Mixing
Matrices: $A^7=I$ and $ABA^-1=B^2$, show that $B^k=I$ for some $k>0$.
Clash Royale CLAN TAG#URR8PPP
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Knowing that $A$ and $B$ are two invertible nxn matrices ($A≠I$, $B≠I$) such that:
$A^7=I$ and $ABA^-1=B^2$
Show that there is an integer $k>0$ such that $B^k=I$ and determine the lowest value for $k$.
I realize that performing $(ABA^-1)^x)$ we get $AB^xA^-1$. I tried to form an argument replacing the $B^x$ to a something like $A^zB^y(A^-1)^z$ and when $z$ is 7 or a multiple, we would be able to get away with $A$ using $A^7=I$. I think this idea of performing $(ABA^-1)^x)$ is right and very useful, but I dont't know if the repetitive ideia of replacing $B^x$ is the best to prove the first part of the exercise. And thinking this way, I don't know if I can formalize a solid argument to find the lowest $k$.
matrices
asked Jul 16 at 14:22
Alexandre Tourinho
1305
add a comment |Â
up vote
-2
down vote
favorite
Knowing that $A$ and $B$ are two invertible nxn matrices ($A≠I$, $B≠I$) such that:
$A^7=I$ and $ABA^-1=B^2$
Show that there is an integer $k>0$ such that $B^k=I$ and determine the lowest value for $k$.
I realize that performing $(ABA^-1)^x)$ we get $AB^xA^-1$. I tried to form an argument replacing the $B^x$ to a something like $A^zB^y(A^-1)^z$ and when $z$ is 7 or a multiple, we would be able to get away with $A$ using $A^7=I$. I think this idea of performing $(ABA^-1)^x)$ is right and very useful, but I dont't know if the repetitive ideia of replacing $B^x$ is the best to prove the first part of the exercise. And thinking this way, I don't know if I can formalize a solid argument to find the lowest $k$.
matrices
asked Jul 16 at 14:22
Alexandre Tourinho
1305
5
What did you try?
– José Carlos Santos
Jul 16 at 14:31
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Knowing that $A$ and $B$ are two invertible nxn matrices ($A≠I$, $B≠I$) such that:
$A^7=I$ and $ABA^-1=B^2$
Show that there is an integer $k>0$ such that $B^k=I$ and determine the lowest value for $k$.
I realize that performing $(ABA^-1)^x)$ we get $AB^xA^-1$. I tried to form an argument replacing the $B^x$ to a something like $A^zB^y(A^-1)^z$ and when $z$ is 7 or a multiple, we would be able to get away with $A$ using $A^7=I$. I think this idea of performing $(ABA^-1)^x)$ is right and very useful, but I dont't know if the repetitive ideia of replacing $B^x$ is the best to prove the first part of the exercise. And thinking this way, I don't know if I can formalize a solid argument to find the lowest $k$.
matrices
asked Jul 16 at 14:22
Alexandre Tourinho
1305
Knowing that $A$ and $B$ are two invertible nxn matrices ($A≠I$, $B≠I$) such that:
$A^7=I$ and $ABA^-1=B^2$
Show that there is an integer $k>0$ such that $B^k=I$ and determine the lowest value for $k$.
I realize that performing $(ABA^-1)^x)$ we get $AB^xA^-1$. I tried to form an argument replacing the $B^x$ to a something like $A^zB^y(A^-1)^z$ and when $z$ is 7 or a multiple, we would be able to get away with $A$ using $A^7=I$. I think this idea of performing $(ABA^-1)^x)$ is right and very useful, but I dont't know if the repetitive ideia of replacing $B^x$ is the best to prove the first part of the exercise. And thinking this way, I don't know if I can formalize a solid argument to find the lowest $k$.
matrices
asked Jul 16 at 14:22
Alexandre Tourinho
1305
edited Jul 16 at 15:14
asked Jul 16 at 14:22
Alexandre Tourinho
1305
asked Jul 16 at 14:22
Alexandre Tourinho
1305
asked Jul 16 at 14:22
Alexandre Tourinho
1305
1305
5
What did you try?
– José Carlos Santos
Jul 16 at 14:31
add a comment |Â
5
What did you try?
– José Carlos Santos
Jul 16 at 14:31
5
5
What did you try?
– José Carlos Santos
Jul 16 at 14:31
What did you try?
– José Carlos Santos
Jul 16 at 14:31
add a comment |Â
1 Answer
1
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5
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Since $A^7 = I$, $$beginalign
B = & A^7BA^-7\
= & A^6(ABA^-1)A^-6 = A^6B^2A^-6\
= & A^5(AB^2A^-1)A^-5 = A^5(ABA^-1)^2A^-5
= A^5(B^2)^2A^-5 = A^5B^4A^-5\
vdots & \
= & B^2^7 = B^128
endalign$$
Since $B$ is invertible, multiply both sides by $B^-1$ gives us $B^127 = I$.
