In symmetric matrix if $A^6=I implies A^2=I$ [on hold]

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Let $A$ be a 3×3 real symmetric matrix such that $A^6 =I.$Then, $A^2 = I.$

Please give me a hint !

linear-algebra matrices

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edited Aug 4 at 4:01

asked Aug 4 at 3:49

Cloud JR

456311

put on hold as off-topic by BLAZE, José Carlos Santos, Jyrki Lahtonen, amWhy, Tyrone 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Jyrki Lahtonen, amWhy, Tyrone

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You've been here long enough to know that you're going to need to show more work than that to get a positive response. What do you know about the properties of real symmetric matrices that might be applicable here? What about matrices that can be squared to produce the identity matrix, in particular?
  – Robert Howard
  Aug 4 at 3:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, did you mean to write $A^6=I$ instead of $A^6=1$?
  – Robert Howard
  Aug 4 at 3:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry i mean $A^6=I$
  – Cloud JR
  Aug 4 at 4:02
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

Let $A$ be a 3×3 real symmetric matrix such that $A^6 =I.$Then, $A^2 = I.$

Please give me a hint !

linear-algebra matrices

share|cite|improve this question

edited Aug 4 at 4:01

asked Aug 4 at 3:49

Cloud JR

456311

put on hold as off-topic by BLAZE, José Carlos Santos, Jyrki Lahtonen, amWhy, Tyrone 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Jyrki Lahtonen, amWhy, Tyrone

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You've been here long enough to know that you're going to need to show more work than that to get a positive response. What do you know about the properties of real symmetric matrices that might be applicable here? What about matrices that can be squared to produce the identity matrix, in particular?
  – Robert Howard
  Aug 4 at 3:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, did you mean to write $A^6=I$ instead of $A^6=1$?
  – Robert Howard
  Aug 4 at 3:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry i mean $A^6=I$
  – Cloud JR
  Aug 4 at 4:02
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

Let $A$ be a 3×3 real symmetric matrix such that $A^6 =I.$Then, $A^2 = I.$

Please give me a hint !

linear-algebra matrices

share|cite|improve this question

edited Aug 4 at 4:01

asked Aug 4 at 3:49

Cloud JR

456311

Let $A$ be a 3×3 real symmetric matrix such that $A^6 =I.$Then, $A^2 = I.$

Please give me a hint !

linear-algebra matrices

share|cite|improve this question

edited Aug 4 at 4:01

asked Aug 4 at 3:49

Cloud JR

456311

share|cite|improve this question

share|cite|improve this question

edited Aug 4 at 4:01

edited Aug 4 at 4:01

edited Aug 4 at 4:01

asked Aug 4 at 3:49

Cloud JR

456311

asked Aug 4 at 3:49

Cloud JR

456311

asked Aug 4 at 3:49

Cloud JR

456311

456311

put on hold as off-topic by BLAZE, José Carlos Santos, Jyrki Lahtonen, amWhy, Tyrone 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Jyrki Lahtonen, amWhy, Tyrone

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by BLAZE, José Carlos Santos, Jyrki Lahtonen, amWhy, Tyrone 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Jyrki Lahtonen, amWhy, Tyrone

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You've been here long enough to know that you're going to need to show more work than that to get a positive response. What do you know about the properties of real symmetric matrices that might be applicable here? What about matrices that can be squared to produce the identity matrix, in particular?
  – Robert Howard
  Aug 4 at 3:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, did you mean to write $A^6=I$ instead of $A^6=1$?
  – Robert Howard
  Aug 4 at 3:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry i mean $A^6=I$
  – Cloud JR
  Aug 4 at 4:02
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You've been here long enough to know that you're going to need to show more work than that to get a positive response. What do you know about the properties of real symmetric matrices that might be applicable here? What about matrices that can be squared to produce the identity matrix, in particular?
  – Robert Howard
  Aug 4 at 3:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, did you mean to write $A^6=I$ instead of $A^6=1$?
  – Robert Howard
  Aug 4 at 3:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry i mean $A^6=I$
  – Cloud JR
  Aug 4 at 4:02
  

  
  
  
  

1

1

You've been here long enough to know that you're going to need to show more work than that to get a positive response. What do you know about the properties of real symmetric matrices that might be applicable here? What about matrices that can be squared to produce the identity matrix, in particular?
– Robert Howard
Aug 4 at 3:53

You've been here long enough to know that you're going to need to show more work than that to get a positive response. What do you know about the properties of real symmetric matrices that might be applicable here? What about matrices that can be squared to produce the identity matrix, in particular?
– Robert Howard
Aug 4 at 3:53

Also, did you mean to write $A^6=I$ instead of $A^6=1$?
– Robert Howard
Aug 4 at 3:53

