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Let $X$ be an n×n real or complex matrix. The exponential of $X$, denoted by $e^X$, is the n×n matrix given by the power series $e^X =\sum_{k=0}^{\infty} X^k/k!$ where $X^{0}$ is defined to be the identity matrix $I$ with the same dimensions as $X$.

If $X$ is zero matrix then $e^X$=$I_{n \times n}$ or $0_{n \times n}$?

ramanujan
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2 Answers2

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It is the identity matrix. The zero matrix to the zeroth power is the identity matrix, $0_{n,n}^0 = I_{n,n}$, for the same reason that $0^0 = 1$ in the numbers. And all the higher terms are zero since $0_{n,n}^k = 0_{n,n}$ for all $k \ge 1$. In symbols: $$e^{0_{n,n}} = 0_{n,n}^0 + 0_{n,n}^1 + \frac{1}{2} 0_{n,n}^2 + \dots = I_{n,n}.$$

Najib Idrissi
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  • $0^0$ doesn't make sense, neither for real numbers nor for matrices. The sequence (with appropriate dimensions) is $e^0 = I + 0^1 + 0^2/2 + \cdots = I$. – Arthur Jun 20 '16 at 11:31
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    @Arthur $0^0$ definitely makes sense and is equal to $1$ in algebra. http://math.stackexchange.com/q/11150/10014 – Najib Idrissi Jun 20 '16 at 11:33
  • @Arthur I suggest that you reread my comment more carefully. It says in algebra. In algebra, there is no question as to what $0^0$ is: it's $1$. The only context where there can be a problem is analysis as $\lim_{(x,y) \to (0,0)} x^y$ does not exist. That's it. – Najib Idrissi Jun 20 '16 at 11:35
  • But this is analysis, and not algebra (or maybe it's a mix of the two). Otherwise, how would you make sense of the existence of infinite, converging sums? – Arthur Jun 20 '16 at 11:39
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    There is a little bit of analysis involved (it is sometimes possible to make sense of infinite sums in algebra, look up projective limits for example, but this is not the case here), but the main point is that the exponents are integers here. In a power series, $x^0$ is always $1$. There is no question about it. Here is another link if you want: https://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_power_of_zero – Najib Idrissi Jun 20 '16 at 11:43
  • In fact you can look at things more-or-less algebraically. $M_n(k)$ is a Lie algebra, look at its universal enveloping algebra and complete it at the augmentation ideal to get $\hat{U}(M_n(k))$, a complete Hopf algebra, and then take grouplike element to get $G = \mathbb{G}(\hat{U}(M_n(k)))$. There is then a well-defined exponential map $\exp : M_n(k) \to G$ as soon as $k$ has characteristic zero... And when $k = \mathbb{R}$, you have $G \cong GL_n(\mathbb{R})$. – Najib Idrissi Jun 20 '16 at 11:48
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$e^X=I_n$ by definition if $X=0_n$. See the wiki https://en.m.wikipedia.org/wiki/Matrix_exponential