Two questions about square matrices and linear combinations

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Let $A$ be an invertible square matrix in $mathbbR^ntimes n$. Let $lambda$ be the unique eigenvalue of $A$ with the largest norm. Assume that we have two good properties, namely $lambda$ is real and positive and its eigenspace has dimension 1. Let $v$ be one of the unit eigenvectors.



We would like to examine whether both of the two properties are correct:



a) For any $1leq ileq n$ $e_i$ can be written as a linear combination of column vectors of column vectors of $A^k$ for positive $k$ and $v$ where all coefficients are nonnegative with the possible exception of the coefficient of $v$?



b) Does there exist a $k$ such that for all $l>k$ any column vector of $A^l$ can be written as a linear combination of column vectors of $A^i$ where $0leq ileq k$ and $v$ where all coefficients are nonnegative with the possible exception of the coefficient of $v$?










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    Let $A$ be an invertible square matrix in $mathbbR^ntimes n$. Let $lambda$ be the unique eigenvalue of $A$ with the largest norm. Assume that we have two good properties, namely $lambda$ is real and positive and its eigenspace has dimension 1. Let $v$ be one of the unit eigenvectors.



    We would like to examine whether both of the two properties are correct:



    a) For any $1leq ileq n$ $e_i$ can be written as a linear combination of column vectors of column vectors of $A^k$ for positive $k$ and $v$ where all coefficients are nonnegative with the possible exception of the coefficient of $v$?



    b) Does there exist a $k$ such that for all $l>k$ any column vector of $A^l$ can be written as a linear combination of column vectors of $A^i$ where $0leq ileq k$ and $v$ where all coefficients are nonnegative with the possible exception of the coefficient of $v$?










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      Let $A$ be an invertible square matrix in $mathbbR^ntimes n$. Let $lambda$ be the unique eigenvalue of $A$ with the largest norm. Assume that we have two good properties, namely $lambda$ is real and positive and its eigenspace has dimension 1. Let $v$ be one of the unit eigenvectors.



      We would like to examine whether both of the two properties are correct:



      a) For any $1leq ileq n$ $e_i$ can be written as a linear combination of column vectors of column vectors of $A^k$ for positive $k$ and $v$ where all coefficients are nonnegative with the possible exception of the coefficient of $v$?



      b) Does there exist a $k$ such that for all $l>k$ any column vector of $A^l$ can be written as a linear combination of column vectors of $A^i$ where $0leq ileq k$ and $v$ where all coefficients are nonnegative with the possible exception of the coefficient of $v$?










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      Let $A$ be an invertible square matrix in $mathbbR^ntimes n$. Let $lambda$ be the unique eigenvalue of $A$ with the largest norm. Assume that we have two good properties, namely $lambda$ is real and positive and its eigenspace has dimension 1. Let $v$ be one of the unit eigenvectors.



      We would like to examine whether both of the two properties are correct:



      a) For any $1leq ileq n$ $e_i$ can be written as a linear combination of column vectors of column vectors of $A^k$ for positive $k$ and $v$ where all coefficients are nonnegative with the possible exception of the coefficient of $v$?



      b) Does there exist a $k$ such that for all $l>k$ any column vector of $A^l$ can be written as a linear combination of column vectors of $A^i$ where $0leq ileq k$ and $v$ where all coefficients are nonnegative with the possible exception of the coefficient of $v$?







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      asked Sep 10 at 19:31









      Ying Zhou

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          Let
          $$A=beginbmatrix 2 & 0 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & k \ 0 & 0 & 1 endbmatrix$$
          Then the largest eigenvalue of $A$ is 2 and its corresponding eigenspace has dimension 1.



          a. One takes $nu=e_1$ and notes that $e_3$ cannot be written as a combination of the second and third columns of $A^k$ with nonnegative coefficients.



          b. Let
          $$B=beginbmatrix 2 & 0 & 0 \ 0 & 1 & -1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & -k \ 0 & 0 & 1 endbmatrix$$



          Again $B$ satisfies the required conditions. However, the third column of $A^l$ cannot be written as a combination of the second and third columns of $A^i$ for $i<l$.






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            Let
            $$A=beginbmatrix 2 & 0 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & k \ 0 & 0 & 1 endbmatrix$$
            Then the largest eigenvalue of $A$ is 2 and its corresponding eigenspace has dimension 1.



            a. One takes $nu=e_1$ and notes that $e_3$ cannot be written as a combination of the second and third columns of $A^k$ with nonnegative coefficients.



            b. Let
            $$B=beginbmatrix 2 & 0 & 0 \ 0 & 1 & -1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & -k \ 0 & 0 & 1 endbmatrix$$



            Again $B$ satisfies the required conditions. However, the third column of $A^l$ cannot be written as a combination of the second and third columns of $A^i$ for $i<l$.






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              Let
              $$A=beginbmatrix 2 & 0 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & k \ 0 & 0 & 1 endbmatrix$$
              Then the largest eigenvalue of $A$ is 2 and its corresponding eigenspace has dimension 1.



              a. One takes $nu=e_1$ and notes that $e_3$ cannot be written as a combination of the second and third columns of $A^k$ with nonnegative coefficients.



              b. Let
              $$B=beginbmatrix 2 & 0 & 0 \ 0 & 1 & -1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & -k \ 0 & 0 & 1 endbmatrix$$



              Again $B$ satisfies the required conditions. However, the third column of $A^l$ cannot be written as a combination of the second and third columns of $A^i$ for $i<l$.






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                Let
                $$A=beginbmatrix 2 & 0 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & k \ 0 & 0 & 1 endbmatrix$$
                Then the largest eigenvalue of $A$ is 2 and its corresponding eigenspace has dimension 1.



                a. One takes $nu=e_1$ and notes that $e_3$ cannot be written as a combination of the second and third columns of $A^k$ with nonnegative coefficients.



                b. Let
                $$B=beginbmatrix 2 & 0 & 0 \ 0 & 1 & -1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & -k \ 0 & 0 & 1 endbmatrix$$



                Again $B$ satisfies the required conditions. However, the third column of $A^l$ cannot be written as a combination of the second and third columns of $A^i$ for $i<l$.






                share|cite|improve this answer












                Let
                $$A=beginbmatrix 2 & 0 & 0 \ 0 & 1 & 1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & k \ 0 & 0 & 1 endbmatrix$$
                Then the largest eigenvalue of $A$ is 2 and its corresponding eigenspace has dimension 1.



                a. One takes $nu=e_1$ and notes that $e_3$ cannot be written as a combination of the second and third columns of $A^k$ with nonnegative coefficients.



                b. Let
                $$B=beginbmatrix 2 & 0 & 0 \ 0 & 1 & -1 \ 0 & 0 & 1 endbmatrix Rightarrow A^k=beginbmatrix 2^k & 0 & 0 \ 0 & 1 & -k \ 0 & 0 & 1 endbmatrix$$



                Again $B$ satisfies the required conditions. However, the third column of $A^l$ cannot be written as a combination of the second and third columns of $A^i$ for $i<l$.







                share|cite|improve this answer












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                answered Sep 11 at 1:31









                Marco

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