Calculating the order of a matrix when speaking of groups












-1














How should I calculate the order of a matrix when speaking of groups?
For example I have the following matrix:



$$ begin{pmatrix}1 & 0 & 0\
0 & -1 & 0\
0 & 0 & -1
end{pmatrix}$$



How should I calculate its order?










share|cite|improve this question





























    -1














    How should I calculate the order of a matrix when speaking of groups?
    For example I have the following matrix:



    $$ begin{pmatrix}1 & 0 & 0\
    0 & -1 & 0\
    0 & 0 & -1
    end{pmatrix}$$



    How should I calculate its order?










    share|cite|improve this question



























      -1












      -1








      -1







      How should I calculate the order of a matrix when speaking of groups?
      For example I have the following matrix:



      $$ begin{pmatrix}1 & 0 & 0\
      0 & -1 & 0\
      0 & 0 & -1
      end{pmatrix}$$



      How should I calculate its order?










      share|cite|improve this question















      How should I calculate the order of a matrix when speaking of groups?
      For example I have the following matrix:



      $$ begin{pmatrix}1 & 0 & 0\
      0 & -1 & 0\
      0 & 0 & -1
      end{pmatrix}$$



      How should I calculate its order?







      matrices group-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 20 at 12:15









      José Carlos Santos

      149k22117219




      149k22117219










      asked Nov 20 at 9:40









      vesii

      635




      635






















          1 Answer
          1






          active

          oldest

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          3














          Hint: Its square is the identity matrix…






          share|cite|improve this answer





















          • but why? should I square it until i get $I_{3times 3}$? what if I have a difficult matrix that we need to square 200 time until we get $I_{3times 3}$? is there a better/faster way?
            – vesii
            Nov 20 at 14:44










          • For any group, if you have an element $gneq e_G$ such $g^2=e_G$, the order of $g$ is $2$. And what you should do is to compute the powers of the matrix, not to keep on squaring them. The use of the Jordan normal form may be helpful here.
            – José Carlos Santos
            Nov 20 at 14:49










          • ok, so for my example the order of the matrix is 2 right (because $M^2=I$)?
            – vesii
            Nov 20 at 15:04










          • Yes, that's correct.
            – José Carlos Santos
            Nov 20 at 15:05










          • Is it possible that the order of a matrix is infinity? by your logic, there is no $nin mathbb{N}$ so $begin{pmatrix}1 & 1 & 0\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix}^n = I$. So we get $o(M)=infty$. is it correct?
            – vesii
            Nov 20 at 15:09











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          Hint: Its square is the identity matrix…






          share|cite|improve this answer





















          • but why? should I square it until i get $I_{3times 3}$? what if I have a difficult matrix that we need to square 200 time until we get $I_{3times 3}$? is there a better/faster way?
            – vesii
            Nov 20 at 14:44










          • For any group, if you have an element $gneq e_G$ such $g^2=e_G$, the order of $g$ is $2$. And what you should do is to compute the powers of the matrix, not to keep on squaring them. The use of the Jordan normal form may be helpful here.
            – José Carlos Santos
            Nov 20 at 14:49










          • ok, so for my example the order of the matrix is 2 right (because $M^2=I$)?
            – vesii
            Nov 20 at 15:04










          • Yes, that's correct.
            – José Carlos Santos
            Nov 20 at 15:05










          • Is it possible that the order of a matrix is infinity? by your logic, there is no $nin mathbb{N}$ so $begin{pmatrix}1 & 1 & 0\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix}^n = I$. So we get $o(M)=infty$. is it correct?
            – vesii
            Nov 20 at 15:09
















          3














          Hint: Its square is the identity matrix…






          share|cite|improve this answer





















          • but why? should I square it until i get $I_{3times 3}$? what if I have a difficult matrix that we need to square 200 time until we get $I_{3times 3}$? is there a better/faster way?
            – vesii
            Nov 20 at 14:44










          • For any group, if you have an element $gneq e_G$ such $g^2=e_G$, the order of $g$ is $2$. And what you should do is to compute the powers of the matrix, not to keep on squaring them. The use of the Jordan normal form may be helpful here.
            – José Carlos Santos
            Nov 20 at 14:49










          • ok, so for my example the order of the matrix is 2 right (because $M^2=I$)?
            – vesii
            Nov 20 at 15:04










          • Yes, that's correct.
            – José Carlos Santos
            Nov 20 at 15:05










