@dlfivefifty, you posted a link above with one (numerical) answer to your own question:
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Not a very convincing answer⌠even the complex explicit example is not convincing as thereâs no proofâŚanyways itâs a silly question 
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Just to be fair, you are asking for â so @Mason computation is a valid âproofâ 
My original question definitely said ==
My fault, I read the tittle not the original question 
For any projection matrix P, P^2=P and e^P=I+(e-1)P. If P=\begin{bmatrix}x_1&y_1\\y_2&x_2\end{bmatrix}, then the first equation implies
\begin{align*} x_i^2+y_1y_2&=x_i\\ y_i(x_1+x_2)&=y_i \end{align*}
and the second implies
\begin{align*} 1+(e-1)x_i&=e^{x_i}\\ (e-1)y_i&=e^{y_i} \end{align*}
Solving these eight equations over the complex numbers is left as an exercise for the reader . Iâd imagine Mathematica could do it with enough computation time.
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With some effort and interval arithmetic you can convert it into a proper proof (you just tweak the proof of convergence of Newton : if you are at a point x where F is small enough, Fâ is inversible and Fââ is bounded in a ball around x, then there is a solution close to x). Agreed itâs a silly question though 
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This vaguely reminds me of a fun math puzzle. Find a function f from R to R such that f^-1(x)=f'(x). (For bonus points, such that f^-1(x)=d^n*f/dx^n(x) ie the nth derivative is the inverse).
Not sure if you are talking about the following
Is there a one-to one and onto function f:(0, \infty) \longrightarrow(0, \infty) such that f^{\prime}=f^{-1}?
The answer is yes, and involves golden ratio (at least the one I found)âŚ
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Itâs not terribly difficult to prove. If A is a matrix of the form \begin{pmatrix} a & b & b \\ b & a & b \\ b & b & a \end{pmatrix} it can be diagonalized (eigenvalues are a - b, a - b, a + 2b) and itâs straightforward to compute e^A in closed form as a matrix of the same form as A with
\begin{align}
a' &= \frac{2e^{a-b} + e^{a+2b}}{3} \\
b' &= \frac{e^{a+2b}-e^{a-b}}{3}
\end{align}
To have exp(A) = exp.(A) we need a' = e^a and b' = e^b which is also straightforward to solve to
\begin{align}
e^a &= -\frac{4}{3} \\
e^b &= -2
\end{align}
(unique up to that point) and the complex solution follows.
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