Also, did you mean to write $A^6=I$ instead of $A^6=1$?
– Robert Howard
Aug 4 at 3:53

Sorry i mean $A^6=I$
– Cloud JR
Aug 4 at 4:02

Sorry i mean $A^6=I$
– Cloud JR
Aug 4 at 4:02

add a comment | 

2 Answers
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accepted

There exists inverse matrix $P$ such that $$PAP^-1=diag(lambda_1,lambda_2,lambda_3).$$

So $$PA^6P^-1=diag(lambda_1^6,lambda_2^6,lambda_3^6)=I.$$
This implies $$lambda_1^6=lambda_2^6=lambda_3^6=1.$$
So $$lambda_1^2=lambda_2^2=lambda_3^2=1;$$
$$PA^2P^-1=diag(lambda_1^2,lambda_2^2,lambda_3^2)=I;$$
and $$A^2=I.$$

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answered Aug 4 at 4:01

Riemann

1,9891217

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up vote
2
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Hint: $$x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$

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answered Aug 4 at 3:57

vadim123

73.6k895183

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

There exists inverse matrix $P$ such that $$PAP^-1=diag(lambda_1,lambda_2,lambda_3).$$

So $$PA^6P^-1=diag(lambda_1^6,lambda_2^6,lambda_3^6)=I.$$
This implies $$lambda_1^6=lambda_2^6=lambda_3^6=1.$$
So $$lambda_1^2=lambda_2^2=lambda_3^2=1;$$
$$PA^2P^-1=diag(lambda_1^2,lambda_2^2,lambda_3^2)=I;$$
and $$A^2=I.$$

share|cite|improve this answer

answered Aug 4 at 4:01

Riemann

1,9891217

add a comment | 

up vote
2
down vote



accepted

There exists inverse matrix $P$ such that $$PAP^-1=diag(lambda_1,lambda_2,lambda_3).$$

So $$PA^6P^-1=diag(lambda_1^6,lambda_2^6,lambda_3^6)=I.$$
This implies $$lambda_1^6=lambda_2^6=lambda_3^6=1.$$
So $$lambda_1^2=lambda_2^2=lambda_3^2=1;$$
$$PA^2P^-1=diag(lambda_1^2,lambda_2^2,lambda_3^2)=I;$$
and $$A^2=I.$$

share|cite|improve this answer

answered Aug 4 at 4:01

Riemann

1,9891217

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

There exists inverse matrix $P$ such that $$PAP^-1=diag(lambda_1,lambda_2,lambda_3).$$

So $$PA^6P^-1=diag(lambda_1^6,lambda_2^6,lambda_3^6)=I.$$
This implies $$lambda_1^6=lambda_2^6=lambda_3^6=1.$$
So $$lambda_1^2=lambda_2^2=lambda_3^2=1;$$
$$PA^2P^-1=diag(lambda_1^2,lambda_2^2,lambda_3^2)=I;$$
and $$A^2=I.$$

share|cite|improve this answer

answered Aug 4 at 4:01

Riemann

1,9891217

There exists inverse matrix $P$ such that $$PAP^-1=diag(lambda_1,lambda_2,lambda_3).$$

So $$PA^6P^-1=diag(lambda_1^6,lambda_2^6,lambda_3^6)=I.$$
This implies $$lambda_1^6=lambda_2^6=lambda_3^6=1.$$
So $$lambda_1^2=lambda_2^2=lambda_3^2=1;$$
$$PA^2P^-1=diag(lambda_1^2,lambda_2^2,lambda_3^2)=I;$$
and $$A^2=I.$$

share|cite|improve this answer

answered Aug 4 at 4:01

Riemann

1,9891217

share|cite|improve this answer

share|cite|improve this answer

answered Aug 4 at 4:01

Riemann

1,9891217

answered Aug 4 at 4:01

Riemann

1,9891217

answered Aug 4 at 4:01

Riemann

1,9891217

1,9891217

add a comment | 

add a comment | 

up vote
2
down vote

Hint: $$x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$

share|cite|improve this answer

answered Aug 4 at 3:57

vadim123

73.6k895183

add a comment | 

up vote
2
down vote

Hint: $$x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$

share|cite|improve this answer

answered Aug 4 at 3:57

vadim123

73.6k895183

add a comment | 

up vote
2
down vote

up vote
2
down vote

Hint: $$x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$

share|cite|improve this answer

answered Aug 4 at 3:57

vadim123

73.6k895183

Hint: $$x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$

share|cite|improve this answer

answered Aug 4 at 3:57

vadim123

73.6k895183

share|cite|improve this answer

share|cite|improve this answer

answered Aug 4 at 3:57

vadim123

73.6k895183

answered Aug 4 at 3:57

vadim123

73.6k895183

answered Aug 4 at 3:57

vadim123

73.6k895183

73.6k895183

add a comment | 

add a comment |