# Relationship between determinant and trace

1. Apr 30, 2010

### krishna mohan

Hi...

We have all seen the equation det(M)=exp(tr(lnM)). I was taught the proof using diagonalisation. I was wondering if there was a proof for non-diagonalisable matrices also.

2. Apr 30, 2010

### Fredrik

Staff Emeritus
Theorem 2.11, page 36. (And I don't think that logarithm is supposed to be there).

3. Apr 30, 2010

### krishna mohan

Thanks... .....the way I have written it, the logarithm is supposed to be there...

4. Apr 30, 2010

### Fredrik

Staff Emeritus
Ah, I see it now. The left-hand side in the book is det(exp(M)), not det(M).

5. Apr 30, 2010

### Simon_Tyler

The book by Hall (linked above) uses the decomposition into diagonalisable + nilpotent which is very important in Lie group theory. As slightly more direct approach is to use http://en.wikipedia.org/wiki/Jordan_normal_form" [Broken].

Schur decomposition: an arbitrary matrix M decomposes as QUQ-1 where U is upper-triangular and (therefore) has the eigenvalues of M on its diagonal.

det(exp(M)) = det(exp(QUQ-1)) = det(Q exp(U) Q-1) = det(exp(U)) = ∏i exp(λi) = exp(∑λi)

exp(tr(M)) = exp(tr(QMQ-1)) = exp(tr(MQ-1Q)) = exp(tr(M)) = exp(∑λi)

btw, in general it is best to not use the logarithm form - because not all matrices will possess a logarithm.

Last edited by a moderator: May 4, 2017