Relationship between determinant and trace

[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.

 

[Fredrik]

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

 

[krishna mohan]

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

 

[Fredrik]

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

 

[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" .

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(λ_i) = exp(∑λ_i)

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

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