The Matrix Exponent of the Identity Matrix, I

[mhsd91]

So, essentially, all I wonder is: What is the The Matrix Exponent of the Identity Matrix, $I$?

Silly question perhaps, but here follows my problem. Per definition, the Matrix Exponent of the matrix $A$ is,

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

as $e^0 = I$. I suspected that, since $I^k = I$ for any integer $k$, we would get

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

such that for an arbitrary constant $a$ we could write

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

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

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

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?

 

[Orodruin]

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

This is correct.

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

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

This is wrong.

 

[Mark44]

So, essentially, all I wonder is: What is the The Matrix Exponent of the Identity Matrix, $I$?

Silly question perhaps, but here follows my problem. Per definition, the Matrix Exponent of the matrix $A$ is,

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

as $e^0 = I$. I suspected that, since $I^k = I$ for any integer $k$, we would get

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

such that for an arbitrary constant $a$ we could write

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

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

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

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?

@mhsd91, when you post a question, please do not delete the three parts of the homework template. The template is required.