Mixing
How to show $text{Tr}(Mlog N)=sum_{i,j}^nlambda_ilog(tilde{lambda_j})(u_i^{top}tilde{u}_j)^2$?
$begingroup$
The above question is the equation $(2.4)$ of the following paper:
MATRIX EXPONENTIATED GRADIENT UPDATES.
Let $M$ and $N$ be two $n times n$ positive definite matrices where $M=ULambda U^{top}$, $N=tilde{U}tilde{Lambda} tilde{U}^{top}
$ and $(lambda_i,v_i)$ are eigenpairs of $M$, likewise for $N$.
How to show the following
$$text{Tr}(Mlog N)=sum_{i,j}lambda_ilog(tilde{lambda_j})(u_i^{top}tilde{u}_j)^2$$
First I do not know what $i,j$ mean in summation, and how we have it using two summations. Second how to get that.
My try:
begin{align}
text{Tr}(Mlog N) &=
text{Tr}(ULambda U^{top}
tilde{U}log(tilde{Lambda})tilde{U}^{top}) \
& = text{Tr}(Lambda U^{top}
tilde{U}log(tilde{Lambda})tilde{U}^{top}U)
end{align}
How can I proceed using matrix calculus to get the result not by expanding? what is the hidden trick?
linear-algebra matrices positive-definite symmetric-matrices
edited Jan 20 at 3:36
loup blanc
23.5k21851
asked Jan 18 at 20:17
SaeedSaeed
1,073310
$endgroup$
add a comment |
$begingroup$
The above question is the equation $(2.4)$ of the following paper:
MATRIX EXPONENTIATED GRADIENT UPDATES.
Let $M$ and $N$ be two $n times n$ positive definite matrices where $M=ULambda U^{top}$, $N=tilde{U}tilde{Lambda} tilde{U}^{top}
$ and $(lambda_i,v_i)$ are eigenpairs of $M$, likewise for $N$.
How to show the following
$$text{Tr}(Mlog N)=sum_{i,j}lambda_ilog(tilde{lambda_j})(u_i^{top}tilde{u}_j)^2$$
First I do not know what $i,j$ mean in summation, and how we have it using two summations. Second how to get that.
My try:
begin{align}
text{Tr}(Mlog N) &=
text{Tr}(ULambda U^{top}
tilde{U}log(tilde{Lambda})tilde{U}^{top}) \
& = text{Tr}(Lambda U^{top}
tilde{U}log(tilde{Lambda})tilde{U}^{top}U)
end{align}
How can I proceed using matrix calculus to get the result not by expanding? what is the hidden trick?
linear-algebra matrices positive-definite symmetric-matrices
edited Jan 20 at 3:36
loup blanc
23.5k21851
asked Jan 18 at 20:17
SaeedSaeed
1,073310
$endgroup$
add a comment |
$begingroup$
The above question is the equation $(2.4)$ of the following paper:
MATRIX EXPONENTIATED GRADIENT UPDATES.
Let $M$ and $N$ be two $n times n$ positive definite matrices where $M=ULambda U^{top}$, $N=tilde{U}tilde{Lambda} tilde{U}^{top}
$ and $(lambda_i,v_i)$ are eigenpairs of $M$, likewise for $N$.
How to show the following
$$text{Tr}(Mlog N)=sum_{i,j}lambda_ilog(tilde{lambda_j})(u_i^{top}tilde{u}_j)^2$$
First I do not know what $i,j$ mean in summation, and how we have it using two summations. Second how to get that.
My try:
begin{align}
text{Tr}(Mlog N) &=
text{Tr}(ULambda U^{top}
tilde{U}log(tilde{Lambda})tilde{U}^{top}) \
& = text{Tr}(Lambda U^{top}
tilde{U}log(tilde{Lambda})tilde{U}^{top}U)
end{align}
How can I proceed using matrix calculus to get the result not by expanding? what is the hidden trick?
linear-algebra matrices positive-definite symmetric-matrices
edited Jan 20 at 3:36
loup blanc
23.5k21851
asked Jan 18 at 20:17
SaeedSaeed
1,073310
$endgroup$
The above question is the equation $(2.4)$ of the following paper:
MATRIX EXPONENTIATED GRADIENT UPDATES.
