Determinant of matrix exponential of $A$

Question

How to find a formula for $\det \left( e^A \right)$ if $\displaystyle\lim_{n \to \infty} \left( I + \frac {A}{n} \right)^n = e^A$?

Answer 1

$e^{A} = T \cdot \Lambda \cdot T^{-1}$, where $\Lambda = \begin{pmatrix} e^\lambda_{1} & 0 & \dots & 0 \\ 0 & e^\lambda_{2} & \dots & 0 \\ \dots & \dots & \ddots & \dots \\ 0 & 0 & \dots & e^\lambda_{n} \end{pmatrix}$ and $\lambda_{i}$ are eigenvalues of matrix A. So $Det(e^A) = Det(\Lambda) = \prod\limits_{i=1}^n e^\lambda_{i}$

Answer 2

Since the determinant is continous we have $det(e^A)=det( lim_{n \rightarrow \infty} (I+A/n)^n)=lim_{n \rightarrow \infty} (det(I+A/n)^n)$.

Let $A$ have eigenvalues $\lambda_1 ,..., \lambda_r$ and assume $A$ is Jordan normal form (you can reduce to this case when the field is algebraically closed).

Then $lim_{n \rightarrow \infty} (det(I+A/n)^n)=lim_{n \rightarrow \infty} (det(B_n)^n)=lim_{n \rightarrow \infty} (1+\frac{\lambda_1}{n})^n...(1+\frac{\lambda_r}{n})^n=e^{\lambda_1}...e^{\lambda_r}= e^{\lambda_1+...+\lambda_r}.$ where $B_n=(I+A/n)$ is an upper triangular matrix with entries $1+\frac{\lambda_i}{n}$ on the diagonal.

Answer 3

Let $A$ be diagonalizable by $P$ and Let the diagonal matrix be $D(\lambda_1,\lambda_2,....,\lambda_n)$ and $e^D=D(e^{\lambda_1},e^{\lambda_2},...e^{\lambda_n}).$ $$P^{-1} A P= D \implies A- P D P^{-1} \implies f(A)= P f(D) P^{-1} \implies e^A= P e^D P^{-1}.$$ Finally, we have $$\det(e^A)=\det(e^D)=e^{\sum_{k=0}^n \lambda_k}$$ But $$\det(e^A) \ne e^{\det(A)},$$ in general.

Because $e^A=I+A+A^2/2!+A^3/3!+...$ and $\det(A+B)\ne \det(A)+ \det(B)$ in general.

Answer 4

That's a well l known thing. As others have noted: If the matrix is complex, it holds that $$ \det(\exp(A))=\exp(trace(A)). $$