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If matrix $A$ is unitary and $B^2 = A$ then $B$ is also unitary [duplicate]

-2

$begingroup$

This question is an exact duplicate of:

• 
  If matrix $A$ is unitary and $B^2=A$, is $B$ necessarily unitary?
  
  1 answer
  
  

I need to prove or give a counterexample:

If matrix $A$ is unitary and $B^2 = A$ then $B$ is also unitary

I think the statement is true since the unitary matrix A can only be Identity matrix I or negative identity matrix $-I$; and $B=A^2$ is an identity matrix which makes sure it is unitary.

linear-algebra orthogonal-matrices

share|cite|improve this question

edited Mar 26 at 16:07

gt6989b

36k22557

asked Mar 26 at 14:24

Denny ShenDenny Shen

82

$endgroup$

marked as duplicate by 5xum, gt6989b, Jyrki Lahtonen, Lord Shark the Unknown, egreg linear-algebra
Users with the  linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.

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Mar 27 at 11:06

This question was marked as an exact duplicate of an existing question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3163142/…
  $endgroup$
  – 5xum
  Mar 26 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "Unitary matrix can only be identity or negative identity" : False, even in the one dimensional case!
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 26 at 14:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Asking the same question twice is a violation of the rules of this site. I suggest you delete this question before the mods do.
  $endgroup$
  – 5xum
  Mar 26 at 14:26
  

  
  
  
  

add a comment | 

-2

$begingroup$

This question is an exact duplicate of:

• 
  If matrix $A$ is unitary and $B^2=A$, is $B$ necessarily unitary?
  
  1 answer
  
  

I need to prove or give a counterexample:

If matrix $A$ is unitary and $B^2 = A$ then $B$ is also unitary

I think the statement is true since the unitary matrix A can only be Identity matrix I or negative identity matrix $-I$; and $B=A^2$ is an identity matrix which makes sure it is unitary.

linear-algebra orthogonal-matrices

share|cite|improve this question

edited Mar 26 at 16:07

gt6989b

36k22557

asked Mar 26 at 14:24

Denny ShenDenny Shen

82

$endgroup$

marked as duplicate by 5xum, gt6989b, Jyrki Lahtonen, Lord Shark the Unknown, egreg linear-algebra
Users with the  linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.

StackExchange.ready(function()
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$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
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Mar 27 at 11:06

This question was marked as an exact duplicate of an existing question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3163142/…
  $endgroup$
  – 5xum
  Mar 26 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "Unitary matrix can only be identity or negative identity" : False, even in the one dimensional case!
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 26 at 14:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Asking the same question twice is a violation of the rules of this site. I suggest you delete this question before the mods do.
  $endgroup$
  – 5xum
  Mar 26 at 14:26
  

  
  
  
  

add a comment | 

$begingroup$

This question is an exact duplicate of:

• 
  If matrix $A$ is unitary and $B^2=A$, is $B$ necessarily unitary?
  
  1 answer
  
  

I need to prove or give a counterexample:

If matrix $A$ is unitary and $B^2 = A$ then $B$ is also unitary

I think the statement is true since the unitary matrix A can only be Identity matrix I or negative identity matrix $-I$; and $B=A^2$ is an identity matrix which makes sure it is unitary.

linear-algebra orthogonal-matrices

share|cite|improve this question

edited Mar 26 at 16:07

gt6989b

36k22557

asked Mar 26 at 14:24

Denny ShenDenny Shen

82

$endgroup$

share|cite|improve this question

edited Mar 26 at 16:07

gt6989b

36k22557

asked Mar 26 at 14:24

Denny ShenDenny Shen

82

share|cite|improve this question

edited Mar 26 at 16:07

gt6989b

36k22557

asked Mar 26 at 14:24

Denny ShenDenny Shen

82

edited Mar 26 at 16:07

gt6989b

36k22557

edited Mar 26 at 16:07

gt6989b

36k22557

asked Mar 26 at 14:24

Denny ShenDenny Shen

82

asked Mar 26 at 14:24

Denny ShenDenny Shen

82

1 Answer
1

active

oldest

votes

0

$begingroup$

HINT

You are incorrect. Both $I$ and $-I$ are, of course, unitary, but there are many such matrices, e.g.
$$
A = beginbmatrix sqrt2/2 & -sqrt2/2 \ sqrt2/2 & sqrt2/2 endbmatrix
$$

But think: matrix $M$ is unitary if and only if $M^T M = I$. What could you say above $(M^2)^T (M^2)$ in such circumstances?

