Eigenvalues and Eigenvectors of Sum of Symmetric Matrix











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Question:



Let A = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}



Find all eigenvalues and eigenvectors of the martrix:



$$sum_{n=1}^{100} A^n = A^{100} +A^{99} +...+A^2+A$$



I know that the eigenvectors of A are begin{bmatrix} 1 \ 1 end{bmatrix} and begin{bmatrix} 1 \ -1 end{bmatrix}
But I do not see any sort of correlation with the sum term and A's eigenvectors.










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  • 3




    Try evaluating $(sum_{n=1}^{100} A^n) begin{bmatrix} 1 \ 1 end{bmatrix}$ and do the same with the other eigenvector. What happens?
    – Giuseppe Negro
    Dec 11 at 19:24

















up vote
5
down vote

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Question:



Let A = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}



Find all eigenvalues and eigenvectors of the martrix:



$$sum_{n=1}^{100} A^n = A^{100} +A^{99} +...+A^2+A$$



I know that the eigenvectors of A are begin{bmatrix} 1 \ 1 end{bmatrix} and begin{bmatrix} 1 \ -1 end{bmatrix}
But I do not see any sort of correlation with the sum term and A's eigenvectors.










share|cite|improve this question







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mhall14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 3




    Try evaluating $(sum_{n=1}^{100} A^n) begin{bmatrix} 1 \ 1 end{bmatrix}$ and do the same with the other eigenvector. What happens?
    – Giuseppe Negro
    Dec 11 at 19:24















up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





Question:



Let A = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}



Find all eigenvalues and eigenvectors of the martrix:



$$sum_{n=1}^{100} A^n = A^{100} +A^{99} +...+A^2+A$$



I know that the eigenvectors of A are begin{bmatrix} 1 \ 1 end{bmatrix} and begin{bmatrix} 1 \ -1 end{bmatrix}
But I do not see any sort of correlation with the sum term and A's eigenvectors.










share|cite|improve this question







New contributor




mhall14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Question:



Let A = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}



Find all eigenvalues and eigenvectors of the martrix:



$$sum_{n=1}^{100} A^n = A^{100} +A^{99} +...+A^2+A$$



I know that the eigenvectors of A are begin{bmatrix} 1 \ 1 end{bmatrix} and begin{bmatrix} 1 \ -1 end{bmatrix}
But I do not see any sort of correlation with the sum term and A's eigenvectors.







linear-algebra eigenvalues-eigenvectors symmetric-matrices






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asked Dec 11 at 19:13









mhall14

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  • 3




    Try evaluating $(sum_{n=1}^{100} A^n) begin{bmatrix} 1 \ 1 end{bmatrix}$ and do the same with the other eigenvector. What happens?
    – Giuseppe Negro
    Dec 11 at 19:24
















  • 3




    Try evaluating $(sum_{n=1}^{100} A^n) begin{bmatrix} 1 \ 1 end{bmatrix}$ and do the same with the other eigenvector. What happens?
    – Giuseppe Negro
    Dec 11 at 19:24










3




3




Try evaluating $(sum_{n=1}^{100} A^n) begin{bmatrix} 1 \ 1 end{bmatrix}$ and do the same with the other eigenvector. What happens?
– Giuseppe Negro
Dec 11 at 19:24






Try evaluating $(sum_{n=1}^{100} A^n) begin{bmatrix} 1 \ 1 end{bmatrix}$ and do the same with the other eigenvector. What happens?
– Giuseppe Negro
Dec 11 at 19:24












6 Answers
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up vote
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Hint: If $$A = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}$$then we have $$A^2 = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}=begin{bmatrix} 2 & 2 \ 2 & 2 \ end{bmatrix}\A^3=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 2&2 \ 2&2 \ end{bmatrix}=begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}\A^4=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}=begin{bmatrix}8&8 \ 8&8 \ end{bmatrix}\.\.\.\.$$and you can prove by induction that $$A^k=begin{bmatrix} 2^{k-1}&2^{k-1} \ 2^{k-1}&2^{k-1}\ end{bmatrix}$$can you finish now?






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    By linearity, given any polynomial $p$ and matrix $A$, the eigenvectors of $p(A)$ are the same as the eigenvectors of $A$, and the associated eigenvalues are $p(lambda)$; see this question.



