Proof of log identity for positive definite matrices
$begingroup$
On Wikipedia, it is claimed that $rm{tr}(log(AB)) = rm{tr}log(A) + rm{tr}log(B)$.
The result is valid only if $A$ and $B$ are positive definite.
I use this result in entropy calculations but I do not know how to prove it - in particular to show that it only works for positive definite matrices. Could someone help me with the proof?
linear-algebra logarithms trace
asked Nov 25 '18 at 22:35
user1936752user1936752
5201513
$endgroup$
add a comment |
$begingroup$
On Wikipedia, it is claimed that $rm{tr}(log(AB)) = rm{tr}log(A) + rm{tr}log(B)$.
The result is valid only if $A$ and $B$ are positive definite.
I use this result in entropy calculations but I do not know how to prove it - in particular to show that it only works for positive definite matrices. Could someone help me with the proof?
linear-algebra logarithms trace
asked Nov 25 '18 at 22:35
user1936752user1936752
5201513
$endgroup$
1
$begingroup$
Assuming that convergence and many-valuedness are non-issues: If you apply the Baker-Campbell-Hausdorff formula to $X = log A$ and $Y = log B$, you get $logleft(ABright) = log A + log B + left(text{a sum of commutators}right)$. Now, take traces on both sides. That said, I don't know why we can assume that convergence and many-valuedness are non-issues.
$endgroup$
– darij grinberg
Nov 25 '18 at 22:50
add a comment |
$begingroup$
On Wikipedia, it is claimed that $rm{tr}(log(AB)) = rm{tr}log(A) + rm{tr}log(B)$.
The result is valid only if $A$ and $B$ are positive definite.
I use this result in entropy calculations but I do not know how to prove it - in particular to show that it only works for positive definite matrices. Could someone help me with the proof?
linear-algebra logarithms trace
asked Nov 25 '18 at 22:35
user1936752user1936752
5201513
$endgroup$
On Wikipedia, it is claimed that $rm{tr}(log(AB)) = rm{tr}log(A) + rm{tr}log(B)$.
The result is valid only if $A$ and $B$ are positive definite.
I use this result in entropy calculations but I do not know how to prove it - in particular to show that it only works for positive definite matrices. Could someone help me with the proof?
linear-algebra logarithms trace
linear-algebra logarithms trace
asked Nov 25 '18 at 22:35
user1936752user1936752
5201513
asked Nov 25 '18 at 22:35
user1936752user1936752
5201513
asked Nov 25 '18 at 22:35
user1936752user1936752
5201513
asked Nov 25 '18 at 22:35
user1936752user1936752
5201513
asked Nov 25 '18 at 22:35
user1936752user1936752
5201513
5201513
1
$begingroup$
Assuming that convergence and many-valuedness are non-issues: If you apply the Baker-Campbell-Hausdorff formula to $X = log A$ and $Y = log B$, you get $logleft(ABright) = log A + log B + left(text{a sum of commutators}right)$. Now, take traces on both sides. That said, I don't know why we can assume that convergence and many-valuedness are non-issues.
$endgroup$
– darij grinberg
Nov 25 '18 at 22:50
add a comment |
1
$begingroup$
Assuming that convergence and many-valuedness are non-issues: If you apply the Baker-Campbell-Hausdorff formula to $X = log A$ and $Y = log B$, you get $logleft(ABright) = log A + log B + left(text{a sum of commutators}right)$. Now, take traces on both sides. That said, I don't know why we can assume that convergence and many-valuedness are non-issues.
$endgroup$
– darij grinberg
Nov 25 '18 at 22:50
1
1
$begingroup$
Assuming that convergence and many-valuedness are non-issues: If you apply the Baker-Campbell-Hausdorff formula to $X = log A$ and $Y = log B$, you get $logleft(ABright) = log A + log B + left(text{a sum of commutators}right)$. Now, take traces on both sides. That said, I don't know why we can assume that convergence and many-valuedness are non-issues.
$endgroup$
– darij grinberg
Nov 25 '18 at 22:50
$begingroup$
Assuming that convergence and many-valuedness are non-issues: If you apply the Baker-Campbell-Hausdorff formula to $X = log A$ and $Y = log B$, you get $logleft(ABright) = log A + log B + left(text{a sum of commutators}right)$. Now, take traces on both sides. That said, I don't know why we can assume that convergence and many-valuedness are non-issues.
$endgroup$
– darij grinberg
Nov 25 '18 at 22:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The spectra of $A,B,AB$ are included in $(0,+infty)$. Then, it suffices to prove the equality of the exponentials of RHS and LHS. Since there are no $leq 0$ eigenvalues, we use the principal logarithm.
Let $U$ be symmetric $>0$.