Let $m$ be the smallest positive integer such that $B^m = I$, we have
$$B^m = I land B^127 = I implies B^gcd(127,m) = I$$
If $m < 127$, the fact $127$ is a primes forces $gcd(127,m) = 1$. This leads to $B = I$ and contradict with what we are told $B ne I$. As a result, $m ge 127 implies m = 127$.
answered Jul 16 at 14:39
achille hui
91k5127246
(+1) nice answer. But please try not to give full solutions when OP has not shown their attempts :)
– TheSimpliFire
Jul 16 at 14:40
@J.G. $A^7=I$ is given
– clark
Jul 16 at 14:45
(+1) for such a neat answer
– Ahmad Bazzi
Jul 16 at 14:47
If $X^A=I$ and $X^B=I$, then $X^mdc(A,B)=I$ is always true or only if $X$ is invertible?
– Alexandre Tourinho
Jul 20 at 1:46
@AlexandreTourinho $X$ is invertible as $X^a = I$ for positive integer $a$ implies $det(X)^a = 1 implies det(X) ne 0$.
– achille hui
Jul 20 at 7:21
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Since $A^7 = I$, $$beginalign
B = & A^7BA^-7\
= & A^6(ABA^-1)A^-6 = A^6B^2A^-6\
= & A^5(AB^2A^-1)A^-5 = A^5(ABA^-1)^2A^-5
= A^5(B^2)^2A^-5 = A^5B^4A^-5\
vdots & \
= & B^2^7 = B^128
endalign$$
Since $B$ is invertible, multiply both sides by $B^-1$ gives us $B^127 = I$.
Let $m$ be the smallest positive integer such that $B^m = I$, we have
$$B^m = I land B^127 = I implies B^gcd(127,m) = I$$
If $m < 127$, the fact $127$ is a primes forces $gcd(127,m) = 1$. This leads to $B = I$ and contradict with what we are told $B ne I$. As a result, $m ge 127 implies m = 127$.
answered Jul 16 at 14:39
achille hui
91k5127246
(+1) nice answer. But please try not to give full solutions when OP has not shown their attempts :)
– TheSimpliFire
Jul 16 at 14:40
@J.G. $A^7=I$ is given
– clark
Jul 16 at 14:45
(+1) for such a neat answer
– Ahmad Bazzi
Jul 16 at 14:47
If $X^A=I$ and $X^B=I$, then $X^mdc(A,B)=I$ is always true or only if $X$ is invertible?
– Alexandre Tourinho
Jul 20 at 1:46
@AlexandreTourinho $X$ is invertible as $X^a = I$ for positive integer $a$ implies $det(X)^a = 1 implies det(X) ne 0$.
– achille hui
Jul 20 at 7:21
 |Â
show 2 more comments
up vote
5
down vote
Since $A^7 = I$, $$beginalign
B = & A^7BA^-7\
= & A^6(ABA^-1)A^-6 = A^6B^2A^-6\
= & A^5(AB^2A^-1)A^-5 = A^5(ABA^-1)^2A^-5
= A^5(B^2)^2A^-5 = A^5B^4A^-5\
vdots & \
= & B^2^7 = B^128
endalign$$
Since $B$ is invertible, multiply both sides by $B^-1$ gives us $B^127 = I$.
Let $m$ be the smallest positive integer such that $B^m = I$, we have
$$B^m = I land B^127 = I implies B^gcd(127,m) = I$$
If $m < 127$, the fact $127$ is a primes forces $gcd(127,m) = 1$. This leads to $B = I$ and contradict with what we are told $B ne I$. As a result, $m ge 127 implies m = 127$.
answered Jul 16 at 14:39
achille hui
91k5127246
(+1) nice answer. But please try not to give full solutions when OP has not shown their attempts :)
– TheSimpliFire
Jul 16 at 14:40
@J.G. $A^7=I$ is given
– clark
Jul 16 at 14:45
(+1) for such a neat answer
– Ahmad Bazzi
Jul 16 at 14:47
If $X^A=I$ and $X^B=I$, then $X^mdc(A,B)=I$ is always true or only if $X$ is invertible?
– Alexandre Tourinho
Jul 20 at 1:46
@AlexandreTourinho $X$ is invertible as $X^a = I$ for positive integer $a$ implies $det(X)^a = 1 implies det(X) ne 0$.
– achille hui
Jul 20 at 7:21
 |Â
show 2 more comments
up vote
5
down vote
up vote
5
down vote
Since $A^7 = I$, $$beginalign
B = & A^7BA^-7\
= & A^6(ABA^-1)A^-6 = A^6B^2A^-6\
= & A^5(AB^2A^-1)A^-5 = A^5(ABA^-1)^2A^-5
= A^5(B^2)^2A^-5 = A^5B^4A^-5\
vdots & \
= & B^2^7 = B^128
endalign$$
Since $B$ is invertible, multiply both sides by $B^-1$ gives us $B^127 = I$.
