Mixing
Is there a name for a matrix whose n-th power is the identity matrix
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I am new to this community so my apologies it this is a duplicate, feel free to flag it.
I am currently working on cyclically symmetric structural mechanics and we exploit the finite group linear representation.
In this theory, properties of rotation matrices are used, specifically the fact that for a rotation matrix $mathbfR$ of angle $2pi/N$, $mathbfR^N = mathbfI$.
Matrices such that $mathbfR^N = mathbf0$ are called nilpotent matrices, but is there a name for the mentioned rotation matrices ?
linear-algebra
asked Jul 24 at 8:52
BambOo
1083
add a comment |Â
up vote
1
down vote
favorite
I am new to this community so my apologies it this is a duplicate, feel free to flag it.
I am currently working on cyclically symmetric structural mechanics and we exploit the finite group linear representation.
In this theory, properties of rotation matrices are used, specifically the fact that for a rotation matrix $mathbfR$ of angle $2pi/N$, $mathbfR^N = mathbfI$.
Matrices such that $mathbfR^N = mathbf0$ are called nilpotent matrices, but is there a name for the mentioned rotation matrices ?
linear-algebra
asked Jul 24 at 8:52
BambOo
1083
2
It would be natural to call it a "torsion matrix", but that doesn't seem to be common according to Google. Perhaps just "matrix of finite order"?
– Henning Makholm
Jul 24 at 9:23
1
@HenningMakholm matrix of finite order seems to be quite used. You should post this as an answer.
– nicomezi
Jul 24 at 9:31
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am new to this community so my apologies it this is a duplicate, feel free to flag it.
I am currently working on cyclically symmetric structural mechanics and we exploit the finite group linear representation.
In this theory, properties of rotation matrices are used, specifically the fact that for a rotation matrix $mathbfR$ of angle $2pi/N$, $mathbfR^N = mathbfI$.
Matrices such that $mathbfR^N = mathbf0$ are called nilpotent matrices, but is there a name for the mentioned rotation matrices ?
linear-algebra
asked Jul 24 at 8:52
BambOo
1083
I am new to this community so my apologies it this is a duplicate, feel free to flag it.
I am currently working on cyclically symmetric structural mechanics and we exploit the finite group linear representation.
In this theory, properties of rotation matrices are used, specifically the fact that for a rotation matrix $mathbfR$ of angle $2pi/N$, $mathbfR^N = mathbfI$.
Matrices such that $mathbfR^N = mathbf0$ are called nilpotent matrices, but is there a name for the mentioned rotation matrices ?
linear-algebra
asked Jul 24 at 8:52
BambOo
1083
asked Jul 24 at 8:52
BambOo
1083
asked Jul 24 at 8:52
BambOo
1083
asked Jul 24 at 8:52
BambOo
1083
1083
2
It would be natural to call it a "torsion matrix", but that doesn't seem to be common according to Google. Perhaps just "matrix of finite order"?
– Henning Makholm
Jul 24 at 9:23
1
@HenningMakholm matrix of finite order seems to be quite used. You should post this as an answer.
– nicomezi
Jul 24 at 9:31
add a comment |Â
2
It would be natural to call it a "torsion matrix", but that doesn't seem to be common according to Google. Perhaps just "matrix of finite order"?
– Henning Makholm
Jul 24 at 9:23
1
@HenningMakholm matrix of finite order seems to be quite used. You should post this as an answer.
– nicomezi
Jul 24 at 9:31
2
2
It would be natural to call it a "torsion matrix", but that doesn't seem to be common according to Google. Perhaps just "matrix of finite order"?
– Henning Makholm
Jul 24 at 9:23
It would be natural to call it a "torsion matrix", but that doesn't seem to be common according to Google. Perhaps just "matrix of finite order"?
– Henning Makholm
Jul 24 at 9:23
1
1
@HenningMakholm matrix of finite order seems to be quite used. You should post this as an answer.
– nicomezi
Jul 24 at 9:31
@HenningMakholm matrix of finite order seems to be quite used. You should post this as an answer.
– nicomezi
Jul 24 at 9:31
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
I would just call it a matrix of finite order.
answered Jul 25 at 8:48
Henning Makholm
225k16290516
add a comment |Â
up vote
0
down vote
I do not know a name for such matrices but ... we have $R^N+1=R$ then $R$ is a periodic matrix of period $N$. See here.
