How many diagonal matrices such that $A^k=I_n$?
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How many diagonal matrices $A in mathbb{C}^{n,n}$ are there, such that $A^k=I_n$ for some $k in mathbb{N} = {1, 2, 3, ...}$?
Edit: I was thinking about $k$-th roots of 1 on the diagonal and then the answer would probably be $infty$. Does that make any sense?
linear-algebra matrices
asked Nov 18 at 11:07
KacperR
295
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up vote
0
down vote
favorite
How many diagonal matrices $A in mathbb{C}^{n,n}$ are there, such that $A^k=I_n$ for some $k in mathbb{N} = {1, 2, 3, ...}$?
Edit: I was thinking about $k$-th roots of 1 on the diagonal and then the answer would probably be $infty$. Does that make any sense?
linear-algebra matrices
asked Nov 18 at 11:07
KacperR
295
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How many diagonal matrices $A in mathbb{C}^{n,n}$ are there, such that $A^k=I_n$ for some $k in mathbb{N} = {1, 2, 3, ...}$?
Edit: I was thinking about $k$-th roots of 1 on the diagonal and then the answer would probably be $infty$. Does that make any sense?
linear-algebra matrices
asked Nov 18 at 11:07
KacperR
295
How many diagonal matrices $A in mathbb{C}^{n,n}$ are there, such that $A^k=I_n$ for some $k in mathbb{N} = {1, 2, 3, ...}$?
Edit: I was thinking about $k$-th roots of 1 on the diagonal and then the answer would probably be $infty$. Does that make any sense?
linear-algebra matrices
linear-algebra matrices
asked Nov 18 at 11:07
KacperR
295
asked Nov 18 at 11:07
KacperR
295
edited Nov 18 at 11:12
asked Nov 18 at 11:07
KacperR
295
asked Nov 18 at 11:07
KacperR
295
asked Nov 18 at 11:07
KacperR
295
295
add a comment |
add a comment |
1 Answer
1
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1
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Hint: Suppose the diagonal elements are ${a_1,a_2,ldots,a_n}$. The determinant of $A$ is $a_1a_2dotsm a_n$. So the determinant of $A^k$ is $a_1^ka_2^kdotsm a_n^k$. But the determinant of $I$ is $1$, so $a_1^ka_2^kdotsm a_n^k=1$.
answered Nov 18 at 11:12
YiFan
1,7741315
And we can go even stronger than that: $a^k = 1$ for all $k$, since (if we denote by $D(b_1,ldots,b_n)$ the diagonal matrix with diagonal values $b_1,ldots,b_n$), $D(a_1,ldots,a_n)^k = D(a_1^k,ldots,a_n^k) = I_n = D(1,ldots,1)$.
– user3482749
Nov 18 at 11:16
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: Suppose the diagonal elements are ${a_1,a_2,ldots,a_n}$. The determinant of $A$ is $a_1a_2dotsm a_n$. So the determinant of $A^k$ is $a_1^ka_2^kdotsm a_n^k$. But the determinant of $I$ is $1$, so $a_1^ka_2^kdotsm a_n^k=1$.
answered Nov 18 at 11:12
YiFan
1,7741315
And we can go even stronger than that: $a^k = 1$ for all $k$, since (if we denote by $D(b_1,ldots,b_n)$ the diagonal matrix with diagonal values $b_1,ldots,b_n$), $D(a_1,ldots,a_n)^k = D(a_1^k,ldots,a_n^k) = I_n = D(1,ldots,1)$.
– user3482749
Nov 18 at 11:16
add a comment |
up vote
1
down vote
accepted
Hint: Suppose the diagonal elements are ${a_1,a_2,ldots,a_n}$. The determinant of $A$ is $a_1a_2dotsm a_n$. So the determinant of $A^k$ is $a_1^ka_2^kdotsm a_n^k$. But the determinant of $I$ is $1$, so $a_1^ka_2^kdotsm a_n^k=1$.
answered Nov 18 at 11:12
YiFan
1,7741315
And we can go even stronger than that: $a^k = 1$ for all $k$, since (if we denote by $D(b_1,ldots,b_n)$ the diagonal matrix with diagonal values $b_1,ldots,b_n$), $D(a_1,ldots,a_n)^k = D(a_1^k,ldots,a_n^k) = I_n = D(1,ldots,1)$.
– user3482749
Nov 18 at 11:16
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: Suppose the diagonal elements are ${a_1,a_2,ldots,a_n}$. The determinant of $A$ is $a_1a_2dotsm a_n$. So the determinant of $A^k$ is $a_1^ka_2^kdotsm a_n^k$. But the determinant of $I$ is $1$, so $a_1^ka_2^kdotsm a_n^k=1$.
answered Nov 18 at 11:12
YiFan
1,7741315
Hint: Suppose the diagonal elements are ${a_1,a_2,ldots,a_n}$. The determinant of $A$ is $a_1a_2dotsm a_n$. So the determinant of $A^k$ is $a_1^ka_2^kdotsm a_n^k$. But the determinant of $I$ is $1$, so $a_1^ka_2^kdotsm a_n^k=1$.
answered Nov 18 at 11:12
YiFan
1,7741315
answered Nov 18 at 11:12
YiFan
1,7741315
answered Nov 18 at 11:12
YiFan
1,7741315
answered Nov 18 at 11:12
YiFan
1,7741315
1,7741315
And we can go even stronger than that: $a^k = 1$ for all $k$, since (if we denote by $D(b_1,ldots,b_n)$ the diagonal matrix with diagonal values $b_1,ldots,b_n$), $D(a_1,ldots,a_n)^k = D(a_1^k,ldots,a_n^k) = I_n = D(1,ldots,1)$.
– user3482749
Nov 18 at 11:16
add a comment |
And we can go even stronger than that: $a^k = 1$ for all $k$, since (if we denote by $D(b_1,ldots,b_n)$ the diagonal matrix with diagonal values $b_1,ldots,b_n$), $D(a_1,ldots,a_n)^k = D(a_1^k,ldots,a_n^k) = I_n = D(1,ldots,1)$.
– user3482749
Nov 18 at 11:16
And we can go even stronger than that: $a^k = 1$ for all $k$, since (if we denote by $D(b_1,ldots,b_n)$ the diagonal matrix with diagonal values $b_1,ldots,b_n$), $D(a_1,ldots,a_n)^k = D(a_1^k,ldots,a_n^k) = I_n = D(1,ldots,1)$.
– user3482749
Nov 18 at 11:16
And we can go even stronger than that: $a^k = 1$ for all $k$, since (if we denote by $D(b_1,ldots,b_n)$ the diagonal matrix with diagonal values $b_1,ldots,b_n$), $D(a_1,ldots,a_n)^k = D(a_1^k,ldots,a_n^k) = I_n = D(1,ldots,1)$.
– user3482749
Nov 18 at 11:16
add a comment |
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