If $A$ is a symmetric matrix, then rank($A^n$) = rank($A$). Is, this statement true?

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If $A$ is a symmetric matrix over $mathbbR$, then rank($A^n$) = rank($A$). Is, this statement true?

For $n=2$ and $n=3$ the statement seems to be true but I have no idea how to prove or disprove for the general case.

Thanks in advance!

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Jul 30 at 9:10

asked Jul 30 at 9:03

reego

787416

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you explain us why the statement is true for $n=2,3$?
  – Babelfish
  Jul 30 at 9:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $n$ an arbitrary (positive) integer? or the dimension of $A$?
  – mr_e_man
  Jul 30 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mr_e_man $n$ is an arbitary positive integer.
  – reego
  Jul 30 at 9:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Spectral theorem should do the trick.
  – nicomezi
  Jul 30 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Symmetric matrix over $mathbbR$ or $mathbbC$?
  – Mathematician 42
  Jul 30 at 9:09
  

  
  
  
  

 | 
show 4 more comments

up vote
-1
down vote

favorite

1

If $A$ is a symmetric matrix over $mathbbR$, then rank($A^n$) = rank($A$). Is, this statement true?

For $n=2$ and $n=3$ the statement seems to be true but I have no idea how to prove or disprove for the general case.

Thanks in advance!

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Jul 30 at 9:10

asked Jul 30 at 9:03

reego

787416

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you explain us why the statement is true for $n=2,3$?
  – Babelfish
  Jul 30 at 9:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $n$ an arbitrary (positive) integer? or the dimension of $A$?
  – mr_e_man
  Jul 30 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mr_e_man $n$ is an arbitary positive integer.
  – reego
  Jul 30 at 9:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Spectral theorem should do the trick.
  – nicomezi
  Jul 30 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Symmetric matrix over $mathbbR$ or $mathbbC$?
  – Mathematician 42
  Jul 30 at 9:09
  

  
  
  
  

 | 
show 4 more comments

up vote
-1
down vote

favorite

1

up vote
-1
down vote

favorite

1

1

If $A$ is a symmetric matrix over $mathbbR$, then rank($A^n$) = rank($A$). Is, this statement true?

For $n=2$ and $n=3$ the statement seems to be true but I have no idea how to prove or disprove for the general case.

Thanks in advance!

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Jul 30 at 9:10

asked Jul 30 at 9:03

reego

787416

If $A$ is a symmetric matrix over $mathbbR$, then rank($A^n$) = rank($A$). Is, this statement true?

For $n=2$ and $n=3$ the statement seems to be true but I have no idea how to prove or disprove for the general case.

Thanks in advance!

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Jul 30 at 9:10

asked Jul 30 at 9:03

reego

787416

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 9:10

edited Jul 30 at 9:10

edited Jul 30 at 9:10

asked Jul 30 at 9:03

reego

787416

asked Jul 30 at 9:03

reego

787416

asked Jul 30 at 9:03

reego

787416

787416

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you explain us why the statement is true for $n=2,3$?
  – Babelfish
  Jul 30 at 9:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $n$ an arbitrary (positive) integer? or the dimension of $A$?
  – mr_e_man
  Jul 30 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mr_e_man $n$ is an arbitary positive integer.
  – reego
  Jul 30 at 9:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Spectral theorem should do the trick.
  – nicomezi
  Jul 30 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Symmetric matrix over $mathbbR$ or $mathbbC$?
  – Mathematician 42
  Jul 30 at 9:09
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you explain us why the statement is true for $n=2,3$?
  – Babelfish
  Jul 30 at 9:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $n$ an arbitrary (positive) integer? or the dimension of $A$?
  – mr_e_man
  Jul 30 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mr_e_man $n$ is an arbitary positive integer.
  – reego
  Jul 30 at 9:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Spectral theorem should do the trick.
  – nicomezi
  Jul 30 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Symmetric matrix over $mathbbR$ or $mathbbC$?
  – Mathematician 42
  Jul 30 at 9:09
  

  
  
  
  

Can you explain us why the statement is true for $n=2,3$?
– Babelfish
Jul 30 at 9:05

