$operatornametrace(B)>0$, and $C$ is a Hermitian positive definite, is $operatornametrace(BC)geq 0$?

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Let $B$ is a Hermitian matrix such that $operatornametrace(B)>0$ (but $B$ is "not" positive definite), and $C$ is a Hermitian positive definite matrix. Can we conclude $operatornametrace(BC)geq 0$? I tried two approaches, but neither worked.

1) We can write
$$operatornametrace(BC)=operatornametrace(C^1/2BC^1/2)$$
where $C^1/2$ is positive definite, and thus $C^1/2BC^1/2$ is congruent with $BC$. But congruence just preserves the number of positive/negative/zero eigenvalues, and doesn't say anything about the magnitude, so apparently this doesn't say anything about the sign of $operatornametrace(BC)$.

2) We can also write eigenvalue decomposition of $B$ and $C$:
$$B=sum_i beta_ib_ib_i^T quadtextandquad C=sum_i gamma_ic_ic_i^T$$
Hence,
$$operatornametrace(BC)=sum_isum_jbeta_igamma_j|b_i^Ta_j|^2$$
We know $gamma_j>0$ (from positive-definiteness of $C$) and $|b_i^Ta_j|^2geq 0$, but for $beta_i$, we only know $sum_ibeta_i>0$. So I don't know how we can conclude $sum_isum_jbeta_igamma_j|b_i^Ta_j|^2geq 0$?

linear-algebra matrices trace positive-definite

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asked Jul 19 at 15:01

Mah

545312

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up vote
0
down vote

favorite

Let $B$ is a Hermitian matrix such that $operatornametrace(B)>0$ (but $B$ is "not" positive definite), and $C$ is a Hermitian positive definite matrix. Can we conclude $operatornametrace(BC)geq 0$? I tried two approaches, but neither worked.

1) We can write
$$operatornametrace(BC)=operatornametrace(C^1/2BC^1/2)$$
where $C^1/2$ is positive definite, and thus $C^1/2BC^1/2$ is congruent with $BC$. But congruence just preserves the number of positive/negative/zero eigenvalues, and doesn't say anything about the magnitude, so apparently this doesn't say anything about the sign of $operatornametrace(BC)$.

2) We can also write eigenvalue decomposition of $B$ and $C$:
$$B=sum_i beta_ib_ib_i^T quadtextandquad C=sum_i gamma_ic_ic_i^T$$
Hence,
$$operatornametrace(BC)=sum_isum_jbeta_igamma_j|b_i^Ta_j|^2$$
We know $gamma_j>0$ (from positive-definiteness of $C$) and $|b_i^Ta_j|^2geq 0$, but for $beta_i$, we only know $sum_ibeta_i>0$. So I don't know how we can conclude $sum_isum_jbeta_igamma_j|b_i^Ta_j|^2geq 0$?

linear-algebra matrices trace positive-definite

share|cite|improve this question

asked Jul 19 at 15:01

Mah

545312

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $B$ is a Hermitian matrix such that $operatornametrace(B)>0$ (but $B$ is "not" positive definite), and $C$ is a Hermitian positive definite matrix. Can we conclude $operatornametrace(BC)geq 0$? I tried two approaches, but neither worked.

1) We can write
$$operatornametrace(BC)=operatornametrace(C^1/2BC^1/2)$$
where $C^1/2$ is positive definite, and thus $C^1/2BC^1/2$ is congruent with $BC$. But congruence just preserves the number of positive/negative/zero eigenvalues, and doesn't say anything about the magnitude, so apparently this doesn't say anything about the sign of $operatornametrace(BC)$.

2) We can also write eigenvalue decomposition of $B$ and $C$:
$$B=sum_i beta_ib_ib_i^T quadtextandquad C=sum_i gamma_ic_ic_i^T$$
Hence,
$$operatornametrace(BC)=sum_isum_jbeta_igamma_j|b_i^Ta_j|^2$$
We know $gamma_j>0$ (from positive-definiteness of $C$) and $|b_i^Ta_j|^2geq 0$, but for $beta_i$, we only know $sum_ibeta_i>0$. So I don't know how we can conclude $sum_isum_jbeta_igamma_j|b_i^Ta_j|^2geq 0$?

linear-algebra matrices trace positive-definite

share|cite|improve this question

asked Jul 19 at 15:01

Mah

545312

Let $B$ is a Hermitian matrix such that $operatornametrace(B)>0$ (but $B$ is "not" positive definite), and $C$ is a Hermitian positive definite matrix. Can we conclude $operatornametrace(BC)geq 0$? I tried two approaches, but neither worked.

1) We can write
$$operatornametrace(BC)=operatornametrace(C^1/2BC^1/2)$$
where $C^1/2$ is positive definite, and thus $C^1/2BC^1/2$ is congruent with $BC$. But congruence just preserves the number of positive/negative/zero eigenvalues, and doesn't say anything about the magnitude, so apparently this doesn't say anything about the sign of $operatornametrace(BC)$.

