How to find the matrix represented by the polynomials $A^{12}-5A^{11}+…+3I$?












0












$begingroup$


I need to find the characteristic equation of the matrix $ A = begin{bmatrix}
2&1&1\
0&1&0\
1&1&2\
end{bmatrix} $ and find the matrix represented by $ A^{12}-5A^{11}+7A^{10}-3A^{9}-A^8+5A^7-7A^6+3A^5+A^4-5A^3+8A^2-4A+3I $



My attempt:



I started by finding the character equation which was $ lambda^3-5lambda^2+7lambda-3=0 $ where $lambda$ is the eigenvalue, and found $lambda = 5,frac{9+sqrt{69}}{2},frac{9-sqrt{69}}{2} $



$lambda = 5$ satisfies the equation and I would take $lambda = 5$ let $lambda = A$ and $ A^3-5 A^2+7A -3=0 $.

I don't know how to proceed further. All help would be appreciated.



Edit: What I can do, is multiply the matrix $A$ by itself and consequently the resultant matrix would be $A^2$ and so on. However, I'm not sure if this will give an accurate answer.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 21:43










  • $begingroup$
    Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
    $endgroup$
    – tNotr
    Dec 9 '18 at 21:48
















0












$begingroup$


I need to find the characteristic equation of the matrix $ A = begin{bmatrix}
2&1&1\
0&1&0\
1&1&2\
end{bmatrix} $ and find the matrix represented by $ A^{12}-5A^{11}+7A^{10}-3A^{9}-A^8+5A^7-7A^6+3A^5+A^4-5A^3+8A^2-4A+3I $



My attempt:



I started by finding the character equation which was $ lambda^3-5lambda^2+7lambda-3=0 $ where $lambda$ is the eigenvalue, and found $lambda = 5,frac{9+sqrt{69}}{2},frac{9-sqrt{69}}{2} $



$lambda = 5$ satisfies the equation and I would take $lambda = 5$ let $lambda = A$ and $ A^3-5 A^2+7A -3=0 $.

I don't know how to proceed further. All help would be appreciated.



Edit: What I can do, is multiply the matrix $A$ by itself and consequently the resultant matrix would be $A^2$ and so on. However, I'm not sure if this will give an accurate answer.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 21:43










  • $begingroup$
    Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
    $endgroup$
    – tNotr
    Dec 9 '18 at 21:48














0












0








0





$begingroup$


I need to find the characteristic equation of the matrix $ A = begin{bmatrix}
2&1&1\
0&1&0\
1&1&2\
end{bmatrix} $ and find the matrix represented by $ A^{12}-5A^{11}+7A^{10}-3A^{9}-A^8+5A^7-7A^6+3A^5+A^4-5A^3+8A^2-4A+3I $



My attempt:



I started by finding the character equation which was $ lambda^3-5lambda^2+7lambda-3=0 $ where $lambda$ is the eigenvalue, and found $lambda = 5,frac{9+sqrt{69}}{2},frac{9-sqrt{69}}{2} $



$lambda = 5$ satisfies the equation and I would take $lambda = 5$ let $lambda = A$ and $ A^3-5 A^2+7A -3=0 $.

I don't know how to proceed further. All help would be appreciated.



Edit: What I can do, is multiply the matrix $A$ by itself and consequently the resultant matrix would be $A^2$ and so on. However, I'm not sure if this will give an accurate answer.






linear-algebra matrices







share|cite|improve this question











$endgroup$




I need to find the characteristic equation of the matrix $ A = begin{bmatrix}
2&1&1\
0&1&0\
1&1&2\
end{bmatrix} $ and find the matrix represented by $ A^{12}-5A^{11}+7A^{10}-3A^{9}-A^8+5A^7-7A^6+3A^5+A^4-5A^3+8A^2-4A+3I $



My attempt:



I started by finding the character equation which was $ lambda^3-5lambda^2+7lambda-3=0 $ where $lambda$ is the eigenvalue, and found $lambda = 5,frac{9+sqrt{69}}{2},frac{9-sqrt{69}}{2} $



$lambda = 5$ satisfies the equation and I would take $lambda = 5$ let $lambda = A$ and $ A^3-5 A^2+7A -3=0 $.

I don't know how to proceed further. All help would be appreciated.



