Characteristic polynomial modulo 12











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Consider the vector space $V =left{a_0+a_1x+cdots+a_{11}x^{11},;a_iinmathbb{R}right}$. Define a linear operator $A$ on $V$ by $A(x^i) = x^{i+4}$ where $i + 4$ is taken modulo $12$.



Find $(a)$ the minimal polynomial of $A$ and $(b)$ the characteristic polynomial of $A$.




My try:



I coulnot find another way so I tried the brute force method. $:$I found the matrix representation of the operator is $$A=begin{bmatrix} 0&0&0&0&0&0&0&0&1&0&0&0\0&0&0&0&0&0&0&0&0&1&0&0\0&0&0&0&0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&0&0&0&0&1\1&0&0&0&0&0&0&0&0&0&0&0\0&1&0&0&0&0&0&0&0&0&0&0\0&0&1&0&0&0&0&0&0&0&0&0\0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0\end{bmatrix}.$$



The characteristic polynomial I found to be $lambda^{12}-4lambda^9+6lambda^6-4lambda^3+1$.$:$(It took me almost 40 minutes. $:$Is there another way to do this problem? Provide hints or suggestions please.



$rule{17cm}{1pt}$



Taking forward the answer provided by $textbf{Servaes}$ "The minimal polynomial of $A|_{U_i}$ is still $X^3−1$." Taking $U_1=span{x_1,x_5,x_9}$ we see $A(x)=x^5,:A^2(x)=x^9,:A^3(x)=ximplies (A^3-I)=0$ and since it factors into linear irredeucible factors, we have the minimal polynomial of $A|_{U_i}$ is $X^3−1$.



Completing the proof: We show that characteristic polynomial of $A$ is the product of characteristic polynomials of $A|_{U_i}$ where $V=oplus U_i$. We have seen that minimal polynomial of $A|_{U_i}$ is $X^3−1$ which is precisely the characteristic polynomial. So let $p_i(lambda)$ is characteristic polynomial corresponding to eigenvalue $lambda_i$ and invariant subspace $U_i$ and $p(lambda)$ is the characteristic polynomial of A. Then $displaystyle p(lambda_i)=0 :forall;i implies p_i(lambda)|p(lambda) :forall;i implies p(lambda)=prod_ip_i(lambda)$.



$Big($$p(lambda)$ is atleast $displaystyleprod_ip_i(lambda)$. If $existslambdaneqlambda_i forall: i$ such that $p(lambda)=0$ then $V$ is not $oplus U_i$ $Big)$



$rule{17cm}{1pt}$



Minimal polynomial : $X^3-1qquad$ Characteristic polynomial : $(X^3-1)^4$.










share|cite|improve this question




























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    Consider the vector space $V =left{a_0+a_1x+cdots+a_{11}x^{11},;a_iinmathbb{R}right}$. Define a linear operator $A$ on $V$ by $A(x^i) = x^{i+4}$ where $i + 4$ is taken modulo $12$.



    Find $(a)$ the minimal polynomial of $A$ and $(b)$ the characteristic polynomial of $A$.




    My try:



    I coulnot find another way so I tried the brute force method. $:$I found the matrix representation of the operator is $$A=begin{bmatrix} 0&0&0&0&0&0&0&0&1&0&0&0\0&0&0&0&0&0&0&0&0&1&0&0\0&0&0&0&0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&0&0&0&0&1\1&0&0&0&0&0&0&0&0&0&0&0\0&1&0&0&0&0&0&0&0&0&0&0\0&0&1&0&0&0&0&0&0&0&0&0\0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0\end{bmatrix}.$$



    The characteristic polynomial I found to be $lambda^{12}-4lambda^9+6lambda^6-4lambda^3+1$.$:$(It took me almost 40 minutes. $:$Is there another way to do this problem? Provide hints or suggestions please.



    $rule{17cm}{1pt}$



    Taking forward the answer provided by $textbf{Servaes}$ "The minimal polynomial of $A|_{U_i}$ is still $X^3−1$." Taking $U_1=span{x_1,x_5,x_9}$ we see $A(x)=x^5,:A^2(x)=x^9,:A^3(x)=ximplies (A^3-I)=0$ and since it factors into linear irredeucible factors, we have the minimal polynomial of $A|_{U_i}$ is $X^3−1$.



