- #1

mhsd91

- 23

- 4

Silly question perhaps, but here follows my problem. Per definition, the Matrix Exponent of the matrix [itex]A[/itex] is,

[itex]

e^{A} = I + A + \frac{A^2}{2} + \ldots = I + \sum_{k=1}^{\infty} \frac{A^k}{k!} = \sum_{k=0}^{\infty} \frac{A^k}{k!}

[/itex]

as [itex]e^0 = I[/itex]. I suspected that, since [itex]I^k = I[/itex] for any integer [itex]k[/itex], we would get

[itex]

e^{I} = I + I + \frac{I}{2} + \ldots = I \cdot \left( \sum_{k=0}^{\infty} \frac{1}{k!} \right) = I \cdot e,\quad e\approx 2.72

[/itex]

such that for an arbitrary constant [itex]a[/itex] we could write

[itex]

e^{aI} = I \left( \sum_{k=0}^{\infty} \frac{a^k}{k!} \right) = I e^{a}

[/itex]

However, apparently this is not the case as a (suggested) solution to some (homework) problem I've been working on claims that

[itex]

e^{aI} = e^{-a} I

[/itex]

With a sign change of a! I think I'm just missing something trivial and fundamental, but I'd really appreciate some help to sort this one out. Might it also be a misprint in the solution?