A neat identity for the determinant of a power of (1+A)

Here we prove a slight variation on the well known identity for a complex matrix A:

\displaystyle {\displaystyle \det (I+A) =\sum _{k=0}^{\infty } {\frac {1}{k!}}\left(-\sum _{j=1}^{\infty }{\frac {(-1)^{j}}{j}}{\rm tr} \left(A^{j}\right)\right)^{k} ~.} \ \ \ \ \ (1)

Here I is the identity matrix. We prove below the following for \alpha \in \Bbb R \setminus \{ 0\}:

\displaystyle {\displaystyle \det [\hspace{1mm}(I+A)^\alpha\hspace{1mm}] =\sum _{k=0}^{\infty } {\frac {1}{k!}}\left(-\alpha\sum _{j=1}^{\infty }{\frac {(-1)^{j}}{j}}{\rm tr} \left(A^{j}\right)\right)^{k} ~.} \ \ \ \ \ (2)

The above formulas are valid in an analytic sense whenever the sum on the right converges. But independently of convergence they are always true if interpreted in terms of formal power series.

First, recall that

\displaystyle \det (\exp(A)) = \exp({\rm tr} (A))~. \ \ \ \ \ (3)

It is actually quite easy to prove (3). We can proceed as follows in 3 steps: (i) The equation clearly holds for diagonal matrices; (ii) since the determinant and trace are invariant with respect to similarity transformations, the formula is also clearly true for all diagonalizable matrices; (iii) finally, the diagonalizable matrices are dense in the set of all square matrices and the determinant and trace are continuous functions, which means that the formula must hold true at all limit points, i.e. for all matrices.

If we replace {A} with {\log (I+A)^\alpha } we get

\displaystyle \det [\hspace{1mm}(I+A)^\alpha\hspace{1mm}] = \exp\big[{\rm tr} [\alpha \log (1+A)]\big]~. \ \ \ \ \ (4)

Recall the Mercator series

\displaystyle \log (1+ x) = \sum_{ j =1}^\infty (-1)^{ j -1} \frac {x^ j } j ~.\ \ \ \ \ (5)

Substituting we get

\displaystyle \det [\hspace{1mm}(I+A)^\alpha\hspace{1mm}] = \exp\bigg[\alpha~{\rm tr} \bigg( \sum_{ j =1}^\infty (-1)^{ j -1} \frac {A^ j } j \bigg) \bigg] = \exp\bigg( {-\alpha} \sum_{ j =1}^\infty \frac{(-1)^{ j }}{ j } {\rm tr}({A^ j })\bigg) ~.\ \ \ \ \ (6)

The claim (2) follows from expanding the exponential in its well-known Taylor series.

{\square}