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Exponential matrix

Let A \in \mathcal{M}_{n \times n} \left( \mathbb{C} \right). We define

    \[e^A = \sum_{m=0}^{\infty} \frac{A^m}{m!}\]

It is known that this series converges. Prove that

    \[\det (e^A) = e^{\Tr (A)}\]

Solution

We triangulise the matrix A , that is A= P^{-1} T P where P is an invertible matrix and T is an upper triangular. This is possible since our matrix is over \mathbb{C} and thus its characteristic polynomial splits. Let \lambda_1, \lambda_2, \dots, \lambda_n be its eigenvalues. Then we note that T^k is upper triangular with \lambda_1^k , \lambda_2^k , \dots, \lambda_n^k in its diagonal. Hence e^T is also upper triangular with e^{\lambda_1}, e^{\lambda_2} , \dots, e^{\lambda_n} in its diagonal. Hence

    \[\det e^T=e^{\lambda_1} e^{\lambda_2} \cdots e^{\lambda_n}=e^{\lambda_1+\lambda_2 +\cdots+\lambda_n}=e^{\Tr (T)}\]

However \Tr (A) = \Tr (T) and P^{-1} T^k P = A^k forall k. Thus P^{-1}e^TP=e^A and finally

    \[\det e^T=\det (P^{-1}e^TP)=\det e^A \]

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2 Comments

  1. Since the space E:=\mathcal M_n(\mathbb C) of all n \times n complex matrices is a finite-dimensional vector space, all norms define the same topology. So we can take a sub-multiplicative norm, that is, a norm \lVert\cdot\rVert such that \lVert AB\rVert\leq \lVert A\rVert \cdot\lVert B\rVert. (For example, we can take \lVert\cdot\rVert to be the operator norm on E.) As a finite dimensional vector space, E is complete, so it’s enough to show normal convergence. We have that, for each integer n \geq 0 ,

        \[0\leq ||A^{n}/n!||\leq \frac{\lVert A\rVert^n}{n!}\]

    and we know that, for each real number x , the series \displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!} converges (it defines the exponential function). Therefore, for any A \in E, the series \displaystyle \sum_{n=0}^{\infty}\frac{A^n}{n!} converges.

    (We also got the additional result that \displaystyle \lVert e^A\rVert\leq e^{\lVert A\rVert} , \;\; \forall A \in E . )

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