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









$endgroup$













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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

    41.1k245106




    41.1k245106












    • $begingroup$
      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

        121k740116




        121k740116






























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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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