$exp(tr(log(U))=det(exp(log(U)))=det(U)$.
We conclude with $det(AB)=det(A)det(B)$.
Remark. i) If ,$A,B$ are symmetric but with $<0$ eigenvalues, then we cannot use the principal logarithm.
ii) if $A,B$ are invertible but not symmetric and have no $<0$ eigenvalues, then $AB$ may have $<0$ eigenvalues
example $A=B=begin{pmatrix}0&1\-1&0end{pmatrix}$.
answered Nov 26 '18 at 10:57
loup blancloup blanc
22.7k21850
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add a comment |
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$begingroup$
The spectra of $A,B,AB$ are included in $(0,+infty)$. Then, it suffices to prove the equality of the exponentials of RHS and LHS. Since there are no $leq 0$ eigenvalues, we use the principal logarithm.
Let $U$ be symmetric $>0$.
$exp(tr(log(U))=det(exp(log(U)))=det(U)$.
We conclude with $det(AB)=det(A)det(B)$.
Remark. i) If ,$A,B$ are symmetric but with $<0$ eigenvalues, then we cannot use the principal logarithm.
ii) if $A,B$ are invertible but not symmetric and have no $<0$ eigenvalues, then $AB$ may have $<0$ eigenvalues
example $A=B=begin{pmatrix}0&1\-1&0end{pmatrix}$.
answered Nov 26 '18 at 10:57
loup blancloup blanc
22.7k21850
$endgroup$
add a comment |
$begingroup$
The spectra of $A,B,AB$ are included in $(0,+infty)$. Then, it suffices to prove the equality of the exponentials of RHS and LHS. Since there are no $leq 0$ eigenvalues, we use the principal logarithm.
Let $U$ be symmetric $>0$.
$exp(tr(log(U))=det(exp(log(U)))=det(U)$.
We conclude with $det(AB)=det(A)det(B)$.
Remark. i) If ,$A,B$ are symmetric but with $<0$ eigenvalues, then we cannot use the principal logarithm.
ii) if $A,B$ are invertible but not symmetric and have no $<0$ eigenvalues, then $AB$ may have $<0$ eigenvalues
example $A=B=begin{pmatrix}0&1\-1&0end{pmatrix}$.
answered Nov 26 '18 at 10:57
loup blancloup blanc
22.7k21850
$endgroup$
add a comment |
$begingroup$
The spectra of $A,B,AB$ are included in $(0,+infty)$. Then, it suffices to prove the equality of the exponentials of RHS and LHS. Since there are no $leq 0$ eigenvalues, we use the principal logarithm.
Let $U$ be symmetric $>0$.
$exp(tr(log(U))=det(exp(log(U)))=det(U)$.
We conclude with $det(AB)=det(A)det(B)$.
Remark. i) If ,$A,B$ are symmetric but with $<0$ eigenvalues, then we cannot use the principal logarithm.
ii) if $A,B$ are invertible but not symmetric and have no $<0$ eigenvalues, then $AB$ may have $<0$ eigenvalues
example $A=B=begin{pmatrix}0&1\-1&0end{pmatrix}$.
answered Nov 26 '18 at 10:57
loup blancloup blanc
22.7k21850
$endgroup$
The spectra of $A,B,AB$ are included in $(0,+infty)$. Then, it suffices to prove the equality of the exponentials of RHS and LHS. Since there are no $leq 0$ eigenvalues, we use the principal logarithm.
Let $U$ be symmetric $>0$.
$exp(tr(log(U))=det(exp(log(U)))=det(U)$.
We conclude with $det(AB)=det(A)det(B)$.
Remark. i) If ,$A,B$ are symmetric but with $<0$ eigenvalues, then we cannot use the principal logarithm.
ii) if $A,B$ are invertible but not symmetric and have no $<0$ eigenvalues, then $AB$ may have $<0$ eigenvalues
example $A=B=begin{pmatrix}0&1\-1&0end{pmatrix}$.
answered Nov 26 '18 at 10:57
loup blancloup blanc
22.7k21850
answered Nov 26 '18 at 10:57
loup blancloup blanc
22.7k21850
answered Nov 26 '18 at 10:57
loup blancloup blanc
22.7k21850
answered Nov 26 '18 at 10:57
loup blancloup blanc
22.7k21850
22.7k21850
add a comment |
add a comment |
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$begingroup$
Assuming that convergence and many-valuedness are non-issues: If you apply the Baker-Campbell-Hausdorff formula to $X = log A$ and $Y = log B$, you get $logleft(ABright) = log A + log B + left(text{a sum of commutators}right)$. Now, take traces on both sides. That said, I don't know why we can assume that convergence and many-valuedness are non-issues.
$endgroup$
– darij grinberg
Nov 25 '18 at 22:50