Let $M$ and $N$ be two $n times n$ positive definite matrices where $M=ULambda U^{top}$, $N=tilde{U}tilde{Lambda} tilde{U}^{top}
$ and $(lambda_i,v_i)$ are eigenpairs of $M$, likewise for $N$.
How to show the following
$$text{Tr}(Mlog N)=sum_{i,j}lambda_ilog(tilde{lambda_j})(u_i^{top}tilde{u}_j)^2$$
First I do not know what $i,j$ mean in summation, and how we have it using two summations. Second how to get that.
My try:
begin{align}
text{Tr}(Mlog N) &=
text{Tr}(ULambda U^{top}
tilde{U}log(tilde{Lambda})tilde{U}^{top}) \
& = text{Tr}(Lambda U^{top}
tilde{U}log(tilde{Lambda})tilde{U}^{top}U)
end{align}
How can I proceed using matrix calculus to get the result not by expanding? what is the hidden trick?
linear-algebra matrices positive-definite symmetric-matrices
linear-algebra matrices positive-definite symmetric-matrices
edited Jan 20 at 3:36
loup blanc
23.5k21851
asked Jan 18 at 20:17
SaeedSaeed
1,073310
edited Jan 20 at 3:36
loup blanc
23.5k21851
asked Jan 18 at 20:17
SaeedSaeed
1,073310
edited Jan 20 at 3:36
loup blanc
23.5k21851
edited Jan 20 at 3:36
loup blanc
23.5k21851
edited Jan 20 at 3:36
loup blanc
23.5k21851
23.5k21851
asked Jan 18 at 20:17
SaeedSaeed
1,073310
asked Jan 18 at 20:17
SaeedSaeed
1,073310
asked Jan 18 at 20:17
SaeedSaeed
1,073310
1,073310
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To easy the notations, let us take
$$
kappa_j=logtilde{lambda}_j
$$
and
$$
v_j=tilde{u}_j.
$$
With these notations, it suffices to show
$$
text{tr}left(Mlog Nright)=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
$$
Now, let us begin our proof. Using
begin{align}
M&=sum_{j=1}^nlambda_ju_ju_j^{top},\
N&=sum_{j=1}^nkappa_jv_jv_j^{top},
end{align}
we have
begin{align}
text{tr}left(Mlog Nright)&=text{tr}left(sum_{i=1}^nlambda_iu_iu_i^{top}sum_{j=1}^nkappa_jv_jv_j^{top}right)\
&=text{tr}left(sum_{i,j=1}^nlambda_ikappa_ju_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(u_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(v_j^{top}u_iu_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(v_j^{top}u_iright)left(u_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
end{align}
This completes the proof.
I would like to thank @loupblanc for kind advices.
answered Jan 20 at 1:42
hypernovahypernova
4,834414
$endgroup$
$begingroup$
Your proof is correct. 2 remarks: i) write $log(N)=cdots$ and ii) using Frobenius is useless because $u_i^Tv_j=v_j^Tu_i$.
$endgroup$
– loup blanc
Jan 20 at 3:35
$begingroup$
@hypernova: Does the summation have $n^2$ terms?
$endgroup$
– Saeed
Jan 20 at 4:27
$begingroup$
@loupblanc: Thank you for your advices. I modified my answer as per your suggestions, credited to you.
$endgroup$
– hypernova
Jan 20 at 5:11
$begingroup$
@Saeed: Yes, the summation contains $n^2$ terms, and each of the indices $i$ and $j$ runs from $1$ to $n$. In other words, $sum_{i,j=1}^n=sum_{i=1}^nsum_{j=1}^n$.
$endgroup$
– hypernova
Jan 20 at 5:11
add a comment |
$begingroup$
The first version of the formula with one index is false. Take $M=begin{pmatrix}50&37\37&61end{pmatrix},N=diag(2,1)$.
answered Jan 19 at 10:04
loup blancloup blanc
23.5k21851
$endgroup$
$begingroup$
It is on page 999 of the following paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. Maybe I am wrong. Because I am confused about $i$ and $j$.
$endgroup$
– Saeed
Jan 20 at 1:29
$begingroup$
Yes, the correct formula is a sum with $2$ indices.