    For instance, in this case, if $Av=lambda v$, then $A^nv=lambda^nv$, and $(sum_{n=1}^{100}A^n)v=sum_{n=1}^{100}(A^nv )=sum_{n=1}^{100}(lambda^nv)=(sum_{n=1}^{100}lambda^n)v$. Thus, $v$ is an eigenvector with eigenvalue $sum_{n=1}^{100}lambda^n$. $A$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$. $p(2)$ is a geometric series, so it is $2^{101}-1$. $p(0)$ is just zero. So $p(A)$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2^{101}-1$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$






    share|cite|improve this answer





















    • This is the simplest way to go.
      – Giuseppe Negro
      Dec 12 at 22:14


















    up vote
    1
    down vote













    Hint :



    Recall the Cayley-Hamilton Theorem (by Wikipedia) :




    For a general n×n invertible matrix $A$, i.e., one with nonzero determinant, $A^{−1}$ can thus be written as an $(n − 1)$-th order polynomial expression in $A$: As indicated, the Cayley–Hamilton theorem amounts to the identity :
    $$p(A) = A^n + c_{n-1}A^{n-1} + dots + cA + (-1)^ndet(A)I_n = O$$
    The coefficients ci are given by the elementary symmetric polynomials of the eigenvalues of $A$. Using Newton identities, the elementary symmetric polynomials can in turn be expressed in terms of power sum symmetric polynomials of the eigenvalues:
    $$s_k = sum_{i=1}^n lambda_i^k = text{tr}(A^k)$$







    share|cite|improve this answer




























      up vote
      1
      down vote













      You can explicitly compute $sum_{i=1}^{100}A^i$. First diagonalize $A$, namely rewrite $A$ as $A=PDP^{-1}$.



      Now
      begin{align}
      sum_{i=1}^{100}A^i&=sum_{i=1}^{100}PD^iP^{-1}\&=Pleft(sum_{i=1}^{100} D^iright)P^{-1}
      end{align}



      Notice that
      $$(D-I)left(sum_{i=1}^{100}D^iright)=D^{101}-I.$$
      SInce $D-I$ is invertible (you can check it)
      $$sum_{i=1}^{100}D^i=(D-I)^{-1}(D^{101}-I).$$
      Therefore
      $$sum_{i=1}^{100}A^i=P(D-I)^{-1}(D^{101}-I)P^{-1}.$$






      share|cite|improve this answer




























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        It is easy to prove that for $kin Bbb{N},$ $$A^k=begin{bmatrix} 2^{k-1} & 2^{k-1} \ 2^{k-1} & 2^{k-1} \ end{bmatrix}.$$ The sum is
        $$Sigma=begin{bmatrix} 2^{100}-1 & 2^{100}-1 \ 2^{100}-1 & 2^{100}-1 \ end{bmatrix},$$ from where the eigenvalues $0$ and $(2^{101}-2).$



        Each matrix $A^k, k=1,dots,100$ has eigenvalues $0$ and $2^k,$ the corresponding eigenvectors are those of $A:$ $(1,-1)^T, (1,1)^T.$

        Thus $(1,-1)^T, (1,1)^T$ are eigenvectors of $Sigma.$






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          Since you have 2 linear independent eigenvectors, $A$ is diagonalizable. You may find useful to replace $A$ in your polynomial expression by its diagonalization because this will simplify the operations you need to do.






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            6 Answers
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            6 Answers
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            up vote
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            Hint: If $$A = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}$$then we have $$A^2 = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}=begin{bmatrix} 2 & 2 \ 2 & 2 \ end{bmatrix}\A^3=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 2&2 \ 2&2 \ end{bmatrix}=begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}\A^4=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}=begin{bmatrix}8&8 \ 8&8 \ end{bmatrix}\.\.\.\.$$and you can prove by induction that $$A^k=begin{bmatrix} 2^{k-1}&2^{k-1} \ 2^{k-1}&2^{k-1}\ end{bmatrix}$$can you finish now?






            share|cite|improve this answer

























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              accepted










              Hint: If $$A = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}$$then we have $$A^2 = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}=begin{bmatrix} 2 & 2 \ 2 & 2 \ end{bmatrix}\A^3=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 2&2 \ 2&2 \ end{bmatrix}=begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}\A^4=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}=begin{bmatrix}8&8 \ 8&8 \ end{bmatrix}\.\.\.\.$$and you can prove by induction that $$A^k=begin{bmatrix} 2^{k-1}&2^{k-1} \ 2^{k-1}&2^{k-1}\ end{bmatrix}$$can you finish now?