Let $m$ be the smallest positive integer such that $B^m = I$, we have
$$B^m = I land B^127 = I implies B^gcd(127,m) = I$$
If $m < 127$, the fact $127$ is a primes forces $gcd(127,m) = 1$. This leads to $B = I$ and contradict with what we are told $B ne I$. As a result, $m ge 127 implies m = 127$.
answered Jul 16 at 14:39
achille hui
91k5127246
Since $A^7 = I$, $$beginalign
B = & A^7BA^-7\
= & A^6(ABA^-1)A^-6 = A^6B^2A^-6\
= & A^5(AB^2A^-1)A^-5 = A^5(ABA^-1)^2A^-5
= A^5(B^2)^2A^-5 = A^5B^4A^-5\
vdots & \
= & B^2^7 = B^128
endalign$$
Since $B$ is invertible, multiply both sides by $B^-1$ gives us $B^127 = I$.
Let $m$ be the smallest positive integer such that $B^m = I$, we have
$$B^m = I land B^127 = I implies B^gcd(127,m) = I$$
If $m < 127$, the fact $127$ is a primes forces $gcd(127,m) = 1$. This leads to $B = I$ and contradict with what we are told $B ne I$. As a result, $m ge 127 implies m = 127$.
answered Jul 16 at 14:39
achille hui
91k5127246
edited Jul 16 at 14:46
answered Jul 16 at 14:39
achille hui
91k5127246
answered Jul 16 at 14:39
achille hui
91k5127246
answered Jul 16 at 14:39
achille hui
91k5127246
91k5127246
(+1) nice answer. But please try not to give full solutions when OP has not shown their attempts :)
– TheSimpliFire
Jul 16 at 14:40
@J.G. $A^7=I$ is given
– clark
Jul 16 at 14:45
(+1) for such a neat answer
– Ahmad Bazzi
Jul 16 at 14:47
If $X^A=I$ and $X^B=I$, then $X^mdc(A,B)=I$ is always true or only if $X$ is invertible?
– Alexandre Tourinho
Jul 20 at 1:46
@AlexandreTourinho $X$ is invertible as $X^a = I$ for positive integer $a$ implies $det(X)^a = 1 implies det(X) ne 0$.
– achille hui
Jul 20 at 7:21
 |Â
show 2 more comments
(+1) nice answer. But please try not to give full solutions when OP has not shown their attempts :)
– TheSimpliFire
Jul 16 at 14:40
@J.G. $A^7=I$ is given
– clark
Jul 16 at 14:45
(+1) for such a neat answer
– Ahmad Bazzi
Jul 16 at 14:47
If $X^A=I$ and $X^B=I$, then $X^mdc(A,B)=I$ is always true or only if $X$ is invertible?
– Alexandre Tourinho
Jul 20 at 1:46
@AlexandreTourinho $X$ is invertible as $X^a = I$ for positive integer $a$ implies $det(X)^a = 1 implies det(X) ne 0$.
– achille hui
Jul 20 at 7:21
(+1) nice answer. But please try not to give full solutions when OP has not shown their attempts :)
– TheSimpliFire
Jul 16 at 14:40
(+1) nice answer. But please try not to give full solutions when OP has not shown their attempts :)
– TheSimpliFire
Jul 16 at 14:40
@J.G. $A^7=I$ is given
– clark
Jul 16 at 14:45
@J.G. $A^7=I$ is given
– clark
Jul 16 at 14:45
(+1) for such a neat answer
– Ahmad Bazzi
Jul 16 at 14:47
(+1) for such a neat answer
– Ahmad Bazzi
Jul 16 at 14:47
If $X^A=I$ and $X^B=I$, then $X^mdc(A,B)=I$ is always true or only if $X$ is invertible?
– Alexandre Tourinho
Jul 20 at 1:46
If $X^A=I$ and $X^B=I$, then $X^mdc(A,B)=I$ is always true or only if $X$ is invertible?
– Alexandre Tourinho
Jul 20 at 1:46
@AlexandreTourinho $X$ is invertible as $X^a = I$ for positive integer $a$ implies $det(X)^a = 1 implies det(X) ne 0$.
– achille hui
Jul 20 at 7:21
@AlexandreTourinho $X$ is invertible as $X^a = I$ for positive integer $a$ implies $det(X)^a = 1 implies det(X) ne 0$.
– achille hui
Jul 20 at 7:21
 |Â
show 2 more comments
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5
What did you try?
– José Carlos Santos
Jul 16 at 14:31