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

$x^TAx=0$ for all $x$ when $A$ is a skew symmetric matrixIs the product of symmetric positive semidefinite matrices positive definite?Linear system with positive semidefinite matrixIf $A$ and $B$ are positive definite, then is $B^{-1} - A^{-1}$ positive semidefinite?Is a sinc-distance matrix positive semidefinite?On one property of positive semidefinite matricesA symmetric real matrix with all diagonal entries unspecified can be completed to be positive semidefiniteSimultaneous Diagonalization of Symmetric Positive Semidefinite matricesAre these positive semidefinite matrices invertible?Showing a matrix positive semidefiniteIs this matrix product positive semidefinite?

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

286

$endgroup$

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

286

$endgroup$

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

286

$endgroup$

share|cite|improve this question

edited Mar 15 at 17:34

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 17:26

honzaikhonzaik

286

share|cite|improve this question

edited Mar 15 at 17:34

Rodrigo de Azevedo

13.2k41960

asked Mar 15 at 17:26

honzaikhonzaik

286

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

286

asked Mar 15 at 17:26

honzaikhonzaik

286

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

1,444311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

1,444311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

1,444311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

1,444311

$endgroup$

share|cite|improve this answer

answered Mar 15 at 17:32

chhrochhro

1,444311

answered Mar 15 at 17:32

chhrochhro

1,444311

answered Mar 15 at 17:32

chhrochhro

1,444311