Your question asks to go the other way. If $M^TM = I$ and $A^2 = M$, can we say that $A^TA = I$ as well?

share|cite|improve this answer

edited Mar 26 at 16:07

answered Mar 26 at 14:27

gt6989bgt6989b

36k22557

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think the question is not "if $A$ is unitary then $A^2$ is", but rather, "if $A$ is unitary, then any square root of $A$ is".
  $endgroup$
  – Jason DeVito
  Mar 26 at 14:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, i think i got it , since in this case (M^2)^T(M^2) will also equal to the Identity matrix, which proves that the statement is true.
  $endgroup$
  – Denny Shen
  Mar 26 at 14:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DennyShen the other comment got it right. You want to go the other way. Say $M^TM = I$, and $A^2 = M$. Can you then say $A^T A = I$ as well?
  $endgroup$
  – gt6989b
  Mar 26 at 16:06
  

  
  
  
  

add a comment | 

0

$begingroup$

HINT

You are incorrect. Both $I$ and $-I$ are, of course, unitary, but there are many such matrices, e.g.
$$
A = beginbmatrix sqrt2/2 & -sqrt2/2 \ sqrt2/2 & sqrt2/2 endbmatrix
$$

But think: matrix $M$ is unitary if and only if $M^T M = I$. What could you say above $(M^2)^T (M^2)$ in such circumstances?

Your question asks to go the other way. If $M^TM = I$ and $A^2 = M$, can we say that $A^TA = I$ as well?

share|cite|improve this answer

edited Mar 26 at 16:07

answered Mar 26 at 14:27

gt6989bgt6989b

36k22557

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think the question is not "if $A$ is unitary then $A^2$ is", but rather, "if $A$ is unitary, then any square root of $A$ is".
  $endgroup$
  – Jason DeVito
  Mar 26 at 14:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, i think i got it , since in this case (M^2)^T(M^2) will also equal to the Identity matrix, which proves that the statement is true.
  $endgroup$
  – Denny Shen
  Mar 26 at 14:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DennyShen the other comment got it right. You want to go the other way. Say $M^TM = I$, and $A^2 = M$. Can you then say $A^T A = I$ as well?
  $endgroup$
  – gt6989b
  Mar 26 at 16:06
  

  
  
  
  

add a comment | 

0

$begingroup$

HINT

You are incorrect. Both $I$ and $-I$ are, of course, unitary, but there are many such matrices, e.g.
$$
A = beginbmatrix sqrt2/2 & -sqrt2/2 \ sqrt2/2 & sqrt2/2 endbmatrix
$$

But think: matrix $M$ is unitary if and only if $M^T M = I$. What could you say above $(M^2)^T (M^2)$ in such circumstances?

Your question asks to go the other way. If $M^TM = I$ and $A^2 = M$, can we say that $A^TA = I$ as well?

share|cite|improve this answer

edited Mar 26 at 16:07

answered Mar 26 at 14:27

gt6989bgt6989b

36k22557

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think the question is not "if $A$ is unitary then $A^2$ is", but rather, "if $A$ is unitary, then any square root of $A$ is".
  $endgroup$
  – Jason DeVito
  Mar 26 at 14:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, i think i got it , since in this case (M^2)^T(M^2) will also equal to the Identity matrix, which proves that the statement is true.
  $endgroup$
  – Denny Shen
  Mar 26 at 14:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DennyShen the other comment got it right. You want to go the other way. Say $M^TM = I$, and $A^2 = M$. Can you then say $A^T A = I$ as well?
  $endgroup$
  – gt6989b
  Mar 26 at 16:06
  

  
  
  
  

add a comment | 

$begingroup$

HINT

You are incorrect. Both $I$ and $-I$ are, of course, unitary, but there are many such matrices, e.g.
$$
A = beginbmatrix sqrt2/2 & -sqrt2/2 \ sqrt2/2 & sqrt2/2 endbmatrix
$$

But think: matrix $M$ is unitary if and only if $M^T M = I$. What could you say above $(M^2)^T (M^2)$ in such circumstances?

Your question asks to go the other way. If $M^TM = I$ and $A^2 = M$, can we say that $A^TA = I$ as well?

share|cite|improve this answer

edited Mar 26 at 16:07

answered Mar 26 at 14:27

gt6989bgt6989b

36k22557

$endgroup$

share|cite|improve this answer

edited Mar 26 at 16:07

answered Mar 26 at 14:27

gt6989bgt6989b

36k22557

answered Mar 26 at 14:27

gt6989bgt6989b

36k22557

answered Mar 26 at 14:27

gt6989bgt6989b

36k22557