Could we simplify the log determinant's concavity proof?
up vote
1
down vote
favorite
The function $f(X) Rightarrow log det X$ is concave as shown here.
However, I was wondering if we could simplify the proof suggested.
When we compute : $g(t) = logdet(Z + tV)$, why not just saying that:
$$begin{aligned}
g(t) &=
logdet(Z)(I+tZ^{-1}V) \
&= sum_i log(1+tlambda_i). + logdet Z
end{aligned}$$
with $(lambda_i)$ the eigenvalues of $Z^{-1}V$?
matrices complex-analysis determinant
asked Oct 10 at 16:34
pl-94
1274
add a comment |
up vote
1
down vote
favorite
The function $f(X) Rightarrow log det X$ is concave as shown here.
However, I was wondering if we could simplify the proof suggested.
When we compute : $g(t) = logdet(Z + tV)$, why not just saying that:
$$begin{aligned}
g(t) &=
logdet(Z)(I+tZ^{-1}V) \
&= sum_i log(1+tlambda_i). + logdet Z
end{aligned}$$
with $(lambda_i)$ the eigenvalues of $Z^{-1}V$?
matrices complex-analysis determinant
asked Oct 10 at 16:34
pl-94
1274
Are your matrices positive semidefinite or something? And yes, I agree that the introduction of $Z^{1/2}$ is unneeded.
– darij grinberg
Oct 10 at 16:37
OK thanks. I think that the problem is not defined if the matrices are not positive definite.
– pl-94
Oct 10 at 16:48
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The function $f(X) Rightarrow log det X$ is concave as shown here.
However, I was wondering if we could simplify the proof suggested.
When we compute : $g(t) = logdet(Z + tV)$, why not just saying that:
$$begin{aligned}
g(t) &=
logdet(Z)(I+tZ^{-1}V) \
&= sum_i log(1+tlambda_i). + logdet Z
end{aligned}$$
with $(lambda_i)$ the eigenvalues of $Z^{-1}V$?
matrices complex-analysis determinant
asked Oct 10 at 16:34
pl-94
1274
The function $f(X) Rightarrow log det X$ is concave as shown here.
However, I was wondering if we could simplify the proof suggested.
When we compute : $g(t) = logdet(Z + tV)$, why not just saying that:
$$begin{aligned}
g(t) &=
logdet(Z)(I+tZ^{-1}V) \
&= sum_i log(1+tlambda_i). + logdet Z
end{aligned}$$
with $(lambda_i)$ the eigenvalues of $Z^{-1}V$?
matrices complex-analysis determinant
matrices complex-analysis determinant
asked Oct 10 at 16:34
pl-94
1274
asked Oct 10 at 16:34
pl-94
1274
asked Oct 10 at 16:34
pl-94
1274
asked Oct 10 at 16:34
pl-94
1274
asked Oct 10 at 16:34
pl-94
1274
1274
Are your matrices positive semidefinite or something? And yes, I agree that the introduction of $Z^{1/2}$ is unneeded.
– darij grinberg
Oct 10 at 16:37
OK thanks. I think that the problem is not defined if the matrices are not positive definite.
– pl-94
Oct 10 at 16:48
add a comment |
Are your matrices positive semidefinite or something? And yes, I agree that the introduction of $Z^{1/2}$ is unneeded.
– darij grinberg
Oct 10 at 16:37
OK thanks. I think that the problem is not defined if the matrices are not positive definite.
– pl-94
Oct 10 at 16:48
Are your matrices positive semidefinite or something? And yes, I agree that the introduction of $Z^{1/2}$ is unneeded.
– darij grinberg
Oct 10 at 16:37
Are your matrices positive semidefinite or something? And yes, I agree that the introduction of $Z^{1/2}$ is unneeded.
– darij grinberg
Oct 10 at 16:37
OK thanks. I think that the problem is not defined if the matrices are not positive definite.
– pl-94
Oct 10 at 16:48
OK thanks. I think that the problem is not defined if the matrices are not positive definite.
– pl-94
Oct 10 at 16:48
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Actually, we can not simplify the proof. Indeed, $Z^{-1}V$ is not necessarily symmetric, and thus, the decomposition to the eigenvalues is not guaranteed.
This explains the use of the trick with $Z^{1/2}$
answered Nov 19 at 10:06
pl-94
1274
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Actually, we can not simplify the proof. Indeed, $Z^{-1}V$ is not necessarily symmetric, and thus, the decomposition to the eigenvalues is not guaranteed.
This explains the use of the trick with $Z^{1/2}$
answered Nov 19 at 10:06
pl-94
1274
add a comment |
up vote
0
down vote
accepted
Actually, we can not simplify the proof. Indeed, $Z^{-1}V$ is not necessarily symmetric, and thus, the decomposition to the eigenvalues is not guaranteed.
This explains the use of the trick with $Z^{1/2}$
answered Nov 19 at 10:06
pl-94
1274
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Actually, we can not simplify the proof. Indeed, $Z^{-1}V$ is not necessarily symmetric, and thus, the decomposition to the eigenvalues is not guaranteed.
This explains the use of the trick with $Z^{1/2}$
answered Nov 19 at 10:06
pl-94
1274
Actually, we can not simplify the proof. Indeed, $Z^{-1}V$ is not necessarily symmetric, and thus, the decomposition to the eigenvalues is not guaranteed.
This explains the use of the trick with $Z^{1/2}$
answered Nov 19 at 10:06
pl-94
1274
answered Nov 19 at 10:06
pl-94
1274
answered Nov 19 at 10:06
pl-94
1274
answered Nov 19 at 10:06
pl-94
1274
1274
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2950262%2fcould-we-simplify-the-log-determinants-concavity-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Are your matrices positive semidefinite or something? And yes, I agree that the introduction of $Z^{1/2}$ is unneeded.
– darij grinberg
Oct 10 at 16:37
OK thanks. I think that the problem is not defined if the matrices are not positive definite.
– pl-94
Oct 10 at 16:48