Could we simplify the log determinant's concavity proof?











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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







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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















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







share|cite|improve this question






















  • 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













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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










1 Answer
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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 $Z^{1/2}$






share|cite|improve this answer





















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    1 Answer
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    active

    oldest

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    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 $Z^{1/2}$






    share|cite|improve this answer

























      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 $Z^{1/2}$






      share|cite|improve this answer























        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 $Z^{1/2}$






        share|cite|improve this answer












        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 $Z^{1/2}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 14 hours ago









        pl-94

        1184




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