    Completing the proof: We show that characteristic polynomial of $A$ is the product of characteristic polynomials of $A|_{U_i}$ where $V=oplus U_i$. We have seen that minimal polynomial of $A|_{U_i}$ is $X^3−1$ which is precisely the characteristic polynomial. So let $p_i(lambda)$ is characteristic polynomial corresponding to eigenvalue $lambda_i$ and invariant subspace $U_i$ and $p(lambda)$ is the characteristic polynomial of A. Then $displaystyle p(lambda_i)=0 :forall;i implies p_i(lambda)|p(lambda) :forall;i implies p(lambda)=prod_ip_i(lambda)$.



    $Big($$p(lambda)$ is atleast $displaystyleprod_ip_i(lambda)$. If $existslambdaneqlambda_i forall: i$ such that $p(lambda)=0$ then $V$ is not $oplus U_i$ $Big)$



    $rule{17cm}{1pt}$



    Minimal polynomial : $X^3-1qquad$ Characteristic polynomial : $(X^3-1)^4$.










    share|cite|improve this question


























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      Consider the vector space $V =left{a_0+a_1x+cdots+a_{11}x^{11},;a_iinmathbb{R}right}$. Define a linear operator $A$ on $V$ by $A(x^i) = x^{i+4}$ where $i + 4$ is taken modulo $12$.



      Find $(a)$ the minimal polynomial of $A$ and $(b)$ the characteristic polynomial of $A$.




      My try:



      I coulnot find another way so I tried the brute force method. $:$I found the matrix representation of the operator is $$A=begin{bmatrix} 0&0&0&0&0&0&0&0&1&0&0&0\0&0&0&0&0&0&0&0&0&1&0&0\0&0&0&0&0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&0&0&0&0&1\1&0&0&0&0&0&0&0&0&0&0&0\0&1&0&0&0&0&0&0&0&0&0&0\0&0&1&0&0&0&0&0&0&0&0&0\0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0\end{bmatrix}.$$



      The characteristic polynomial I found to be $lambda^{12}-4lambda^9+6lambda^6-4lambda^3+1$.$:$(It took me almost 40 minutes. $:$Is there another way to do this problem? Provide hints or suggestions please.



      $rule{17cm}{1pt}$



      Taking forward the answer provided by $textbf{Servaes}$ "The minimal polynomial of $A|_{U_i}$ is still $X^3−1$." Taking $U_1=span{x_1,x_5,x_9}$ we see $A(x)=x^5,:A^2(x)=x^9,:A^3(x)=ximplies (A^3-I)=0$ and since it factors into linear irredeucible factors, we have the minimal polynomial of $A|_{U_i}$ is $X^3−1$.



      Completing the proof: We show that characteristic polynomial of $A$ is the product of characteristic polynomials of $A|_{U_i}$ where $V=oplus U_i$. We have seen that minimal polynomial of $A|_{U_i}$ is $X^3−1$ which is precisely the characteristic polynomial. So let $p_i(lambda)$ is characteristic polynomial corresponding to eigenvalue $lambda_i$ and invariant subspace $U_i$ and $p(lambda)$ is the characteristic polynomial of A. Then $displaystyle p(lambda_i)=0 :forall;i implies p_i(lambda)|p(lambda) :forall;i implies p(lambda)=prod_ip_i(lambda)$.



      $Big($$p(lambda)$ is atleast $displaystyleprod_ip_i(lambda)$. If $existslambdaneqlambda_i forall: i$ such that $p(lambda)=0$ then $V$ is not $oplus U_i$ $Big)$



      $rule{17cm}{1pt}$



      Minimal polynomial : $X^3-1qquad$ Characteristic polynomial : $(X^3-1)^4$.










      share|cite|improve this question
















      Consider the vector space $V =left{a_0+a_1x+cdots+a_{11}x^{11},;a_iinmathbb{R}right}$. Define a linear operator $A$ on $V$ by $A(x^i) = x^{i+4}$ where $i + 4$ is taken modulo $12$.



      Find $(a)$ the minimal polynomial of $A$ and $(b)$ the characteristic polynomial of $A$.




