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If both $A-B$ and $B-A$ are positive semidefinite, then $A = B$

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1

$begingroup$

Let $A, B$ be two positive semidefinite matrices. Prove that if both $A-B$ and $B-A$ are positive semidefinite, then $A = B$.

I can show that their diagonal elements are the same but for others I have no idea. Any help will be appreciated.

linear-algebra matrices symmetric-matrices positive-semidefinite

share|cite|improve this question

edited Mar 15 at 17:34

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 17:26

honzaikhonzaik

266

$endgroup$

add a comment | 

1

$begingroup$

Let $A, B$ be two positive semidefinite matrices. Prove that if both $A-B$ and $B-A$ are positive semidefinite, then $A = B$.

I can show that their diagonal elements are the same but for others I have no idea. Any help will be appreciated.

linear-algebra matrices symmetric-matrices positive-semidefinite

share|cite|improve this question

edited Mar 15 at 17:34

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 17:26

honzaikhonzaik

266

$endgroup$

add a comment | 

$begingroup$

Let $A, B$ be two positive semidefinite matrices. Prove that if both $A-B$ and $B-A$ are positive semidefinite, then $A = B$.

I can show that their diagonal elements are the same but for others I have no idea. Any help will be appreciated.

linear-algebra matrices symmetric-matrices positive-semidefinite

share|cite|improve this question

edited Mar 15 at 17:34

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 17:26

honzaikhonzaik

266

$endgroup$

share|cite|improve this question

edited Mar 15 at 17:34

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 17:26

honzaikhonzaik

266

share|cite|improve this question

edited Mar 15 at 17:34

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 17:26

honzaikhonzaik

266

edited Mar 15 at 17:34

Rodrigo de Azevedo

13.2k41960

edited Mar 15 at 17:34

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 17:26

honzaikhonzaik

266

asked Mar 15 at 17:26

honzaikhonzaik

266

2 Answers
2

active

oldest

votes

2

$begingroup$

One route is to observe that if $lambda,v$ is an eigenvalue/eigenvector pair of $M$ with , then $(-M)v=-lambda v$, meaning that $-lambda$ is an eigenvalue of $(-M)$. Since $M:=A-B$ is positive semidefinite, conclude that any eigenvalue of $M$ satisfies $0leq lambdaleq 0$, so $lambda=0$, so that $M=0$.

share|cite|improve this answer

answered Mar 15 at 17:45

Alex R.Alex R.

25.1k12452

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much! Thats a neat proof
  $endgroup$
  – honzaik
  Mar 15 at 18:15
  

  
  
  
  

add a comment | 

1

$begingroup$

By assumption $A-B=-(B-A)$ is both positive semidefinite and negative semidefinite. This can only happen if $A-B=0$.

share|cite|improve this answer

answered Mar 15 at 17:32

chhrochhro

1,434311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  its this equivalent to saying that $x^TAx=0 forall x implies A = 0$ which is wrong (math.stackexchange.com/questions/358281/…) ?
  $endgroup$
  – honzaik
  Mar 15 at 17:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assumed we are working over $mathbb C$ and so we are using $x^*Ax$. So the link is not relevant.
  $endgroup$
  – chhro
  Mar 15 at 18:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  actually, the argument still works since the real positive semidefinite matrices=real symmetric matrices with nonnegative eigenvalues. I don't see the point of mentioning the link. I didn't use that in the argument.
  $endgroup$
  – chhro
  Mar 15 at 18:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  well i assumed you meant $0 geq -x^T (B-A) x = x^T (A-B) x geq 0$. I guess when I hear positive semidef etc I assume this definition with vectors.
  $endgroup$
  – honzaik
  Mar 15 at 18:10
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

One route is to observe that if $lambda,v$ is an eigenvalue/eigenvector pair of $M$ with , then $(-M)v=-lambda v$, meaning that $-lambda$ is an eigenvalue of $(-M)$. Since $M:=A-B$ is positive semidefinite, conclude that any eigenvalue of $M$ satisfies $0leq lambdaleq 0$, so $lambda=0$, so that $M=0$.

share|cite|improve this answer

answered Mar 15 at 17:45

Alex R.Alex R.