Then $R^N$ is idempotent ( $(R^N)^2=R^2N=R^N+1R^N-1=R^N)$ and the only invertible idempotent matrix is identity. Then an equivalent notion is "invertible periodic matrix".
answered Jul 24 at 9:18
nicomezi
3,4121819
It seems that for $N = 3$, such matrices are called $k$-matrices
– BambOo
Jul 24 at 9:32
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I would just call it a matrix of finite order.
answered Jul 25 at 8:48
Henning Makholm
225k16290516
add a comment |Â
up vote
2
down vote
accepted
I would just call it a matrix of finite order.
answered Jul 25 at 8:48
Henning Makholm
225k16290516
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I would just call it a matrix of finite order.
answered Jul 25 at 8:48
Henning Makholm
225k16290516
I would just call it a matrix of finite order.
answered Jul 25 at 8:48
Henning Makholm
225k16290516
answered Jul 25 at 8:48
Henning Makholm
225k16290516
answered Jul 25 at 8:48
Henning Makholm
225k16290516
answered Jul 25 at 8:48
Henning Makholm
225k16290516
225k16290516
add a comment |Â
add a comment |Â
up vote
0
down vote
I do not know a name for such matrices but ... we have $R^N+1=R$ then $R$ is a periodic matrix of period $N$. See here.
Then $R^N$ is idempotent ( $(R^N)^2=R^2N=R^N+1R^N-1=R^N)$ and the only invertible idempotent matrix is identity. Then an equivalent notion is "invertible periodic matrix".
answered Jul 24 at 9:18
nicomezi
3,4121819
It seems that for $N = 3$, such matrices are called $k$-matrices
– BambOo
Jul 24 at 9:32
add a comment |Â
up vote
0
down vote
I do not know a name for such matrices but ... we have $R^N+1=R$ then $R$ is a periodic matrix of period $N$. See here.
Then $R^N$ is idempotent ( $(R^N)^2=R^2N=R^N+1R^N-1=R^N)$ and the only invertible idempotent matrix is identity. Then an equivalent notion is "invertible periodic matrix".
answered Jul 24 at 9:18
nicomezi
3,4121819
It seems that for $N = 3$, such matrices are called $k$-matrices
– BambOo
Jul 24 at 9:32
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I do not know a name for such matrices but ... we have $R^N+1=R$ then $R$ is a periodic matrix of period $N$. See here.
Then $R^N$ is idempotent ( $(R^N)^2=R^2N=R^N+1R^N-1=R^N)$ and the only invertible idempotent matrix is identity. Then an equivalent notion is "invertible periodic matrix".
answered Jul 24 at 9:18
nicomezi
3,4121819
I do not know a name for such matrices but ... we have $R^N+1=R$ then $R$ is a periodic matrix of period $N$. See here.
Then $R^N$ is idempotent ( $(R^N)^2=R^2N=R^N+1R^N-1=R^N)$ and the only invertible idempotent matrix is identity. Then an equivalent notion is "invertible periodic matrix".
answered Jul 24 at 9:18
nicomezi
3,4121819
answered Jul 24 at 9:18
nicomezi
3,4121819
answered Jul 24 at 9:18
nicomezi
3,4121819
answered Jul 24 at 9:18
nicomezi
3,4121819
3,4121819
It seems that for $N = 3$, such matrices are called $k$-matrices
– BambOo
Jul 24 at 9:32
add a comment |Â
It seems that for $N = 3$, such matrices are called $k$-matrices
– BambOo
Jul 24 at 9:32
It seems that for $N = 3$, such matrices are called $k$-matrices
– BambOo
Jul 24 at 9:32
It seems that for $N = 3$, such matrices are called $k$-matrices
– BambOo
Jul 24 at 9:32
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861132%2fis-there-a-name-for-a-matrix-whose-n-th-power-is-the-identity-matrix%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
It would be natural to call it a "torsion matrix", but that doesn't seem to be common according to Google. Perhaps just "matrix of finite order"?
– Henning Makholm
Jul 24 at 9:23
1
@HenningMakholm matrix of finite order seems to be quite used. You should post this as an answer.
– nicomezi
Jul 24 at 9:31