Can you explain us why the statement is true for $n=2,3$?
– Babelfish
Jul 30 at 9:05

Is $n$ an arbitrary (positive) integer? or the dimension of $A$?
– mr_e_man
Jul 30 at 9:06

Is $n$ an arbitrary (positive) integer? or the dimension of $A$?
– mr_e_man
Jul 30 at 9:06

@mr_e_man $n$ is an arbitary positive integer.
– reego
Jul 30 at 9:07

@mr_e_man $n$ is an arbitary positive integer.
– reego
Jul 30 at 9:07

1

1

Spectral theorem should do the trick.
– nicomezi
Jul 30 at 9:07

Spectral theorem should do the trick.
– nicomezi
Jul 30 at 9:07

Symmetric matrix over $mathbbR$ or $mathbbC$?
– Mathematician 42
Jul 30 at 9:09

Symmetric matrix over $mathbbR$ or $mathbbC$?
– Mathematician 42
Jul 30 at 9:09

 | 
show 4 more comments

1 Answer
1

active

oldest

votes

up vote
5
down vote



accepted

Let $A$ be a symmetric $m times m$ - Matrix.

If $x in ker(A^2)$, then $(Ax|Ax)=(A^2x|x)=0$, hence $x in ker(A)$.

This gives $ker(A)=ker(A^2)$, hence $ker(A)=ker(A^k)$ for all $k ge 1$.

It follows (why ?): $im(A)=im(A^k)$ for all $k ge 1$.

Hence $rank(A)=rank(A^k)$ for all $k ge 1$.

share|cite|improve this answer

answered Jul 30 at 9:10

Fred

37k1237

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good elementary proof. (+1)
  – nicomezi
  Jul 30 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi: Thanks !
  – Fred
  Jul 30 at 9:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What nicomezi said! I was about to invoke the theorem of symmetric matrices being diagonalizable etc :-)
  – Jyrki Lahtonen
  Jul 30 at 9:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)$ is the standard inner product on $mathbbR^N$ of $Ax$ with itself.
  – nicomezi
  Jul 30 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)=(A^TAx|x)$ and $A^T=A$.
  – Fred
  Jul 30 at 9:35
  

  
  
  
  

 | 
show 2 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

Let $A$ be a symmetric $m times m$ - Matrix.

If $x in ker(A^2)$, then $(Ax|Ax)=(A^2x|x)=0$, hence $x in ker(A)$.

This gives $ker(A)=ker(A^2)$, hence $ker(A)=ker(A^k)$ for all $k ge 1$.

It follows (why ?): $im(A)=im(A^k)$ for all $k ge 1$.

Hence $rank(A)=rank(A^k)$ for all $k ge 1$.

share|cite|improve this answer

answered Jul 30 at 9:10

Fred

37k1237

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good elementary proof. (+1)
  – nicomezi
  Jul 30 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi: Thanks !
  – Fred
  Jul 30 at 9:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What nicomezi said! I was about to invoke the theorem of symmetric matrices being diagonalizable etc :-)
  – Jyrki Lahtonen
  Jul 30 at 9:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)$ is the standard inner product on $mathbbR^N$ of $Ax$ with itself.
  – nicomezi
  Jul 30 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)=(A^TAx|x)$ and $A^T=A$.
  – Fred
  Jul 30 at 9:35
  

  
  
  
  

 | 
show 2 more comments

up vote
5
down vote



accepted

Let $A$ be a symmetric $m times m$ - Matrix.

If $x in ker(A^2)$, then $(Ax|Ax)=(A^2x|x)=0$, hence $x in ker(A)$.

This gives $ker(A)=ker(A^2)$, hence $ker(A)=ker(A^k)$ for all $k ge 1$.

It follows (why ?): $im(A)=im(A^k)$ for all $k ge 1$.