2) We can also write eigenvalue decomposition of $B$ and $C$:
$$B=sum_i beta_ib_ib_i^T quadtextandquad C=sum_i gamma_ic_ic_i^T$$
Hence,
$$operatornametrace(BC)=sum_isum_jbeta_igamma_j|b_i^Ta_j|^2$$
We know $gamma_j>0$ (from positive-definiteness of $C$) and $|b_i^Ta_j|^2geq 0$, but for $beta_i$, we only know $sum_ibeta_i>0$. So I don't know how we can conclude $sum_isum_jbeta_igamma_j|b_i^Ta_j|^2geq 0$?

linear-algebra matrices trace positive-definite

share|cite|improve this question

asked Jul 19 at 15:01

Mah

545312

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 15:01

Mah

545312

asked Jul 19 at 15:01

Mah

545312

asked Jul 19 at 15:01

Mah

545312

545312

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

Not necessarily. For instance, consider
$$
B = pmatrix2&0\0&-1, quad C = pmatrix1&0\0&3
$$

share|cite|improve this answer

answered Jul 19 at 15:03

Omnomnomnom

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might be interested in the following fact: if $B$ is Hermitian, then we can only have $ operatornametrace(BC) geq 0 $ for all positive (semi)definite $C$ if $B$ is itself positive semidefinite.
  – Omnomnomnom
  Jul 19 at 15:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is interesting fact, I didn't know about the "only" part of it.
  – Mah
  Jul 19 at 15:10
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Not necessarily. For instance, consider
$$
B = pmatrix2&0\0&-1, quad C = pmatrix1&0\0&3
$$

share|cite|improve this answer

answered Jul 19 at 15:03

Omnomnomnom

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might be interested in the following fact: if $B$ is Hermitian, then we can only have $ operatornametrace(BC) geq 0 $ for all positive (semi)definite $C$ if $B$ is itself positive semidefinite.
  – Omnomnomnom
  Jul 19 at 15:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is interesting fact, I didn't know about the "only" part of it.
  – Mah
  Jul 19 at 15:10
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Not necessarily. For instance, consider
$$
B = pmatrix2&0\0&-1, quad C = pmatrix1&0\0&3
$$

share|cite|improve this answer

answered Jul 19 at 15:03

Omnomnomnom

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might be interested in the following fact: if $B$ is Hermitian, then we can only have $ operatornametrace(BC) geq 0 $ for all positive (semi)definite $C$ if $B$ is itself positive semidefinite.
  – Omnomnomnom
  Jul 19 at 15:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is interesting fact, I didn't know about the "only" part of it.
  – Mah
  Jul 19 at 15:10
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Not necessarily. For instance, consider
$$
B = pmatrix2&0\0&-1, quad C = pmatrix1&0\0&3
$$

share|cite|improve this answer

answered Jul 19 at 15:03

Omnomnomnom

121k784170

Not necessarily. For instance, consider
$$
B = pmatrix2&0\0&-1, quad C = pmatrix1&0\0&3
$$

share|cite|improve this answer

answered Jul 19 at 15:03

Omnomnomnom

121k784170

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 15:03

Omnomnomnom

121k784170

answered Jul 19 at 15:03

Omnomnomnom

121k784170

answered Jul 19 at 15:03

Omnomnomnom

121k784170

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might be interested in the following fact: if $B$ is Hermitian, then we can only have $ operatornametrace(BC) geq 0 $ for all positive (semi)definite $C$ if $B$ is itself positive semidefinite.
  – Omnomnomnom
  Jul 19 at 15:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is interesting fact, I didn't know about the "only" part of it.
  – Mah
  Jul 19 at 15:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might be interested in the following fact: if $B$ is Hermitian, then we can only have $ operatornametrace(BC) geq 0 $ for all positive (semi)definite $C$ if $B$ is itself positive semidefinite.
  – Omnomnomnom
  Jul 19 at 15:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is interesting fact, I didn't know about the "only" part of it.
  – Mah
  Jul 19 at 15:10
  

  
  
  
  

You might be interested in the following fact: if $B$ is Hermitian, then we can only have $ operatornametrace(BC) geq 0 $ for all positive (semi)definite $C$ if $B$ is itself positive semidefinite.
– Omnomnomnom
Jul 19 at 15:07

You might be interested in the following fact: if $B$ is Hermitian, then we can only have $ operatornametrace(BC) geq 0 $ for all positive (semi)definite $C$ if $B$ is itself positive semidefinite.
– Omnomnomnom
Jul 19 at 15:07

That is interesting fact, I didn't know about the "only" part of it.
– Mah
Jul 19 at 15:10

That is interesting fact, I didn't know about the "only" part of it.
– Mah
Jul 19 at 15:10

add a comment | 

 

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