      My try:



      I coulnot find another way so I tried the brute force method. $:$I found the matrix representation of the operator is $$A=begin{bmatrix} 0&0&0&0&0&0&0&0&1&0&0&0\0&0&0&0&0&0&0&0&0&1&0&0\0&0&0&0&0&0&0&0&0&0&1&0\0&0&0&0&0&0&0&0&0&0&0&1\1&0&0&0&0&0&0&0&0&0&0&0\0&1&0&0&0&0&0&0&0&0&0&0\0&0&1&0&0&0&0&0&0&0&0&0\0&0&0&1&0&0&0&0&0&0&0&0\0&0&0&0&1&0&0&0&0&0&0&0\0&0&0&0&0&1&0&0&0&0&0&0\0&0&0&0&0&0&1&0&0&0&0&0\0&0&0&0&0&0&0&1&0&0&0&0\end{bmatrix}.$$



      The characteristic polynomial I found to be $lambda^{12}-4lambda^9+6lambda^6-4lambda^3+1$.$:$(It took me almost 40 minutes. $:$Is there another way to do this problem? Provide hints or suggestions please.



      $rule{17cm}{1pt}$



      Taking forward the answer provided by $textbf{Servaes}$ "The minimal polynomial of $A|_{U_i}$ is still $X^3−1$." Taking $U_1=span{x_1,x_5,x_9}$ we see $A(x)=x^5,:A^2(x)=x^9,:A^3(x)=ximplies (A^3-I)=0$ and since it factors into linear irredeucible factors, we have the minimal polynomial of $A|_{U_i}$ is $X^3−1$.



      Completing the proof: We show that characteristic polynomial of $A$ is the product of characteristic polynomials of $A|_{U_i}$ where $V=oplus U_i$. We have seen that minimal polynomial of $A|_{U_i}$ is $X^3−1$ which is precisely the characteristic polynomial. So let $p_i(lambda)$ is characteristic polynomial corresponding to eigenvalue $lambda_i$ and invariant subspace $U_i$ and $p(lambda)$ is the characteristic polynomial of A. Then $displaystyle p(lambda_i)=0 :forall;i implies p_i(lambda)|p(lambda) :forall;i implies p(lambda)=prod_ip_i(lambda)$.



      $Big($$p(lambda)$ is atleast $displaystyleprod_ip_i(lambda)$. If $existslambdaneqlambda_i forall: i$ such that $p(lambda)=0$ then $V$ is not $oplus U_i$ $Big)$



      $rule{17cm}{1pt}$



      Minimal polynomial : $X^3-1qquad$ Characteristic polynomial : $(X^3-1)^4$.







      linear-algebra matrices minimal-polynomials






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      edited Nov 17 at 8:24

























      asked Nov 16 at 17:00









      Yadati Kiran

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          It usually helps to find some nontrivial relation that the given operator satisfies. Clearly
          $$A^3(x^i)=x^{i+12}=x^i,$$
          for all $i$, so $A$ is a zero of $X^3-I$. This factors as
          $$X^3-I=(X-I)(X^2+X+I),$$
          which has no repeated factors, so this is the minimal polynomial of $A$. The characteristic polynomial has the same irreducible factors and has degree $12$, so the minimal polynomial equals
          $$(X-I)^a(X^2+X+I)^b,$$
          for some positive integers $a$ and $b$ satisfying $a+2b=12$.



          Note that $a$ is the algebraic multiplicity of the eigenvalue $1$, which is at least $4$ because
          $$1+x^4+x^8,qquad x+x^5+x^9,qquad x^2+x^6+x^{10},qquad x^3+x^7+x^{11},$$
          are four linearly independent eigenvectors with eigenvalue $1$. So $(a,b)$ is either $(4,4)$, $(6,3)$, $(8,2)$ or $(10,1)$.





          The following is a bit contrived and implicitly assumes some slightly advanced ideas, but it is an (almost) computation-free way of determining the characteristic polynomial:



          For $iin{1,2,3,4}$ let $U_i:=operatorname{span}(x^i,A(x^i),A^2(x^i))$. Then $U_icap U_j=0$ whenever $ineq j$, and the $U_i$ together span $V$ and are invariant under $A$. This yields a decomposition
          $$V=U_1oplus U_2oplus U_3oplus U_4,$$
          of $A$-invariant subspaces. The minimal polynomial of $Avert_{U_i}$ is still $X^3-1$ (verify this!), hence it is the characteristic polynomial of the restriction. The characteristic polynomial of $A$ is the product of the characteristic polynomials of the $Avert_{U_i}$, so it is $(X^3-1)^4$.






