Mixing
![Does $sum_k=1^infty frac 1 k^1+frac 1 k $ converge or diverge? [duplicate]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
An upper bound for $trace(XYXZ)$ in terms of $trace(X^2YZ)$ [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I want to find an upper bound for $trace(XYXZ)$ in terms of $trace(X^2YZ)$. $X, Y, Z$ are Hermitian positive definite matrices.
Any idea/help is appreciated.
Update: Assume that $Y$ and $Z$ commute: $YZ=ZY$.
linear-algebra matrices trace positive-definite
asked Jul 16 at 23:50
Mah
550312
closed as off-topic by Brian Borchers, Isaac Browne, Mostafa Ayaz, Claude Leibovici, Parcly Taxel Jul 17 at 11:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, Isaac Browne, Mostafa Ayaz, Claude Leibovici, Parcly Taxel
 |Â
show 2 more comments
up vote
1
down vote
favorite
I want to find an upper bound for $trace(XYXZ)$ in terms of $trace(X^2YZ)$. $X, Y, Z$ are Hermitian positive definite matrices.
Any idea/help is appreciated.
Update: Assume that $Y$ and $Z$ commute: $YZ=ZY$.
linear-algebra matrices trace positive-definite
asked Jul 16 at 23:50
Mah
550312
closed as off-topic by Brian Borchers, Isaac Browne, Mostafa Ayaz, Claude Leibovici, Parcly Taxel Jul 17 at 11:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, Isaac Browne, Mostafa Ayaz, Claude Leibovici, Parcly Taxel
There cannot be such a bound because $X^2$ may be equal to zero while your first trace may not be
– lcv
Jul 17 at 4:21
@lcv but X is positive definite, so X^2 can’t be zero.
– Mah
Jul 17 at 4:26
Oh I missed that. Still why should it be true?
– lcv
Jul 17 at 4:31
1
This result comes to mind, but I'm not sure if it's directly useful.
– Omnomnomnom
Jul 17 at 4:41
4 people gave meaningful answers and comments on the question, so why it is marked as off-topic?!
– Mah
Jul 17 at 15:33
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to find an upper bound for $trace(XYXZ)$ in terms of $trace(X^2YZ)$. $X, Y, Z$ are Hermitian positive definite matrices.
Any idea/help is appreciated.
Update: Assume that $Y$ and $Z$ commute: $YZ=ZY$.
linear-algebra matrices trace positive-definite
asked Jul 16 at 23:50
Mah
550312
I want to find an upper bound for $trace(XYXZ)$ in terms of $trace(X^2YZ)$. $X, Y, Z$ are Hermitian positive definite matrices.
Any idea/help is appreciated.
Update: Assume that $Y$ and $Z$ commute: $YZ=ZY$.
linear-algebra matrices trace positive-definite
asked Jul 16 at 23:50
Mah
550312
edited Jul 21 at 16:19
asked Jul 16 at 23:50
Mah
550312
asked Jul 16 at 23:50
Mah
550312
asked Jul 16 at 23:50
Mah
550312
550312
closed as off-topic by Brian Borchers, Isaac Browne, Mostafa Ayaz, Claude Leibovici, Parcly Taxel Jul 17 at 11:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, Isaac Browne, Mostafa Ayaz, Claude Leibovici, Parcly Taxel
closed as off-topic by Brian Borchers, Isaac Browne, Mostafa Ayaz, Claude Leibovici, Parcly Taxel Jul 17 at 11:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, Isaac Browne, Mostafa Ayaz, Claude Leibovici, Parcly Taxel
There cannot be such a bound because $X^2$ may be equal to zero while your first trace may not be
– lcv
Jul 17 at 4:21
@lcv but X is positive definite, so X^2 can’t be zero.
– Mah
Jul 17 at 4:26
Oh I missed that. Still why should it be true?
– lcv
Jul 17 at 4:31
1
This result comes to mind, but I'm not sure if it's directly useful.
– Omnomnomnom
Jul 17 at 4:41
4 people gave meaningful answers and comments on the question, so why it is marked as off-topic?!
– Mah
Jul 17 at 15:33
 |Â
show 2 more comments
There cannot be such a bound because $X^2$ may be equal to zero while your first trace may not be
– lcv
Jul 17 at 4:21
@lcv but X is positive definite, so X^2 can’t be zero.
– Mah
Jul 17 at 4:26
Oh I missed that. Still why should it be true?
