Mixing
Could we simplify the log determinant's concavity proof?
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1
down vote
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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
1184
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
1184
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
1184
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
1184
asked Oct 10 at 16:34
pl-94
1184
asked Oct 10 at 16:34
pl-94
1184
asked Oct 10 at 16:34
pl-94
1184
asked Oct 10 at 16:34
pl-94
1184
1184
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
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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 14 hours ago
pl-94
1184
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 14 hours ago
pl-94
1184
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 14 hours ago
pl-94
1184
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 14 hours ago
pl-94
1184
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 14 hours ago
pl-94
1184
answered 14 hours ago
pl-94
1184
answered 14 hours ago
pl-94
1184
answered 14 hours ago
pl-94
1184
1184
add a comment |
add a comment |
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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