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                up vote
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                Hint: If $$A = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}$$then we have $$A^2 = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}=begin{bmatrix} 2 & 2 \ 2 & 2 \ end{bmatrix}\A^3=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 2&2 \ 2&2 \ end{bmatrix}=begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}\A^4=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}=begin{bmatrix}8&8 \ 8&8 \ end{bmatrix}\.\.\.\.$$and you can prove by induction that $$A^k=begin{bmatrix} 2^{k-1}&2^{k-1} \ 2^{k-1}&2^{k-1}\ end{bmatrix}$$can you finish now?






                share|cite|improve this answer












                Hint: If $$A = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}$$then we have $$A^2 = begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}=begin{bmatrix} 2 & 2 \ 2 & 2 \ end{bmatrix}\A^3=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 2&2 \ 2&2 \ end{bmatrix}=begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}\A^4=begin{bmatrix} 1 & 1 \ 1 & 1 \ end{bmatrix}begin{bmatrix} 4&4 \ 4&4 \ end{bmatrix}=begin{bmatrix}8&8 \ 8&8 \ end{bmatrix}\.\.\.\.$$and you can prove by induction that $$A^k=begin{bmatrix} 2^{k-1}&2^{k-1} \ 2^{k-1}&2^{k-1}\ end{bmatrix}$$can you finish now?







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                answered Dec 11 at 19:27









                Mostafa Ayaz

                13.6k3836




                13.6k3836






















                    up vote
                    3
                    down vote













                    By linearity, given any polynomial $p$ and matrix $A$, the eigenvectors of $p(A)$ are the same as the eigenvectors of $A$, and the associated eigenvalues are $p(lambda)$; see this question.



                    For instance, in this case, if $Av=lambda v$, then $A^nv=lambda^nv$, and $(sum_{n=1}^{100}A^n)v=sum_{n=1}^{100}(A^nv )=sum_{n=1}^{100}(lambda^nv)=(sum_{n=1}^{100}lambda^n)v$. Thus, $v$ is an eigenvector with eigenvalue $sum_{n=1}^{100}lambda^n$. $A$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$. $p(2)$ is a geometric series, so it is $2^{101}-1$. $p(0)$ is just zero. So $p(A)$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2^{101}-1$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$






                    share|cite|improve this answer





















                    • This is the simplest way to go.
                      – Giuseppe Negro
                      Dec 12 at 22:14















                    up vote
                    3
                    down vote













                    By linearity, given any polynomial $p$ and matrix $A$, the eigenvectors of $p(A)$ are the same as the eigenvectors of $A$, and the associated eigenvalues are $p(lambda)$; see this question.



                    For instance, in this case, if $Av=lambda v$, then $A^nv=lambda^nv$, and $(sum_{n=1}^{100}A^n)v=sum_{n=1}^{100}(A^nv )=sum_{n=1}^{100}(lambda^nv)=(sum_{n=1}^{100}lambda^n)v$. Thus, $v$ is an eigenvector with eigenvalue $sum_{n=1}^{100}lambda^n$. $A$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$. $p(2)$ is a geometric series, so it is $2^{101}-1$. $p(0)$ is just zero. So $p(A)$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2^{101}-1$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$






                    share|cite|improve this answer





















                    • This is the simplest way to go.
                      – Giuseppe Negro
                      Dec 12 at 22:14













                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    By linearity, given any polynomial $p$ and matrix $A$, the eigenvectors of $p(A)$ are the same as the eigenvectors of $A$, and the associated eigenvalues are $p(lambda)$; see this question.



                    For instance, in this case, if $Av=lambda v$, then $A^nv=lambda^nv$, and $(sum_{n=1}^{100}A^n)v=sum_{n=1}^{100}(A^nv )=sum_{n=1}^{100}(lambda^nv)=(sum_{n=1}^{100}lambda^n)v$. Thus, $v$ is an eigenvector with eigenvalue $sum_{n=1}^{100}lambda^n$. $A$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$. $p(2)$ is a geometric series, so it is $2^{101}-1$. $p(0)$ is just zero. So $p(A)$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2^{101}-1$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$






                    share|cite|improve this answer












                    By linearity, given any polynomial $p$ and matrix $A$, the eigenvectors of $p(A)$ are the same as the eigenvectors of $A$, and the associated eigenvalues are $p(lambda)$; see this question.