– lcv
Jul 17 at 4:31
1
This result comes to mind, but I'm not sure if it's directly useful.
– Omnomnomnom
Jul 17 at 4:41
4 people gave meaningful answers and comments on the question, so why it is marked as off-topic?!
– Mah
Jul 17 at 15:33
There cannot be such a bound because $X^2$ may be equal to zero while your first trace may not be
– lcv
Jul 17 at 4:21
There cannot be such a bound because $X^2$ may be equal to zero while your first trace may not be
– lcv
Jul 17 at 4:21
@lcv but X is positive definite, so X^2 can’t be zero.
– Mah
Jul 17 at 4:26
@lcv but X is positive definite, so X^2 can’t be zero.
– Mah
Jul 17 at 4:26
Oh I missed that. Still why should it be true?
– lcv
Jul 17 at 4:31
Oh I missed that. Still why should it be true?
– lcv
Jul 17 at 4:31
1
1
This result comes to mind, but I'm not sure if it's directly useful.
– Omnomnomnom
Jul 17 at 4:41
This result comes to mind, but I'm not sure if it's directly useful.
– Omnomnomnom
Jul 17 at 4:41
4 people gave meaningful answers and comments on the question, so why it is marked as off-topic?!
– Mah
Jul 17 at 15:33
4 people gave meaningful answers and comments on the question, so why it is marked as off-topic?!
– Mah
Jul 17 at 15:33
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Short answer: None of the two has a reason to be larger than the other. Some bounds in terms of the condition numbers of $Y,Z$ are
$$trace(XYXZ) leq min(kappa(Y), kappa(Z)) cdot trace(X^2YZ)$$
$$trace(X^2YZ) leq min(kappa(Y), kappa(Z)) cdot trace(XYXZ)$$
First, we can simultaneously diagonalize $Y$ and $Z$, say $Y = PY'P^-1$, $Z = PZ'P^-1$ with $P$ unitary, then we get
$$trace(X'Y'X'Z') quad, quad trace(X'^2Y'Z')$$
with $X' = P^-1XP$ still hermitian. Thus we may suppose $Y,Z$ are diagonal. Then
$$beginalign*
trace(XYXZ) &= sum_i,j|X_ij|^2Y_jjZ_ii \
trace(X^2YZ) &= sum_i,j|X_ij|^2Y_iiZ_ii
endalign*$$
The $Y_ii,Z_ii>0$ are arbitrary, so all we can say is that one is bounded by $$kappa(Y) = max_i neq j|Y_ii/Y_jj|$$ (or $kappa(Z) = max_i neq j|Z_ii/Z_jj|$) times the other.
answered Jul 17 at 7:01
barto
13k32479
add a comment |Â
up vote
4
down vote
It is possible for $texttrace(X^2 Y Z) = 0$ with $X,Y,Z$ positive definite Hermitian. Try
$$ X^2 = pmatrix2 & -1cr -1 & 1cr, Y = pmatrix1/10 & 0cr 0 & 1cr, Z = pmatrix2 & 1cr 1 & 7/10$$
However, this will not work with the added condition that $Y$ and $Z$ commute.
If $Y$ and $Z$ commute, so do $Y$ and $Z^1/2$, and $YZ = Z^1/2 Y Z^1/2$ is positive definite. Then
$$texttrace(X^2 Y Z) = texttrace((YZ)^1/2 X^2 (YZ)^1/2) > 0$$
answered Jul 17 at 6:09
Robert Israel
304k22201442
I see. Nice observation. Thanks!
– Mah
Jul 17 at 13:10
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Short answer: None of the two has a reason to be larger than the other. Some bounds in terms of the condition numbers of $Y,Z$ are
$$trace(XYXZ) leq min(kappa(Y), kappa(Z)) cdot trace(X^2YZ)$$
$$trace(X^2YZ) leq min(kappa(Y), kappa(Z)) cdot trace(XYXZ)$$
First, we can simultaneously diagonalize $Y$ and $Z$, say $Y = PY'P^-1$, $Z = PZ'P^-1$ with $P$ unitary, then we get
$$trace(X'Y'X'Z') quad, quad trace(X'^2Y'Z')$$
with $X' = P^-1XP$ still hermitian. Thus we may suppose $Y,Z$ are diagonal. Then
$$beginalign*
trace(XYXZ) &= sum_i,j|X_ij|^2Y_jjZ_ii \
trace(X^2YZ) &= sum_i,j|X_ij|^2Y_iiZ_ii
endalign*$$
The $Y_ii,Z_ii>0$ are arbitrary, so all we can say is that one is bounded by $$kappa(Y) = max_i neq j|Y_ii/Y_jj|$$ (or $kappa(Z) = max_i neq j|Z_ii/Z_jj|$) times the other.