25.1k12452

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much! Thats a neat proof
  $endgroup$
  – honzaik
  Mar 15 at 18:15
  

  
  
  
  

add a comment | 

2

$begingroup$

One route is to observe that if $lambda,v$ is an eigenvalue/eigenvector pair of $M$ with , then $(-M)v=-lambda v$, meaning that $-lambda$ is an eigenvalue of $(-M)$. Since $M:=A-B$ is positive semidefinite, conclude that any eigenvalue of $M$ satisfies $0leq lambdaleq 0$, so $lambda=0$, so that $M=0$.

share|cite|improve this answer

answered Mar 15 at 17:45

Alex R.Alex R.

25.1k12452

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much! Thats a neat proof
  $endgroup$
  – honzaik
  Mar 15 at 18:15
  

  
  
  
  

add a comment | 

$begingroup$

One route is to observe that if $lambda,v$ is an eigenvalue/eigenvector pair of $M$ with , then $(-M)v=-lambda v$, meaning that $-lambda$ is an eigenvalue of $(-M)$. Since $M:=A-B$ is positive semidefinite, conclude that any eigenvalue of $M$ satisfies $0leq lambdaleq 0$, so $lambda=0$, so that $M=0$.

share|cite|improve this answer

answered Mar 15 at 17:45

Alex R.Alex R.

25.1k12452

$endgroup$

share|cite|improve this answer

answered Mar 15 at 17:45

Alex R.Alex R.

25.1k12452

answered Mar 15 at 17:45

Alex R.Alex R.

25.1k12452

answered Mar 15 at 17:45

Alex R.Alex R.

25.1k12452

1

$begingroup$

By assumption $A-B=-(B-A)$ is both positive semidefinite and negative semidefinite. This can only happen if $A-B=0$.

share|cite|improve this answer

answered Mar 15 at 17:32

chhrochhro

1,434311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  its this equivalent to saying that $x^TAx=0 forall x implies A = 0$ which is wrong (math.stackexchange.com/questions/358281/…) ?
  $endgroup$
  – honzaik
  Mar 15 at 17:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assumed we are working over $mathbb C$ and so we are using $x^*Ax$. So the link is not relevant.
  $endgroup$
  – chhro
  Mar 15 at 18:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  actually, the argument still works since the real positive semidefinite matrices=real symmetric matrices with nonnegative eigenvalues. I don't see the point of mentioning the link. I didn't use that in the argument.
  $endgroup$
  – chhro
  Mar 15 at 18:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  well i assumed you meant $0 geq -x^T (B-A) x = x^T (A-B) x geq 0$. I guess when I hear positive semidef etc I assume this definition with vectors.
  $endgroup$
  – honzaik
  Mar 15 at 18:10
  
  

  
  
  
  

add a comment | 

1

$begingroup$

By assumption $A-B=-(B-A)$ is both positive semidefinite and negative semidefinite. This can only happen if $A-B=0$.

share|cite|improve this answer

answered Mar 15 at 17:32

chhrochhro

1,434311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  its this equivalent to saying that $x^TAx=0 forall x implies A = 0$ which is wrong (math.stackexchange.com/questions/358281/…) ?
  $endgroup$
  – honzaik
  Mar 15 at 17:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assumed we are working over $mathbb C$ and so we are using $x^*Ax$. So the link is not relevant.
  $endgroup$
  – chhro
  Mar 15 at 18:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  actually, the argument still works since the real positive semidefinite matrices=real symmetric matrices with nonnegative eigenvalues. I don't see the point of mentioning the link. I didn't use that in the argument.
  $endgroup$
  – chhro
  Mar 15 at 18:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  well i assumed you meant $0 geq -x^T (B-A) x = x^T (A-B) x geq 0$. I guess when I hear positive semidef etc I assume this definition with vectors.
  $endgroup$
  – honzaik
  Mar 15 at 18:10
  
  

  
  
  
  

add a comment | 

$begingroup$

By assumption $A-B=-(B-A)$ is both positive semidefinite and negative semidefinite. This can only happen if $A-B=0$.

share|cite|improve this answer

answered Mar 15 at 17:32

chhrochhro

1,434311

$endgroup$

share|cite|improve this answer

answered Mar 15 at 17:32

chhrochhro

1,434311

answered Mar 15 at 17:32

chhrochhro

1,434311

answered Mar 15 at 17:32

chhrochhro

1,434311