$endgroup$
– loup blanc
Jan 20 at 3:38
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To easy the notations, let us take
$$
kappa_j=logtilde{lambda}_j
$$
and
$$
v_j=tilde{u}_j.
$$
With these notations, it suffices to show
$$
text{tr}left(Mlog Nright)=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
$$
Now, let us begin our proof. Using
begin{align}
M&=sum_{j=1}^nlambda_ju_ju_j^{top},\
N&=sum_{j=1}^nkappa_jv_jv_j^{top},
end{align}
we have
begin{align}
text{tr}left(Mlog Nright)&=text{tr}left(sum_{i=1}^nlambda_iu_iu_i^{top}sum_{j=1}^nkappa_jv_jv_j^{top}right)\
&=text{tr}left(sum_{i,j=1}^nlambda_ikappa_ju_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(u_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(v_j^{top}u_iu_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(v_j^{top}u_iright)left(u_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
end{align}
This completes the proof.
I would like to thank @loupblanc for kind advices.
answered Jan 20 at 1:42
hypernovahypernova
4,834414
$endgroup$
$begingroup$
Your proof is correct. 2 remarks: i) write $log(N)=cdots$ and ii) using Frobenius is useless because $u_i^Tv_j=v_j^Tu_i$.
$endgroup$
– loup blanc
Jan 20 at 3:35
$begingroup$
@hypernova: Does the summation have $n^2$ terms?
$endgroup$
– Saeed
Jan 20 at 4:27
$begingroup$
@loupblanc: Thank you for your advices. I modified my answer as per your suggestions, credited to you.
$endgroup$
– hypernova
Jan 20 at 5:11
$begingroup$
@Saeed: Yes, the summation contains $n^2$ terms, and each of the indices $i$ and $j$ runs from $1$ to $n$. In other words, $sum_{i,j=1}^n=sum_{i=1}^nsum_{j=1}^n$.
$endgroup$
– hypernova
Jan 20 at 5:11
add a comment |
$begingroup$
To easy the notations, let us take
$$
kappa_j=logtilde{lambda}_j
$$
and
$$
v_j=tilde{u}_j.
$$
With these notations, it suffices to show
$$
text{tr}left(Mlog Nright)=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
$$
Now, let us begin our proof. Using
begin{align}
M&=sum_{j=1}^nlambda_ju_ju_j^{top},\
N&=sum_{j=1}^nkappa_jv_jv_j^{top},
end{align}
we have
begin{align}
text{tr}left(Mlog Nright)&=text{tr}left(sum_{i=1}^nlambda_iu_iu_i^{top}sum_{j=1}^nkappa_jv_jv_j^{top}right)\
&=text{tr}left(sum_{i,j=1}^nlambda_ikappa_ju_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(u_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(v_j^{top}u_iu_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(v_j^{top}u_iright)left(u_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
end{align}
This completes the proof.
I would like to thank @loupblanc for kind advices.
answered Jan 20 at 1:42
hypernovahypernova
4,834414
$endgroup$
$begingroup$
Your proof is correct. 2 remarks: i) write $log(N)=cdots$ and ii) using Frobenius is useless because $u_i^Tv_j=v_j^Tu_i$.
$endgroup$
– loup blanc
Jan 20 at 3:35
$begingroup$
@hypernova: Does the summation have $n^2$ terms?
$endgroup$
– Saeed
Jan 20 at 4:27
$begingroup$
@loupblanc: Thank you for your advices. I modified my answer as per your suggestions, credited to you.
$endgroup$
– hypernova
Jan 20 at 5:11
$begingroup$
@Saeed: Yes, the summation contains $n^2$ terms, and each of the indices $i$ and $j$ runs from $1$ to $n$. In other words, $sum_{i,j=1}^n=sum_{i=1}^nsum_{j=1}^n$.
$endgroup$
– hypernova
Jan 20 at 5:11
add a comment |
$begingroup$
To easy the notations, let us take
$$
kappa_j=logtilde{lambda}_j
$$
and
$$
v_j=tilde{u}_j.
$$
With these notations, it suffices to show
$$
text{tr}left(Mlog Nright)=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
$$
Now, let us begin our proof. Using
begin{align}
M&=sum_{j=1}^nlambda_ju_ju_j^{top},\
N&=sum_{j=1}^nkappa_jv_jv_j^{top},
end{align}
we have
begin{align}
text{tr}left(Mlog Nright)&=text{tr}left(sum_{i=1}^nlambda_iu_iu_i^{top}sum_{j=1}^nkappa_jv_jv_j^{top}right)\
&=text{tr}left(sum_{i,j=1}^nlambda_ikappa_ju_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(u_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(v_j^{top}u_iu_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(v_j^{top}u_iright)left(u_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
end{align}
This completes the proof.