answered Jul 17 at 7:01
barto
13k32479
add a comment |Â
up vote
3
down vote
accepted
Short answer: None of the two has a reason to be larger than the other. Some bounds in terms of the condition numbers of $Y,Z$ are
$$trace(XYXZ) leq min(kappa(Y), kappa(Z)) cdot trace(X^2YZ)$$
$$trace(X^2YZ) leq min(kappa(Y), kappa(Z)) cdot trace(XYXZ)$$
First, we can simultaneously diagonalize $Y$ and $Z$, say $Y = PY'P^-1$, $Z = PZ'P^-1$ with $P$ unitary, then we get
$$trace(X'Y'X'Z') quad, quad trace(X'^2Y'Z')$$
with $X' = P^-1XP$ still hermitian. Thus we may suppose $Y,Z$ are diagonal. Then
$$beginalign*
trace(XYXZ) &= sum_i,j|X_ij|^2Y_jjZ_ii \
trace(X^2YZ) &= sum_i,j|X_ij|^2Y_iiZ_ii
endalign*$$
The $Y_ii,Z_ii>0$ are arbitrary, so all we can say is that one is bounded by $$kappa(Y) = max_i neq j|Y_ii/Y_jj|$$ (or $kappa(Z) = max_i neq j|Z_ii/Z_jj|$) times the other.
answered Jul 17 at 7:01
barto
13k32479
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Short answer: None of the two has a reason to be larger than the other. Some bounds in terms of the condition numbers of $Y,Z$ are
$$trace(XYXZ) leq min(kappa(Y), kappa(Z)) cdot trace(X^2YZ)$$
$$trace(X^2YZ) leq min(kappa(Y), kappa(Z)) cdot trace(XYXZ)$$
First, we can simultaneously diagonalize $Y$ and $Z$, say $Y = PY'P^-1$, $Z = PZ'P^-1$ with $P$ unitary, then we get
$$trace(X'Y'X'Z') quad, quad trace(X'^2Y'Z')$$
with $X' = P^-1XP$ still hermitian. Thus we may suppose $Y,Z$ are diagonal. Then
$$beginalign*
trace(XYXZ) &= sum_i,j|X_ij|^2Y_jjZ_ii \
trace(X^2YZ) &= sum_i,j|X_ij|^2Y_iiZ_ii
endalign*$$
The $Y_ii,Z_ii>0$ are arbitrary, so all we can say is that one is bounded by $$kappa(Y) = max_i neq j|Y_ii/Y_jj|$$ (or $kappa(Z) = max_i neq j|Z_ii/Z_jj|$) times the other.
answered Jul 17 at 7:01
barto
13k32479
Short answer: None of the two has a reason to be larger than the other. Some bounds in terms of the condition numbers of $Y,Z$ are
$$trace(XYXZ) leq min(kappa(Y), kappa(Z)) cdot trace(X^2YZ)$$
$$trace(X^2YZ) leq min(kappa(Y), kappa(Z)) cdot trace(XYXZ)$$
First, we can simultaneously diagonalize $Y$ and $Z$, say $Y = PY'P^-1$, $Z = PZ'P^-1$ with $P$ unitary, then we get
$$trace(X'Y'X'Z') quad, quad trace(X'^2Y'Z')$$
with $X' = P^-1XP$ still hermitian. Thus we may suppose $Y,Z$ are diagonal. Then
$$beginalign*
trace(XYXZ) &= sum_i,j|X_ij|^2Y_jjZ_ii \
trace(X^2YZ) &= sum_i,j|X_ij|^2Y_iiZ_ii
endalign*$$
The $Y_ii,Z_ii>0$ are arbitrary, so all we can say is that one is bounded by $$kappa(Y) = max_i neq j|Y_ii/Y_jj|$$ (or $kappa(Z) = max_i neq j|Z_ii/Z_jj|$) times the other.
answered Jul 17 at 7:01
barto
13k32479
edited Jul 17 at 14:53
answered Jul 17 at 7:01
barto
13k32479
answered Jul 17 at 7:01
barto
13k32479
answered Jul 17 at 7:01
barto
13k32479
13k32479
add a comment |Â
add a comment |Â
up vote
4
down vote
It is possible for $texttrace(X^2 Y Z) = 0$ with $X,Y,Z$ positive definite Hermitian. Try
$$ X^2 = pmatrix2 & -1cr -1 & 1cr, Y = pmatrix1/10 & 0cr 0 & 1cr, Z = pmatrix2 & 1cr 1 & 7/10$$
However, this will not work with the added condition that $Y$ and $Z$ commute.