          • Is it possible that the order of a matrix is infinity? by your logic, there is no $nin mathbb{N}$ so $begin{pmatrix}1 & 1 & 0\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix}^n = I$. So we get $o(M)=infty$. is it correct?
            – vesii
            Nov 20 at 15:09














          3












          3








          3






          Hint: Its square is the identity matrix…






          share|cite|improve this answer












          Hint: Its square is the identity matrix…







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 9:42









          José Carlos Santos

          149k22117219




          149k22117219












          • but why? should I square it until i get $I_{3times 3}$? what if I have a difficult matrix that we need to square 200 time until we get $I_{3times 3}$? is there a better/faster way?
            – vesii
            Nov 20 at 14:44










          • For any group, if you have an element $gneq e_G$ such $g^2=e_G$, the order of $g$ is $2$. And what you should do is to compute the powers of the matrix, not to keep on squaring them. The use of the Jordan normal form may be helpful here.
            – José Carlos Santos
            Nov 20 at 14:49










          • ok, so for my example the order of the matrix is 2 right (because $M^2=I$)?
            – vesii
            Nov 20 at 15:04










          • Yes, that's correct.
            – José Carlos Santos
            Nov 20 at 15:05










          • Is it possible that the order of a matrix is infinity? by your logic, there is no $nin mathbb{N}$ so $begin{pmatrix}1 & 1 & 0\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix}^n = I$. So we get $o(M)=infty$. is it correct?
            – vesii
            Nov 20 at 15:09


















          • but why? should I square it until i get $I_{3times 3}$? what if I have a difficult matrix that we need to square 200 time until we get $I_{3times 3}$? is there a better/faster way?
            – vesii
            Nov 20 at 14:44










          • For any group, if you have an element $gneq e_G$ such $g^2=e_G$, the order of $g$ is $2$. And what you should do is to compute the powers of the matrix, not to keep on squaring them. The use of the Jordan normal form may be helpful here.
            – José Carlos Santos
            Nov 20 at 14:49










          • ok, so for my example the order of the matrix is 2 right (because $M^2=I$)?
            – vesii
            Nov 20 at 15:04










          • Yes, that's correct.
            – José Carlos Santos
            Nov 20 at 15:05










          • Is it possible that the order of a matrix is infinity? by your logic, there is no $nin mathbb{N}$ so $begin{pmatrix}1 & 1 & 0\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix}^n = I$. So we get $o(M)=infty$. is it correct?
            – vesii
            Nov 20 at 15:09
















          but why? should I square it until i get $I_{3times 3}$? what if I have a difficult matrix that we need to square 200 time until we get $I_{3times 3}$? is there a better/faster way?
          – vesii
          Nov 20 at 14:44




          but why? should I square it until i get $I_{3times 3}$? what if I have a difficult matrix that we need to square 200 time until we get $I_{3times 3}$? is there a better/faster way?
          – vesii
          Nov 20 at 14:44












          For any group, if you have an element $gneq e_G$ such $g^2=e_G$, the order of $g$ is $2$. And what you should do is to compute the powers of the matrix, not to keep on squaring them. The use of the Jordan normal form may be helpful here.
          – José Carlos Santos
          Nov 20 at 14:49




          For any group, if you have an element $gneq e_G$ such $g^2=e_G$, the order of $g$ is $2$. And what you should do is to compute the powers of the matrix, not to keep on squaring them. The use of the Jordan normal form may be helpful here.
          – José Carlos Santos
          Nov 20 at 14:49












          ok, so for my example the order of the matrix is 2 right (because $M^2=I$)?
          – vesii
          Nov 20 at 15:04




          ok, so for my example the order of the matrix is 2 right (because $M^2=I$)?
          – vesii
          Nov 20 at 15:04












          Yes, that's correct.
          – José Carlos Santos
          Nov 20 at 15:05




          Yes, that's correct.
          – José Carlos Santos
          Nov 20 at 15:05












          Is it possible that the order of a matrix is infinity? by your logic, there is no $nin mathbb{N}$ so $begin{pmatrix}1 & 1 & 0\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix}^n = I$. So we get $o(M)=infty$. is it correct?
          – vesii
          Nov 20 at 15:09




          Is it possible that the order of a matrix is infinity? by your logic, there is no $nin mathbb{N}$ so $begin{pmatrix}1 & 1 & 0\ 0 & 1 & 0\ 0 & 0 & 1 end{pmatrix}^n = I$. So we get $o(M)=infty$. is it correct?
          – vesii
          Nov 20 at 15:09


















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