Calculating the order of a matrix when speaking of groups
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-1
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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
add a comment |
up vote
-1
down vote
favorite
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
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
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
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
matrices group-theory
edited Nov 20 at 12:15
José Carlos Santos
140k19111204
140k19111204
asked Nov 20 at 9:40
vesii
525
525
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
Hint: Its square is the identity matrix…
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint: Its square is the identity matrix…
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
|
show 1 more comment
up vote
3
down vote
Hint: Its square is the identity matrix…
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
|
show 1 more comment
up vote
3
down vote
up vote
3
down vote
Hint: Its square is the identity matrix…
Hint: Its square is the identity matrix…
answered Nov 20 at 9:42
José Carlos Santos
140k19111204
140k19111204
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
|
show 1 more comment
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
|
show 1 more comment
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