Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working through a derivation in someone's thesis at the moment to understand an important result, but I am more than a bit rusty on matrices. Could anyone give me some tips on these identities? They are stated without proof and I'm having a hard time finding a derivation online.

Below, X is a matrix and E is a scalar, and X is a function of E.

1) $Tr(X' X^{-1}) = \frac{d}{dE} Tr(ln(X))$

When I first saw this I thought it would be the same as treating X as a scalar, then by the definition of the ln function the above would be true. Is the fact that there is a trace and that X is a matrix important in the derivation?

He does mention that "" $Tr(log X)' = Tr[X' X] = Tr[X X']$ "" but I think this was probably a typo, since an expression of the form $Tr[X' X]$ does not appear in his calculation.

2) $Tr(ln(X)) = ln(det(X))$

This one I am a bit stuck on, I would guess that it has something to do with the definitions of the trace and determinant but not sure where to go from there. I haven't done anything with matrices in about 3 years, and I'm a physicist, so keep it basic :)

EDIT OK here is my working for proof 1 using Robert's guide below:

$$ X' = \frac{d}{dE} \sum_{n=0}^{\infty} \frac{L^n}{n!} $$ Using the chain rule, $$ X' = \sum_{n=0}^{\infty} \frac{n L' L^{n-1}}{n!} = \sum_{n=0}^{\infty} \frac{L' L^{n-1}}{(n-1)!} = \sum_{n=1}^{\infty} \frac{L' L^{n-1}}{n!}$$ Here is the bit I don't quite follow, regarding the introduction of the dummy j which seems to cancel later on in the calculation without being used. $$ \sum_{j=0}^{n-1} \sum_{n=1}^{\infty} \frac{L' L^j L^{n-1 - j}}{n!} $$ Now using this expression for X': $$ X' X^{-1} = X' e^{-L} = \sum_{j=0}^{n-1} \sum_{n=1}^{\infty} \frac{L' L^j L^{n-1 - j}}{n!} e^{-L} $$ Here is where I find a problem now, since $$ \sum_{j=0}^{n-1} (\sum_{n=1}^{\infty} \frac{L^{n-1}}{n!}) L' e^{-L} $$ Now my part in the brackets in that last expression isn't $e^L$ so doesn't cancel nicely. I am pretty sure I am missing something with commutativity and when you introduced the sum over j!? EDIT 2 Just realised my last step on the chain rule, changing the sum from n=0 to n=1 doesn't make much sense.

share|cite|improve this question
The first error is that if $L'$ and $L$ don't commute, $\dfrac{d}{dE} L^n$ is not $n L' L^{n-1}$. You have to use Leibniz's rule in the form $(AB)' = A'B + AB'$. Thus $(L^2)' = L'L + LL'$, which is not the same as $2L'L$. – Robert Israel Jun 6 '12 at 20:34
I tried doing $$\frac{d}{dE} L^n = \frac{d}{dE} (L L^{n-1})$$ which gave $$ \sum_{n=0}^{\infty} \frac{L' L^{n-1} + (n-1)L L^{n-2}}{n!}$$, I'm not really sure where I would go from that point though – Josh Jun 6 '12 at 20:47
So $(ABC)' = A'BC + A(BC)' = A'BC + AB'C + ABC'$, and $\dfrac{d}{dE} L^n = L' L^{n-1} + L L' L^{n-2} + \ldots + L^{n-1} L' = \sum_{j=0}^{n-1} L^j L' L^{n-1-j}$ – Robert Israel Jun 6 '12 at 23:45
up vote 3 down vote accepted

1) is a bit tricky if $X'$ and $X$ don't commute. $\log(X)$ is a matrix $L$ such that $\exp(L) = X$. Now $$\dfrac{d}{dE} X = \dfrac{d}{dE} \sum_{n=0}^\infty \dfrac{L^n}{n!} = \sum_{n=1}^\infty \sum_{j=0}^{n-1} \dfrac{L^j L' L^{n-1-j}}{n!} $$ and so $$ X' X^{-1} = X' \exp(-L) = \sum_{n=1}^\infty \sum_{j=0}^{n-1}\frac{L^j L' L^{n-1-j}}{n!} \exp(-L)$$ but since $\text{Tr}(AB) = \text{Tr}(BA)$ and $L$ commutes with $\exp(-L)$, $$ \text{Tr}(X' X^{-1}) = \sum_{n=1}^\infty \sum_{j=0}^{n-1} \dfrac{\text{Tr}\left(L^j L' L^{n-1-j} \exp(-L) \right)}{n!} = \sum_{n=1}^\infty \dfrac{\text{Tr}\left( L'L^{n-1} \exp(-L)\right)}{(n-1)!} = \text{Tr}(L' \exp(L) \exp(-L)) = \text{Tr}(L')$$

2) If $X$ has eigenvalues $\lambda_j$ (counted by algebraic multiplicity), $\log(X)$ has eigenvalues $\log(\lambda_j)$. Then $\text{Tr}(\log(X)) = \sum_j \log(\lambda_j) = \log(\prod_j \lambda_j) = \log(\det(X))$. Actually there is a question of which branches of the logarithm to use when there are non-positive eigenvalues, so it is more accurate to say that $\text{Tr}(\log(X))$ is one of the branches of $\log(\det(X))$.

share|cite|improve this answer
Thanks, should have mentioned that X is a square matrix which as I understand your answer is only valid for? – Josh Jun 6 '12 at 19:06
Yes, there's no way to define a logarithm or exponential of a non-square matrix. – Robert Israel Jun 6 '12 at 19:14
In 1) what is the point of introducing the $L^{j} L^{-j}$ part, the way I have worked it through it seems to cancel later without being used at all? Also I don't quite get how the $n!$ in the denominator near the end changes back to $n-1!$ on the next step. I will edit in my working at the top in a moment. – Josh Jun 6 '12 at 20:01
See my comment on your attempted proof above. $n/(n!) = 1/(n-1)!$. – Robert Israel Jun 6 '12 at 20:36
On the last part of your calculation, do I understand right that due to the cyclical property of the Trace (and the commutation of L with $e^{-L}$) you can cancel out the j entirely, however this would leave me with $$\sum_{n=1}^{\infty} \frac{Tr(L' L^{n-1} e^{-L})}{n!}$$, so I'm not sure where the n in the numerator would come from to do $n/{n!} = 1/(n-1)!$ as you suggest – Josh Jun 6 '12 at 21:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.