          share|cite|improve this answer























          • How did you determine the four independent eigenvectors?
            – Yadati Kiran
            Nov 16 at 17:23






          • 2




            Because $A$ acts so nicely on the $x^i$, it is not hard to construct eigenvectors; because for example $A(1)=x^4$ and $A(x^4)=x^8$ and $A(x^8)=1$, their sum is invariant under $A$.
            – Servaes
            Nov 16 at 17:25






          • 1




            I've added a method to compute the characteristic polynomial using similar ideas.
            – Servaes
            Nov 16 at 17:36










          • Now I get it! Thanks.
            – Yadati Kiran
            Nov 16 at 17:42






          • 1




            I was initially tried using invariant subspaces. But could not accurately define the linearly independent vectors. Now I understood.
            – Yadati Kiran
            Nov 16 at 17:44













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          up vote
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          It usually helps to find some nontrivial relation that the given operator satisfies. Clearly
          $$A^3(x^i)=x^{i+12}=x^i,$$
          for all $i$, so $A$ is a zero of $X^3-I$. This factors as
          $$X^3-I=(X-I)(X^2+X+I),$$
          which has no repeated factors, so this is the minimal polynomial of $A$. The characteristic polynomial has the same irreducible factors and has degree $12$, so the minimal polynomial equals
          $$(X-I)^a(X^2+X+I)^b,$$
          for some positive integers $a$ and $b$ satisfying $a+2b=12$.



          Note that $a$ is the algebraic multiplicity of the eigenvalue $1$, which is at least $4$ because
          $$1+x^4+x^8,qquad x+x^5+x^9,qquad x^2+x^6+x^{10},qquad x^3+x^7+x^{11},$$
          are four linearly independent eigenvectors with eigenvalue $1$. So $(a,b)$ is either $(4,4)$, $(6,3)$, $(8,2)$ or $(10,1)$.





          The following is a bit contrived and implicitly assumes some slightly advanced ideas, but it is an (almost) computation-free way of determining the characteristic polynomial:



          For $iin{1,2,3,4}$ let $U_i:=operatorname{span}(x^i,A(x^i),A^2(x^i))$. Then $U_icap U_j=0$ whenever $ineq j$, and the $U_i$ together span $V$ and are invariant under $A$. This yields a decomposition
          $$V=U_1oplus U_2oplus U_3oplus U_4,$$
          of $A$-invariant subspaces. The minimal polynomial of $Avert_{U_i}$ is still $X^3-1$ (verify this!), hence it is the characteristic polynomial of the restriction. The characteristic polynomial of $A$ is the product of the characteristic polynomials of the $Avert_{U_i}$, so it is $(X^3-1)^4$.






          share|cite|improve this answer























          • How did you determine the four independent eigenvectors?
            – Yadati Kiran
            Nov 16 at 17:23






          • 2




            Because $A$ acts so nicely on the $x^i$, it is not hard to construct eigenvectors; because for example $A(1)=x^4$ and $A(x^4)=x^8$ and $A(x^8)=1$, their sum is invariant under $A$.
            – Servaes
            Nov 16 at 17:25






          • 1




            I've added a method to compute the characteristic polynomial using similar ideas.
            – Servaes
            Nov 16 at 17:36










          • Now I get it! Thanks.
            – Yadati Kiran
            Nov 16 at 17:42






          • 1




            I was initially tried using invariant subspaces. But could not accurately define the linearly independent vectors. Now I understood.
            – Yadati Kiran
            Nov 16 at 17:44

















          up vote
          5
          down vote



          accepted










          It usually helps to find some nontrivial relation that the given operator satisfies. Clearly
          $$A^3(x^i)=x^{i+12}=x^i,$$
          for all $i$, so $A$ is a zero of $X^3-I$. This factors as
          $$X^3-I=(X-I)(X^2+X+I),$$
          which has no repeated factors, so this is the minimal polynomial of $A$. The characteristic polynomial has the same irreducible factors and has degree $12$, so the minimal polynomial equals
          $$(X-I)^a(X^2+X+I)^b,$$
          for some positive integers $a$ and $b$ satisfying $a+2b=12$.



          Note that $a$ is the algebraic multiplicity of the eigenvalue $1$, which is at least $4$ because
          $$1+x^4+x^8,qquad x+x^5+x^9,qquad x^2+x^6+x^{10},qquad x^3+x^7+x^{11},$$
          are four linearly independent eigenvectors with eigenvalue $1$. So $(a,b)$ is either $(4,4)$, $(6,3)$, $(8,2)$ or $(10,1)$.