I would like to thank @loupblanc for kind advices.
answered Jan 20 at 1:42
hypernovahypernova
4,834414
$endgroup$
To easy the notations, let us take
$$
kappa_j=logtilde{lambda}_j
$$
and
$$
v_j=tilde{u}_j.
$$
With these notations, it suffices to show
$$
text{tr}left(Mlog Nright)=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
$$
Now, let us begin our proof. Using
begin{align}
M&=sum_{j=1}^nlambda_ju_ju_j^{top},\
N&=sum_{j=1}^nkappa_jv_jv_j^{top},
end{align}
we have
begin{align}
text{tr}left(Mlog Nright)&=text{tr}left(sum_{i=1}^nlambda_iu_iu_i^{top}sum_{j=1}^nkappa_jv_jv_j^{top}right)\
&=text{tr}left(sum_{i,j=1}^nlambda_ikappa_ju_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(u_iu_i^{top}v_jv_j^{top}right)\
&=sum_{i,j=1}^nlambda_ikappa_jtext{tr}left(v_j^{top}u_iu_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(v_j^{top}u_iright)left(u_i^{top}v_jright)\
&=sum_{i,j=1}^nlambda_ikappa_jleft(u_i^{top}v_jright)^2.
end{align}
This completes the proof.
I would like to thank @loupblanc for kind advices.
answered Jan 20 at 1:42
hypernovahypernova
4,834414
edited Jan 20 at 5:10
answered Jan 20 at 1:42
hypernovahypernova
4,834414
answered Jan 20 at 1:42
hypernovahypernova
4,834414
answered Jan 20 at 1:42
hypernovahypernova
4,834414
4,834414
$begingroup$
Your proof is correct. 2 remarks: i) write $log(N)=cdots$ and ii) using Frobenius is useless because $u_i^Tv_j=v_j^Tu_i$.
$endgroup$
– loup blanc
Jan 20 at 3:35
$begingroup$
@hypernova: Does the summation have $n^2$ terms?
$endgroup$
– Saeed
Jan 20 at 4:27
$begingroup$
@loupblanc: Thank you for your advices. I modified my answer as per your suggestions, credited to you.
$endgroup$
– hypernova
Jan 20 at 5:11
$begingroup$
@Saeed: Yes, the summation contains $n^2$ terms, and each of the indices $i$ and $j$ runs from $1$ to $n$. In other words, $sum_{i,j=1}^n=sum_{i=1}^nsum_{j=1}^n$.
$endgroup$
– hypernova
Jan 20 at 5:11
add a comment |
$begingroup$
Your proof is correct. 2 remarks: i) write $log(N)=cdots$ and ii) using Frobenius is useless because $u_i^Tv_j=v_j^Tu_i$.
$endgroup$
– loup blanc
Jan 20 at 3:35
$begingroup$
@hypernova: Does the summation have $n^2$ terms?
$endgroup$
– Saeed
Jan 20 at 4:27
$begingroup$
@loupblanc: Thank you for your advices. I modified my answer as per your suggestions, credited to you.
$endgroup$
– hypernova
Jan 20 at 5:11
$begingroup$
@Saeed: Yes, the summation contains $n^2$ terms, and each of the indices $i$ and $j$ runs from $1$ to $n$. In other words, $sum_{i,j=1}^n=sum_{i=1}^nsum_{j=1}^n$.
$endgroup$
– hypernova
Jan 20 at 5:11
$begingroup$
Your proof is correct. 2 remarks: i) write $log(N)=cdots$ and ii) using Frobenius is useless because $u_i^Tv_j=v_j^Tu_i$.
$endgroup$
– loup blanc
Jan 20 at 3:35
$begingroup$
Your proof is correct. 2 remarks: i) write $log(N)=cdots$ and ii) using Frobenius is useless because $u_i^Tv_j=v_j^Tu_i$.
$endgroup$
– loup blanc
Jan 20 at 3:35
$begingroup$
@hypernova: Does the summation have $n^2$ terms?
$endgroup$
– Saeed
Jan 20 at 4:27
$begingroup$
@hypernova: Does the summation have $n^2$ terms?