Edit: What I can do, is multiply the matrix $A$ by itself and consequently the resultant matrix would be $A^2$ and so on. However, I'm not sure if this will give an accurate answer.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 21:59







tNotr

















asked Dec 9 '18 at 21:30









tNotrtNotr

142




142








  • 1




    $begingroup$
    I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 21:43










  • $begingroup$
    Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
    $endgroup$
    – tNotr
    Dec 9 '18 at 21:48














  • 1




    $begingroup$
    I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 21:43










  • $begingroup$
    Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
    $endgroup$
    – tNotr
    Dec 9 '18 at 21:48








1




1




$begingroup$
I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
$endgroup$
– JimmyK4542
Dec 9 '18 at 21:43




$begingroup$
I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
$endgroup$
– JimmyK4542
Dec 9 '18 at 21:43












$begingroup$
Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
$endgroup$
– tNotr
Dec 9 '18 at 21:48




$begingroup$
Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
$endgroup$
– tNotr
Dec 9 '18 at 21:48










2 Answers
2






active

oldest

votes


















4












$begingroup$

The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.



So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).



Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).



Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).



Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).



If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$



Can you take it from here?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I find ${}+8lambda$ for the characteristic polynomial.
    $endgroup$
    – Bernard
    Dec 9 '18 at 21:53










  • $begingroup$
    yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
    $endgroup$
    – tNotr
    Dec 9 '18 at 21:55












  • $begingroup$
    You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 22:25



















2












$begingroup$

Check you computation for the characteristic polynomial $chi_A$.



Hint:



Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
$$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
$$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.



    So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).



    Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).



    Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).



    Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).



    If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$



    Can you take it from here?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I find ${}+8lambda$ for the characteristic polynomial.
      $endgroup$
      – Bernard
      Dec 9 '18 at 21:53










    • $begingroup$
      yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
      $endgroup$
      – tNotr
      Dec 9 '18 at 21:55












    • $begingroup$
      You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
      $endgroup$
      – JimmyK4542
      Dec 9 '18 at 22:25
















    4












    $begingroup$

    The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.



    So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).



    Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).



    Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).



    Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).



    If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$



    Can you take it from here?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I find ${}+8lambda$ for the characteristic polynomial.
      $endgroup$
      – Bernard
      Dec 9 '18 at 21:53










    • $begingroup$
      yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
      $endgroup$
      – tNotr
      Dec 9 '18 at 21:55












    • $begingroup$
      You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
      $endgroup$
      – JimmyK4542
      Dec 9 '18 at 22:25














    4












    4








    4





    $begingroup$

    The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.



    So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).



    Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).



    Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).



    Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).



    If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$



    Can you take it from here?






    share|cite|improve this answer









    $endgroup$



    The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.



    So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).



    Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).



    Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).



    Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).



    If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$



    Can you take it from here?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 9 '18 at 21:39









    JimmyK4542JimmyK4542

    40.8k245105




    40.8k245105












    • $begingroup$
      I find ${}+8lambda$ for the characteristic polynomial.
      $endgroup$
      – Bernard
      Dec 9 '18 at 21:53










    • $begingroup$
      yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
      $endgroup$
      – tNotr
      Dec 9 '18 at 21:55












    • $begingroup$
      You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
      $endgroup$
      – JimmyK4542
      Dec 9 '18 at 22:25


















    • $begingroup$
      I find ${}+8lambda$ for the characteristic polynomial.
      $endgroup$
      – Bernard
      Dec 9 '18 at 21:53










    • $begingroup$
      yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
      $endgroup$
      – tNotr
      Dec 9 '18 at 21:55












    • $begingroup$
      You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
      $endgroup$
      – JimmyK4542
      Dec 9 '18 at 22:25
















    $begingroup$
    I find ${}+8lambda$ for the characteristic polynomial.
    $endgroup$
    – Bernard
    Dec 9 '18 at 21:53




    $begingroup$
    I find ${}+8lambda$ for the characteristic polynomial.
    $endgroup$
    – Bernard
    Dec 9 '18 at 21:53












    $begingroup$
    yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
    $endgroup$
    – tNotr
    Dec 9 '18 at 21:55






    $begingroup$
    yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
    $endgroup$
    – tNotr
    Dec 9 '18 at 21:55














    $begingroup$
    You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 22:25




    $begingroup$
    You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
    $endgroup$
    – JimmyK4542
    Dec 9 '18 at 22:25











    2












    $begingroup$

    Check you computation for the characteristic polynomial $chi_A$.



    Hint:



    Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
    $$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
    $$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Check you computation for the characteristic polynomial $chi_A$.



      Hint:



      Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
      $$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
      $$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Check you computation for the characteristic polynomial $chi_A$.



        Hint:



        Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
        $$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
        $$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:






        share|cite|improve this answer









        $endgroup$



        Check you computation for the characteristic polynomial $chi_A$.



        Hint:



        Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
        $$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
        $$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 22:00









        BernardBernard

        120k740114




        120k740114






























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