          The following is a bit contrived and implicitly assumes some slightly advanced ideas, but it is an (almost) computation-free way of determining the characteristic polynomial:



          For $iin{1,2,3,4}$ let $U_i:=operatorname{span}(x^i,A(x^i),A^2(x^i))$. Then $U_icap U_j=0$ whenever $ineq j$, and the $U_i$ together span $V$ and are invariant under $A$. This yields a decomposition
          $$V=U_1oplus U_2oplus U_3oplus U_4,$$
          of $A$-invariant subspaces. The minimal polynomial of $Avert_{U_i}$ is still $X^3-1$ (verify this!), hence it is the characteristic polynomial of the restriction. The characteristic polynomial of $A$ is the product of the characteristic polynomials of the $Avert_{U_i}$, so it is $(X^3-1)^4$.






          share|cite|improve this answer























          • How did you determine the four independent eigenvectors?
            – Yadati Kiran
            Nov 16 at 17:23






          • 2




            Because $A$ acts so nicely on the $x^i$, it is not hard to construct eigenvectors; because for example $A(1)=x^4$ and $A(x^4)=x^8$ and $A(x^8)=1$, their sum is invariant under $A$.
            – Servaes
            Nov 16 at 17:25






          • 1




            I've added a method to compute the characteristic polynomial using similar ideas.
            – Servaes
            Nov 16 at 17:36










          • Now I get it! Thanks.
            – Yadati Kiran
            Nov 16 at 17:42






          • 1




            I was initially tried using invariant subspaces. But could not accurately define the linearly independent vectors. Now I understood.
            – Yadati Kiran
            Nov 16 at 17:44















          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          It usually helps to find some nontrivial relation that the given operator satisfies. Clearly
          $$A^3(x^i)=x^{i+12}=x^i,$$
          for all $i$, so $A$ is a zero of $X^3-I$. This factors as
          $$X^3-I=(X-I)(X^2+X+I),$$
          which has no repeated factors, so this is the minimal polynomial of $A$. The characteristic polynomial has the same irreducible factors and has degree $12$, so the minimal polynomial equals
          $$(X-I)^a(X^2+X+I)^b,$$
          for some positive integers $a$ and $b$ satisfying $a+2b=12$.



          Note that $a$ is the algebraic multiplicity of the eigenvalue $1$, which is at least $4$ because
          $$1+x^4+x^8,qquad x+x^5+x^9,qquad x^2+x^6+x^{10},qquad x^3+x^7+x^{11},$$
          are four linearly independent eigenvectors with eigenvalue $1$. So $(a,b)$ is either $(4,4)$, $(6,3)$, $(8,2)$ or $(10,1)$.





          The following is a bit contrived and implicitly assumes some slightly advanced ideas, but it is an (almost) computation-free way of determining the characteristic polynomial:



          For $iin{1,2,3,4}$ let $U_i:=operatorname{span}(x^i,A(x^i),A^2(x^i))$. Then $U_icap U_j=0$ whenever $ineq j$, and the $U_i$ together span $V$ and are invariant under $A$. This yields a decomposition
          $$V=U_1oplus U_2oplus U_3oplus U_4,$$
          of $A$-invariant subspaces. The minimal polynomial of $Avert_{U_i}$ is still $X^3-1$ (verify this!), hence it is the characteristic polynomial of the restriction. The characteristic polynomial of $A$ is the product of the characteristic polynomials of the $Avert_{U_i}$, so it is $(X^3-1)^4$.






          share|cite|improve this answer














          It usually helps to find some nontrivial relation that the given operator satisfies. Clearly
          $$A^3(x^i)=x^{i+12}=x^i,$$
          for all $i$, so $A$ is a zero of $X^3-I$. This factors as
          $$X^3-I=(X-I)(X^2+X+I),$$
          which has no repeated factors, so this is the minimal polynomial of $A$. The characteristic polynomial has the same irreducible factors and has degree $12$, so the minimal polynomial equals
          $$(X-I)^a(X^2+X+I)^b,$$
          for some positive integers $a$ and $b$ satisfying $a+2b=12$.