Hence $rank(A)=rank(A^k)$ for all $k ge 1$.

share|cite|improve this answer

answered Jul 30 at 9:10

Fred

37k1237

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good elementary proof. (+1)
  – nicomezi
  Jul 30 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi: Thanks !
  – Fred
  Jul 30 at 9:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What nicomezi said! I was about to invoke the theorem of symmetric matrices being diagonalizable etc :-)
  – Jyrki Lahtonen
  Jul 30 at 9:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)$ is the standard inner product on $mathbbR^N$ of $Ax$ with itself.
  – nicomezi
  Jul 30 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)=(A^TAx|x)$ and $A^T=A$.
  – Fred
  Jul 30 at 9:35
  

  
  
  
  

 | 
show 2 more comments

up vote
5
down vote



accepted

up vote
5
down vote



accepted

Let $A$ be a symmetric $m times m$ - Matrix.

If $x in ker(A^2)$, then $(Ax|Ax)=(A^2x|x)=0$, hence $x in ker(A)$.

This gives $ker(A)=ker(A^2)$, hence $ker(A)=ker(A^k)$ for all $k ge 1$.

It follows (why ?): $im(A)=im(A^k)$ for all $k ge 1$.

Hence $rank(A)=rank(A^k)$ for all $k ge 1$.

share|cite|improve this answer

answered Jul 30 at 9:10

Fred

37k1237

Let $A$ be a symmetric $m times m$ - Matrix.

If $x in ker(A^2)$, then $(Ax|Ax)=(A^2x|x)=0$, hence $x in ker(A)$.

This gives $ker(A)=ker(A^2)$, hence $ker(A)=ker(A^k)$ for all $k ge 1$.

It follows (why ?): $im(A)=im(A^k)$ for all $k ge 1$.

Hence $rank(A)=rank(A^k)$ for all $k ge 1$.

share|cite|improve this answer

answered Jul 30 at 9:10

Fred

37k1237

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 9:10

Fred

37k1237

answered Jul 30 at 9:10

Fred

37k1237

answered Jul 30 at 9:10

Fred

37k1237

37k1237

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good elementary proof. (+1)
  – nicomezi
  Jul 30 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi: Thanks !
  – Fred
  Jul 30 at 9:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What nicomezi said! I was about to invoke the theorem of symmetric matrices being diagonalizable etc :-)
  – Jyrki Lahtonen
  Jul 30 at 9:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)$ is the standard inner product on $mathbbR^N$ of $Ax$ with itself.
  – nicomezi
  Jul 30 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)=(A^TAx|x)$ and $A^T=A$.
  – Fred
  Jul 30 at 9:35
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good elementary proof. (+1)
  – nicomezi
  Jul 30 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi: Thanks !
  – Fred
  Jul 30 at 9:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What nicomezi said! I was about to invoke the theorem of symmetric matrices being diagonalizable etc :-)
  – Jyrki Lahtonen
  Jul 30 at 9:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)$ is the standard inner product on $mathbbR^N$ of $Ax$ with itself.
  – nicomezi
  Jul 30 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(Ax|Ax)=(A^TAx|x)$ and $A^T=A$.
  – Fred
  Jul 30 at 9:35
  

  
  
  
  

Good elementary proof. (+1)
– nicomezi
Jul 30 at 9:11

Good elementary proof. (+1)
– nicomezi
Jul 30 at 9:11

@nicomezi: Thanks !
– Fred
Jul 30 at 9:12

@nicomezi: Thanks !
– Fred
Jul 30 at 9:12

What nicomezi said! I was about to invoke the theorem of symmetric matrices being diagonalizable etc :-)
– Jyrki Lahtonen
Jul 30 at 9:13

What nicomezi said! I was about to invoke the theorem of symmetric matrices being diagonalizable etc :-)
– Jyrki Lahtonen
Jul 30 at 9:13

1

1

$(Ax|Ax)$ is the standard inner product on $mathbbR^N$ of $Ax$ with itself.
– nicomezi
Jul 30 at 9:26

$(Ax|Ax)$ is the standard inner product on $mathbbR^N$ of $Ax$ with itself.
– nicomezi
Jul 30 at 9:26

1

1

$(Ax|Ax)=(A^TAx|x)$ and $A^T=A$.
– Fred
Jul 30 at 9:35

$(Ax|Ax)=(A^TAx|x)$ and $A^T=A$.
– Fred
Jul 30 at 9:35

 | 
show 2 more comments

 

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