If $Y$ and $Z$ commute, so do $Y$ and $Z^1/2$, and $YZ = Z^1/2 Y Z^1/2$ is positive definite. Then
$$texttrace(X^2 Y Z) = texttrace((YZ)^1/2 X^2 (YZ)^1/2) > 0$$
answered Jul 17 at 6:09
Robert Israel
304k22201442
I see. Nice observation. Thanks!
– Mah
Jul 17 at 13:10
add a comment |Â
up vote
4
down vote
It is possible for $texttrace(X^2 Y Z) = 0$ with $X,Y,Z$ positive definite Hermitian. Try
$$ X^2 = pmatrix2 & -1cr -1 & 1cr, Y = pmatrix1/10 & 0cr 0 & 1cr, Z = pmatrix2 & 1cr 1 & 7/10$$
However, this will not work with the added condition that $Y$ and $Z$ commute.
If $Y$ and $Z$ commute, so do $Y$ and $Z^1/2$, and $YZ = Z^1/2 Y Z^1/2$ is positive definite. Then
$$texttrace(X^2 Y Z) = texttrace((YZ)^1/2 X^2 (YZ)^1/2) > 0$$
answered Jul 17 at 6:09
Robert Israel
304k22201442
I see. Nice observation. Thanks!
– Mah
Jul 17 at 13:10
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It is possible for $texttrace(X^2 Y Z) = 0$ with $X,Y,Z$ positive definite Hermitian. Try
$$ X^2 = pmatrix2 & -1cr -1 & 1cr, Y = pmatrix1/10 & 0cr 0 & 1cr, Z = pmatrix2 & 1cr 1 & 7/10$$
However, this will not work with the added condition that $Y$ and $Z$ commute.
If $Y$ and $Z$ commute, so do $Y$ and $Z^1/2$, and $YZ = Z^1/2 Y Z^1/2$ is positive definite. Then
$$texttrace(X^2 Y Z) = texttrace((YZ)^1/2 X^2 (YZ)^1/2) > 0$$
answered Jul 17 at 6:09
Robert Israel
304k22201442
It is possible for $texttrace(X^2 Y Z) = 0$ with $X,Y,Z$ positive definite Hermitian. Try
$$ X^2 = pmatrix2 & -1cr -1 & 1cr, Y = pmatrix1/10 & 0cr 0 & 1cr, Z = pmatrix2 & 1cr 1 & 7/10$$
However, this will not work with the added condition that $Y$ and $Z$ commute.
If $Y$ and $Z$ commute, so do $Y$ and $Z^1/2$, and $YZ = Z^1/2 Y Z^1/2$ is positive definite. Then
$$texttrace(X^2 Y Z) = texttrace((YZ)^1/2 X^2 (YZ)^1/2) > 0$$
answered Jul 17 at 6:09
Robert Israel
304k22201442
edited Jul 17 at 6:38
answered Jul 17 at 6:09
Robert Israel
304k22201442
answered Jul 17 at 6:09
Robert Israel
304k22201442
answered Jul 17 at 6:09
Robert Israel
304k22201442
304k22201442
I see. Nice observation. Thanks!
– Mah
Jul 17 at 13:10
add a comment |Â
I see. Nice observation. Thanks!
– Mah
Jul 17 at 13:10
I see. Nice observation. Thanks!
– Mah
Jul 17 at 13:10
I see. Nice observation. Thanks!
– Mah
Jul 17 at 13:10
add a comment |Â
There cannot be such a bound because $X^2$ may be equal to zero while your first trace may not be
– lcv
Jul 17 at 4:21
@lcv but X is positive definite, so X^2 can’t be zero.
– Mah
Jul 17 at 4:26
Oh I missed that. Still why should it be true?
– lcv
Jul 17 at 4:31
1
This result comes to mind, but I'm not sure if it's directly useful.
– Omnomnomnom
Jul 17 at 4:41
4 people gave meaningful answers and comments on the question, so why it is marked as off-topic?!
– Mah
Jul 17 at 15:33