                    For instance, in this case, if $Av=lambda v$, then $A^nv=lambda^nv$, and $(sum_{n=1}^{100}A^n)v=sum_{n=1}^{100}(A^nv )=sum_{n=1}^{100}(lambda^nv)=(sum_{n=1}^{100}lambda^n)v$. Thus, $v$ is an eigenvector with eigenvalue $sum_{n=1}^{100}lambda^n$. $A$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$. $p(2)$ is a geometric series, so it is $2^{101}-1$. $p(0)$ is just zero. So $p(A)$ has eigenvectors, eigenvalues of $v=begin{bmatrix} 1 \ 1 end{bmatrix} $ $lambda=2^{101}-1$ and $v=begin{bmatrix} 1 \ -1 end{bmatrix} $ $lambda=0$







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                    answered Dec 11 at 22:46









                    Acccumulation

                    6,6512616




                    6,6512616












                    • This is the simplest way to go.
                      – Giuseppe Negro
                      Dec 12 at 22:14


















                    • This is the simplest way to go.
                      – Giuseppe Negro
                      Dec 12 at 22:14
















                    This is the simplest way to go.
                    – Giuseppe Negro
                    Dec 12 at 22:14




                    This is the simplest way to go.
                    – Giuseppe Negro
                    Dec 12 at 22:14










                    up vote
                    1
                    down vote













                    Hint :



                    Recall the Cayley-Hamilton Theorem (by Wikipedia) :




                    For a general n×n invertible matrix $A$, i.e., one with nonzero determinant, $A^{−1}$ can thus be written as an $(n − 1)$-th order polynomial expression in $A$: As indicated, the Cayley–Hamilton theorem amounts to the identity :
                    $$p(A) = A^n + c_{n-1}A^{n-1} + dots + cA + (-1)^ndet(A)I_n = O$$
                    The coefficients ci are given by the elementary symmetric polynomials of the eigenvalues of $A$. Using Newton identities, the elementary symmetric polynomials can in turn be expressed in terms of power sum symmetric polynomials of the eigenvalues:
                    $$s_k = sum_{i=1}^n lambda_i^k = text{tr}(A^k)$$







                    share|cite|improve this answer

























                      up vote
                      1
                      down vote













                      Hint :



                      Recall the Cayley-Hamilton Theorem (by Wikipedia) :




                      For a general n×n invertible matrix $A$, i.e., one with nonzero determinant, $A^{−1}$ can thus be written as an $(n − 1)$-th order polynomial expression in $A$: As indicated, the Cayley–Hamilton theorem amounts to the identity :
                      $$p(A) = A^n + c_{n-1}A^{n-1} + dots + cA + (-1)^ndet(A)I_n = O$$
                      The coefficients ci are given by the elementary symmetric polynomials of the eigenvalues of $A$. Using Newton identities, the elementary symmetric polynomials can in turn be expressed in terms of power sum symmetric polynomials of the eigenvalues:
                      $$s_k = sum_{i=1}^n lambda_i^k = text{tr}(A^k)$$







                      share|cite|improve this answer























                        up vote
                        1
                        down vote










                        up vote
                        1
                        down vote









                        Hint :



                        Recall the Cayley-Hamilton Theorem (by Wikipedia) :




                        For a general n×n invertible matrix $A$, i.e., one with nonzero determinant, $A^{−1}$ can thus be written as an $(n − 1)$-th order polynomial expression in $A$: As indicated, the Cayley–Hamilton theorem amounts to the identity :
                        $$p(A) = A^n + c_{n-1}A^{n-1} + dots + cA + (-1)^ndet(A)I_n = O$$
                        The coefficients ci are given by the elementary symmetric polynomials of the eigenvalues of $A$. Using Newton identities, the elementary symmetric polynomials can in turn be expressed in terms of power sum symmetric polynomials of the eigenvalues:
                        $$s_k = sum_{i=1}^n lambda_i^k = text{tr}(A^k)$$







                        share|cite|improve this answer












                        Hint :



                        Recall the Cayley-Hamilton Theorem (by Wikipedia) :