$endgroup$
– Saeed
Jan 20 at 4:27
$begingroup$
@loupblanc: Thank you for your advices. I modified my answer as per your suggestions, credited to you.
$endgroup$
– hypernova
Jan 20 at 5:11
$begingroup$
@loupblanc: Thank you for your advices. I modified my answer as per your suggestions, credited to you.
$endgroup$
– hypernova
Jan 20 at 5:11
$begingroup$
@Saeed: Yes, the summation contains $n^2$ terms, and each of the indices $i$ and $j$ runs from $1$ to $n$. In other words, $sum_{i,j=1}^n=sum_{i=1}^nsum_{j=1}^n$.
$endgroup$
– hypernova
Jan 20 at 5:11
$begingroup$
@Saeed: Yes, the summation contains $n^2$ terms, and each of the indices $i$ and $j$ runs from $1$ to $n$. In other words, $sum_{i,j=1}^n=sum_{i=1}^nsum_{j=1}^n$.
$endgroup$
– hypernova
Jan 20 at 5:11
add a comment |
$begingroup$
The first version of the formula with one index is false. Take $M=begin{pmatrix}50&37\37&61end{pmatrix},N=diag(2,1)$.
answered Jan 19 at 10:04
loup blancloup blanc
23.5k21851
$endgroup$
$begingroup$
It is on page 999 of the following paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. Maybe I am wrong. Because I am confused about $i$ and $j$.
$endgroup$
– Saeed
Jan 20 at 1:29
$begingroup$
Yes, the correct formula is a sum with $2$ indices.
$endgroup$
– loup blanc
Jan 20 at 3:38
add a comment |
$begingroup$
The first version of the formula with one index is false. Take $M=begin{pmatrix}50&37\37&61end{pmatrix},N=diag(2,1)$.
answered Jan 19 at 10:04
loup blancloup blanc
23.5k21851
$endgroup$
$begingroup$
It is on page 999 of the following paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. Maybe I am wrong. Because I am confused about $i$ and $j$.
$endgroup$
– Saeed
Jan 20 at 1:29
$begingroup$
Yes, the correct formula is a sum with $2$ indices.
$endgroup$
– loup blanc
Jan 20 at 3:38
add a comment |
$begingroup$
The first version of the formula with one index is false. Take $M=begin{pmatrix}50&37\37&61end{pmatrix},N=diag(2,1)$.
answered Jan 19 at 10:04
loup blancloup blanc
23.5k21851
$endgroup$
The first version of the formula with one index is false. Take $M=begin{pmatrix}50&37\37&61end{pmatrix},N=diag(2,1)$.
answered Jan 19 at 10:04
loup blancloup blanc
23.5k21851
edited Jan 20 at 3:41
answered Jan 19 at 10:04
loup blancloup blanc
23.5k21851
answered Jan 19 at 10:04
loup blancloup blanc
23.5k21851
answered Jan 19 at 10:04
loup blancloup blanc
23.5k21851
23.5k21851
$begingroup$
It is on page 999 of the following paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. Maybe I am wrong. Because I am confused about $i$ and $j$.
$endgroup$
– Saeed
Jan 20 at 1:29
$begingroup$
Yes, the correct formula is a sum with $2$ indices.
$endgroup$
– loup blanc
Jan 20 at 3:38
add a comment |
$begingroup$
It is on page 999 of the following paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. Maybe I am wrong. Because I am confused about $i$ and $j$.
$endgroup$
– Saeed
Jan 20 at 1:29
$begingroup$
Yes, the correct formula is a sum with $2$ indices.
$endgroup$
– loup blanc
Jan 20 at 3:38
$begingroup$
It is on page 999 of the following paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. Maybe I am wrong. Because I am confused about $i$ and $j$.
$endgroup$
– Saeed
Jan 20 at 1:29
$begingroup$
It is on page 999 of the following paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. Maybe I am wrong. Because I am confused about $i$ and $j$.
$endgroup$
– Saeed
Jan 20 at 1:29
$begingroup$
Yes, the correct formula is a sum with $2$ indices.
$endgroup$
– loup blanc
Jan 20 at 3:38
$begingroup$
Yes, the correct formula is a sum with $2$ indices.
$endgroup$
– loup blanc
Jan 20 at 3:38
add a comment |
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