Does $exists$ a matrix $A$ so that ${A^p:pgeq 1}$ spans $M_n(mathbb R)?$












0












$begingroup$


The main question is :




(TRUE/FALSE)



For any $ngeq 2,$ there exists an $ntimes n$ real matrix $A$ such thatthe set ${A^p|pgeq 1}$ spans the $mathbb R$ vector space $M_n(mathbb R).$




I tried whether we can find such $A$ that the above set spans $left(begin{array}{cc} 1 & 0\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 1\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 0\ 1 & 0 end{array}right)$ and $left(begin{array}{cc} 0 & 0\ 0 & 1 end{array}right)$. But I could not go further.



Any help is appreciated. Thank you






linear-algebra matrices







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$endgroup$








  • 4




    $begingroup$
    Hint: use the Cayley-Hamilton theorem.
    $endgroup$
    – Mindlack
    Jan 6 at 13:35






  • 2




    $begingroup$
    Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
    $endgroup$
    – Song
    Jan 6 at 13:37












  • $begingroup$
    Yes we can't find $n^2$ linearly independent matrices.
    $endgroup$
    – nurun nesha
    Jan 6 at 13:38
















0












$begingroup$


The main question is :




(TRUE/FALSE)



For any $ngeq 2,$ there exists an $ntimes n$ real matrix $A$ such thatthe set ${A^p|pgeq 1}$ spans the $mathbb R$ vector space $M_n(mathbb R).$




I tried whether we can find such $A$ that the above set spans $left(begin{array}{cc} 1 & 0\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 1\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 0\ 1 & 0 end{array}right)$ and $left(begin{array}{cc} 0 & 0\ 0 & 1 end{array}right)$. But I could not go further.



Any help is appreciated. Thank you






linear-algebra matrices







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Hint: use the Cayley-Hamilton theorem.
    $endgroup$
    – Mindlack
    Jan 6 at 13:35






  • 2




    $begingroup$
    Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
    $endgroup$
    – Song
    Jan 6 at 13:37












  • $begingroup$
    Yes we can't find $n^2$ linearly independent matrices.
    $endgroup$
    – nurun nesha
    Jan 6 at 13:38














0












0








0


1



$begingroup$


The main question is :




(TRUE/FALSE)



For any $ngeq 2,$ there exists an $ntimes n$ real matrix $A$ such thatthe set ${A^p|pgeq 1}$ spans the $mathbb R$ vector space $M_n(mathbb R).$




I tried whether we can find such $A$ that the above set spans $left(begin{array}{cc} 1 & 0\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 1\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 0\ 1 & 0 end{array}right)$ and $left(begin{array}{cc} 0 & 0\ 0 & 1 end{array}right)$. But I could not go further.



Any help is appreciated. Thank you






linear-algebra matrices







share|cite|improve this question









$endgroup$




The main question is :




(TRUE/FALSE)



For any $ngeq 2,$ there exists an $ntimes n$ real matrix $A$ such thatthe set ${A^p|pgeq 1}$ spans the $mathbb R$ vector space $M_n(mathbb R).$




I tried whether we can find such $A$ that the above set spans $left(begin{array}{cc} 1 & 0\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 1\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 0\ 1 & 0 end{array}right)$ and $left(begin{array}{cc} 0 & 0\ 0 & 1 end{array}right)$. But I could not go further.



Any help is appreciated. Thank you






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










asked Jan 6 at 13:28









nurun neshanurun nesha

1,0242623




1,0242623








  • 4




    $begingroup$
    Hint: use the Cayley-Hamilton theorem.
    $endgroup$
    – Mindlack
    Jan 6 at 13:35






  • 2




    $begingroup$
    Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
    $endgroup$
    – Song
    Jan 6 at 13:37












  • $begingroup$
    Yes we can't find $n^2$ linearly independent matrices.
    $endgroup$
    – nurun nesha
    Jan 6 at 13:38














  • 4




    $begingroup$
    Hint: use the Cayley-Hamilton theorem.
    $endgroup$
    – Mindlack
    Jan 6 at 13:35






  • 2




    $begingroup$
    Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
    $endgroup$
    – Song
    Jan 6 at 13:37












  • $begingroup$
    Yes we can't find $n^2$ linearly independent matrices.
    $endgroup$
    – nurun nesha
    Jan 6 at 13:38








4




4




$begingroup$
Hint: use the Cayley-Hamilton theorem.
$endgroup$
– Mindlack
Jan 6 at 13:35




$begingroup$
Hint: use the Cayley-Hamilton theorem.
$endgroup$
– Mindlack
Jan 6 at 13:35




2




2




$begingroup$
Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
$endgroup$
– Song
Jan 6 at 13:37






$begingroup$
Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
$endgroup$
– Song
Jan 6 at 13:37














$begingroup$
Yes we can't find $n^2$ linearly independent matrices.
$endgroup$
– nurun nesha
Jan 6 at 13:38




$begingroup$
Yes we can't find $n^2$ linearly independent matrices.
$endgroup$
– nurun nesha
Jan 6 at 13:38










3 Answers
3






active

oldest

votes


















7












$begingroup$

It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    In fact it is false for all $nge2$ (an even stronger statement).



    By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.



    So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.



    This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.



    So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.



    But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7












        $begingroup$

        It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.






        share|cite|improve this answer









        $endgroup$


















          7












          $begingroup$

          It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.






          share|cite|improve this answer









          $endgroup$
















            7












            7








            7





            $begingroup$

            It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.






            share|cite|improve this answer









            $endgroup$



            It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 13:44









            Dietrich BurdeDietrich Burde

            79.1k647103




            79.1k647103























                2












                $begingroup$

                In fact it is false for all $nge2$ (an even stronger statement).



                By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.



                So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.



                This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.



                So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.



                But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  In fact it is false for all $nge2$ (an even stronger statement).



                  By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.



                  So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.



                  This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.



                  So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.



                  But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    In fact it is false for all $nge2$ (an even stronger statement).



                    By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.



                    So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.



                    This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.



                    So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.



                    But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.






                    share|cite|improve this answer











                    $endgroup$



                    In fact it is false for all $nge2$ (an even stronger statement).



                    By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.



                    So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.



                    This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.



                    So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.



                    But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 6 at 15:47

























                    answered Jan 6 at 14:56









                    Chris CusterChris Custer

                    13.1k3827




                    13.1k3827























                        2












                        $begingroup$

                        Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.






                            share|cite|improve this answer









                            $endgroup$



                            Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 7 at 16:20









                            user1551user1551

                            72.7k566127




                            72.7k566127






























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