                        For a general n×n invertible matrix $A$, i.e., one with nonzero determinant, $A^{−1}$ can thus be written as an $(n − 1)$-th order polynomial expression in $A$: As indicated, the Cayley–Hamilton theorem amounts to the identity :
                        $$p(A) = A^n + c_{n-1}A^{n-1} + dots + cA + (-1)^ndet(A)I_n = O$$
                        The coefficients ci are given by the elementary symmetric polynomials of the eigenvalues of $A$. Using Newton identities, the elementary symmetric polynomials can in turn be expressed in terms of power sum symmetric polynomials of the eigenvalues:
                        $$s_k = sum_{i=1}^n lambda_i^k = text{tr}(A^k)$$








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                        answered Dec 11 at 19:20









                        Rebellos

                        13.9k31244




                        13.9k31244






















                            up vote
                            1
                            down vote













                            You can explicitly compute $sum_{i=1}^{100}A^i$. First diagonalize $A$, namely rewrite $A$ as $A=PDP^{-1}$.



                            Now
                            begin{align}
                            sum_{i=1}^{100}A^i&=sum_{i=1}^{100}PD^iP^{-1}\&=Pleft(sum_{i=1}^{100} D^iright)P^{-1}
                            end{align}



                            Notice that
                            $$(D-I)left(sum_{i=1}^{100}D^iright)=D^{101}-I.$$
                            SInce $D-I$ is invertible (you can check it)
                            $$sum_{i=1}^{100}D^i=(D-I)^{-1}(D^{101}-I).$$
                            Therefore
                            $$sum_{i=1}^{100}A^i=P(D-I)^{-1}(D^{101}-I)P^{-1}.$$






                            share|cite|improve this answer

























                              up vote
                              1
                              down vote













                              You can explicitly compute $sum_{i=1}^{100}A^i$. First diagonalize $A$, namely rewrite $A$ as $A=PDP^{-1}$.



                              Now
                              begin{align}
                              sum_{i=1}^{100}A^i&=sum_{i=1}^{100}PD^iP^{-1}\&=Pleft(sum_{i=1}^{100} D^iright)P^{-1}
                              end{align}



                              Notice that
                              $$(D-I)left(sum_{i=1}^{100}D^iright)=D^{101}-I.$$
                              SInce $D-I$ is invertible (you can check it)
                              $$sum_{i=1}^{100}D^i=(D-I)^{-1}(D^{101}-I).$$
                              Therefore
                              $$sum_{i=1}^{100}A^i=P(D-I)^{-1}(D^{101}-I)P^{-1}.$$






                              share|cite|improve this answer























                                up vote
                                1
                                down vote










                                up vote
                                1
                                down vote









                                You can explicitly compute $sum_{i=1}^{100}A^i$. First diagonalize $A$, namely rewrite $A$ as $A=PDP^{-1}$.



                                Now
                                begin{align}
                                sum_{i=1}^{100}A^i&=sum_{i=1}^{100}PD^iP^{-1}\&=Pleft(sum_{i=1}^{100} D^iright)P^{-1}
                                end{align}



                                Notice that
                                $$(D-I)left(sum_{i=1}^{100}D^iright)=D^{101}-I.$$
                                SInce $D-I$ is invertible (you can check it)
                                $$sum_{i=1}^{100}D^i=(D-I)^{-1}(D^{101}-I).$$
                                Therefore
                                $$sum_{i=1}^{100}A^i=P(D-I)^{-1}(D^{101}-I)P^{-1}.$$






                                share|cite|improve this answer












                                You can explicitly compute $sum_{i=1}^{100}A^i$. First diagonalize $A$, namely rewrite $A$ as $A=PDP^{-1}$.



                                Now
                                begin{align}
                                sum_{i=1}^{100}A^i&=sum_{i=1}^{100}PD^iP^{-1}\&=Pleft(sum_{i=1}^{100} D^iright)P^{-1}
                                end{align}



                                Notice that
                                $$(D-I)left(sum_{i=1}^{100}D^iright)=D^{101}-I.$$
                                SInce $D-I$ is invertible (you can check it)
                                $$sum_{i=1}^{100}D^i=(D-I)^{-1}(D^{101}-I).$$
                                Therefore
                                $$sum_{i=1}^{100}A^i=P(D-I)^{-1}(D^{101}-I)P^{-1}.$$







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                                answered Dec 11 at 19:37









                                user9077

                                1,254612




                                1,254612






















                                    up vote
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                                    It is easy to prove that for $kin Bbb{N},$ $$A^k=begin{bmatrix} 2^{k-1} & 2^{k-1} \ 2^{k-1} & 2^{k-1} \ end{bmatrix}.$$ The sum is
                                    $$Sigma=begin{bmatrix} 2^{100}-1 & 2^{100}-1 \ 2^{100}-1 & 2^{100}-1 \ end{bmatrix},$$ from where the eigenvalues $0$ and $(2^{101}-2).$