          Note that $a$ is the algebraic multiplicity of the eigenvalue $1$, which is at least $4$ because
          $$1+x^4+x^8,qquad x+x^5+x^9,qquad x^2+x^6+x^{10},qquad x^3+x^7+x^{11},$$
          are four linearly independent eigenvectors with eigenvalue $1$. So $(a,b)$ is either $(4,4)$, $(6,3)$, $(8,2)$ or $(10,1)$.





          The following is a bit contrived and implicitly assumes some slightly advanced ideas, but it is an (almost) computation-free way of determining the characteristic polynomial:



          For $iin{1,2,3,4}$ let $U_i:=operatorname{span}(x^i,A(x^i),A^2(x^i))$. Then $U_icap U_j=0$ whenever $ineq j$, and the $U_i$ together span $V$ and are invariant under $A$. This yields a decomposition
          $$V=U_1oplus U_2oplus U_3oplus U_4,$$
          of $A$-invariant subspaces. The minimal polynomial of $Avert_{U_i}$ is still $X^3-1$ (verify this!), hence it is the characteristic polynomial of the restriction. The characteristic polynomial of $A$ is the product of the characteristic polynomials of the $Avert_{U_i}$, so it is $(X^3-1)^4$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 16 at 17:43

























          answered Nov 16 at 17:03









          Servaes

          20.6k33789




          20.6k33789












          • How did you determine the four independent eigenvectors?
            – Yadati Kiran
            Nov 16 at 17:23






          • 2




            Because $A$ acts so nicely on the $x^i$, it is not hard to construct eigenvectors; because for example $A(1)=x^4$ and $A(x^4)=x^8$ and $A(x^8)=1$, their sum is invariant under $A$.
            – Servaes
            Nov 16 at 17:25






          • 1




            I've added a method to compute the characteristic polynomial using similar ideas.
            – Servaes
            Nov 16 at 17:36










          • Now I get it! Thanks.
            – Yadati Kiran
            Nov 16 at 17:42






          • 1




            I was initially tried using invariant subspaces. But could not accurately define the linearly independent vectors. Now I understood.
            – Yadati Kiran
            Nov 16 at 17:44




















          • How did you determine the four independent eigenvectors?
            – Yadati Kiran
            Nov 16 at 17:23






          • 2




            Because $A$ acts so nicely on the $x^i$, it is not hard to construct eigenvectors; because for example $A(1)=x^4$ and $A(x^4)=x^8$ and $A(x^8)=1$, their sum is invariant under $A$.
            – Servaes
            Nov 16 at 17:25






          • 1




            I've added a method to compute the characteristic polynomial using similar ideas.
            – Servaes
            Nov 16 at 17:36










          • Now I get it! Thanks.
            – Yadati Kiran
            Nov 16 at 17:42






          • 1




            I was initially tried using invariant subspaces. But could not accurately define the linearly independent vectors. Now I understood.
            – Yadati Kiran
            Nov 16 at 17:44


















          How did you determine the four independent eigenvectors?
          – Yadati Kiran
          Nov 16 at 17:23




          How did you determine the four independent eigenvectors?
          – Yadati Kiran
          Nov 16 at 17:23




          2




          2




          Because $A$ acts so nicely on the $x^i$, it is not hard to construct eigenvectors; because for example $A(1)=x^4$ and $A(x^4)=x^8$ and $A(x^8)=1$, their sum is invariant under $A$.
          – Servaes
          Nov 16 at 17:25




          Because $A$ acts so nicely on the $x^i$, it is not hard to construct eigenvectors; because for example $A(1)=x^4$ and $A(x^4)=x^8$ and $A(x^8)=1$, their sum is invariant under $A$.
          – Servaes
          Nov 16 at 17:25




          1




          1




          I've added a method to compute the characteristic polynomial using similar ideas.
          – Servaes
          Nov 16 at 17:36




          I've added a method to compute the characteristic polynomial using similar ideas.
          – Servaes
          Nov 16 at 17:36












          Now I get it! Thanks.
          – Yadati Kiran
          Nov 16 at 17:42




          Now I get it! Thanks.
          – Yadati Kiran
          Nov 16 at 17:42




          1




          1




          I was initially tried using invariant subspaces. But could not accurately define the linearly independent vectors. Now I understood.
          – Yadati Kiran
          Nov 16 at 17:44






          I was initially tried using invariant subspaces. But could not accurately define the linearly independent vectors. Now I understood.
          – Yadati Kiran
          Nov 16 at 17:44




















           

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