=begin{pmatrix}-1&0&ldots&0&1\0&-1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&-1end{pmatrix}$$. Similarly, $$A+I=begin{pmatrix}1&0&ldots&0&1\0&1&ldots&1&0\ldots&ldots&ldots&ldots&ldots\1&0&ldots&0&1end{pmatrix}$$. Now multiplying would naturally give us the zero matrix as the $1$s and $-1$s cancel out in pairs at each non zero product entry .







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 9 at 11:59









              vidyarthividyarthi

              2,9311832




              2,9311832












              • $begingroup$
                You'd need to distinguish odd/even order of the matrix (for the form of the sum and difference).
                $endgroup$
                – user376343
                Jan 9 at 19:25




















              • $begingroup$
                You'd need to distinguish odd/even order of the matrix (for the form of the sum and difference).
                $endgroup$
                – user376343
                Jan 9 at 19:25


















              $begingroup$
              You'd need to distinguish odd/even order of the matrix (for the form of the sum and difference).
              $endgroup$
              – user376343
              Jan 9 at 19:25






              $begingroup$
              You'd need to distinguish odd/even order of the matrix (for the form of the sum and difference).
              $endgroup$
              – user376343
              Jan 9 at 19:25













              1












              $begingroup$

              First by some computation, we can quickly find that its characteristic polynomial is



              $ch_A(x) = (x-1)^{big{lceil}frac{n}{2}big{rceil}}(x+1)^{big{lfloor}frac{n}{2}big{rfloor}}$,



              which means that $1$ and $-1$ are the only eigenvalues of $A$.



              It turns out that geometric multiplicity of both eigenvalues are only one (i.e. $m(x) = x^2-1$), since $A^2 = I$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                First by some computation, we can quickly find that its characteristic polynomial is



                $ch_A(x) = (x-1)^{big{lceil}frac{n}{2}big{rceil}}(x+1)^{big{lfloor}frac{n}{2}big{rfloor}}$,



                which means that $1$ and $-1$ are the only eigenvalues of $A$.



                It turns out that geometric multiplicity of both eigenvalues are only one (i.e. $m(x) = x^2-1$), since $A^2 = I$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  First by some computation, we can quickly find that its characteristic polynomial is



                  $ch_A(x) = (x-1)^{big{lceil}frac{n}{2}big{rceil}}(x+1)^{big{lfloor}frac{n}{2}big{rfloor}}$,



                  which means that $1$ and $-1$ are the only eigenvalues of $A$.



                  It turns out that geometric multiplicity of both eigenvalues are only one (i.e. $m(x) = x^2-1$), since $A^2 = I$.






                  share|cite|improve this answer









                  $endgroup$



                  First by some computation, we can quickly find that its characteristic polynomial is



                  $ch_A(x) = (x-1)^{big{lceil}frac{n}{2}big{rceil}}(x+1)^{big{lfloor}frac{n}{2}big{rfloor}}$,



                  which means that $1$ and $-1$ are the only eigenvalues of $A$.



                  It turns out that geometric multiplicity of both eigenvalues are only one (i.e. $m(x) = x^2-1$), since $A^2 = I$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 9 at 12:04









                  Yanger MaYanger Ma

                  1014




                  1014






























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