                                    Each matrix $A^k, k=1,dots,100$ has eigenvalues $0$ and $2^k,$ the corresponding eigenvectors are those of $A:$ $(1,-1)^T, (1,1)^T.$

                                    Thus $(1,-1)^T, (1,1)^T$ are eigenvectors of $Sigma.$






                                    share|cite|improve this answer



























                                      up vote
                                      1
                                      down vote













                                      It is easy to prove that for $kin Bbb{N},$ $$A^k=begin{bmatrix} 2^{k-1} & 2^{k-1} \ 2^{k-1} & 2^{k-1} \ end{bmatrix}.$$ The sum is
                                      $$Sigma=begin{bmatrix} 2^{100}-1 & 2^{100}-1 \ 2^{100}-1 & 2^{100}-1 \ end{bmatrix},$$ from where the eigenvalues $0$ and $(2^{101}-2).$



                                      Each matrix $A^k, k=1,dots,100$ has eigenvalues $0$ and $2^k,$ the corresponding eigenvectors are those of $A:$ $(1,-1)^T, (1,1)^T.$

                                      Thus $(1,-1)^T, (1,1)^T$ are eigenvectors of $Sigma.$






                                      share|cite|improve this answer

























                                        up vote
                                        1
                                        down vote










                                        up vote
                                        1
                                        down vote









                                        It is easy to prove that for $kin Bbb{N},$ $$A^k=begin{bmatrix} 2^{k-1} & 2^{k-1} \ 2^{k-1} & 2^{k-1} \ end{bmatrix}.$$ The sum is
                                        $$Sigma=begin{bmatrix} 2^{100}-1 & 2^{100}-1 \ 2^{100}-1 & 2^{100}-1 \ end{bmatrix},$$ from where the eigenvalues $0$ and $(2^{101}-2).$



                                        Each matrix $A^k, k=1,dots,100$ has eigenvalues $0$ and $2^k,$ the corresponding eigenvectors are those of $A:$ $(1,-1)^T, (1,1)^T.$

                                        Thus $(1,-1)^T, (1,1)^T$ are eigenvectors of $Sigma.$






                                        share|cite|improve this answer














                                        It is easy to prove that for $kin Bbb{N},$ $$A^k=begin{bmatrix} 2^{k-1} & 2^{k-1} \ 2^{k-1} & 2^{k-1} \ end{bmatrix}.$$ The sum is
                                        $$Sigma=begin{bmatrix} 2^{100}-1 & 2^{100}-1 \ 2^{100}-1 & 2^{100}-1 \ end{bmatrix},$$ from where the eigenvalues $0$ and $(2^{101}-2).$



                                        Each matrix $A^k, k=1,dots,100$ has eigenvalues $0$ and $2^k,$ the corresponding eigenvectors are those of $A:$ $(1,-1)^T, (1,1)^T.$

                                        Thus $(1,-1)^T, (1,1)^T$ are eigenvectors of $Sigma.$







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Dec 11 at 23:13

























                                        answered Dec 11 at 19:55









                                        user376343

                                        2,7682822




                                        2,7682822






















                                            up vote
                                            0
                                            down vote













                                            Since you have 2 linear independent eigenvectors, $A$ is diagonalizable. You may find useful to replace $A$ in your polynomial expression by its diagonalization because this will simplify the operations you need to do.






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                                              up vote
                                              0
                                              down vote













                                              Since you have 2 linear independent eigenvectors, $A$ is diagonalizable. You may find useful to replace $A$ in your polynomial expression by its diagonalization because this will simplify the operations you need to do.






                                              share|cite|improve this answer























                                                up vote
                                                0
                                                down vote










                                                up vote
                                                0
                                                down vote









                                                Since you have 2 linear independent eigenvectors, $A$ is diagonalizable. You may find useful to replace $A$ in your polynomial expression by its diagonalization because this will simplify the operations you need to do.






                                                share|cite|improve this answer












                                                Since you have 2 linear independent eigenvectors, $A$ is diagonalizable. You may find useful to replace $A$ in your polynomial expression by its diagonalization because this will simplify the operations you need to do.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Dec 11 at 19:23









                                                Javi

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