Is there any complex number $a$ solving $\exp (X)=1+aX$ for given square matrix $X$? I am doing exercises in Rossmann's book on Lie groups. Exercise 1.2.12 goes like this:
$X \in M_n (\mathbb{C})$, $L$ is a subspace of $M_n (\mathbb{C})$, s.t. $[X, Y]\in L$ for $Y\in L$. Prove that $\exp (-X)\exp (X+Y)\in 1+L $.
There is a hint that we consider derivative of $\exp (-X)\exp (X+tY)$. That is 
$$\frac{I-e^{-\mathrm{ad}X}}{\mathrm{ad}X}Y,$$
which 'seems' to sit in $L$.
However, if you let $L=\mathbb{C} X$, one claims that it contradicts for $\exp (X)$ may not be expressed as $1+aX$.
My question is whether the claim is true or not. Any help is appreciated.
 A: I would like to note that you incorrectly used Theorem 5 in the book.  I restate it here (with minor changes to make things clearer and less confusing).

Theorem 5 [section 1.2, page 15, in Lie Groups: An Introduction Through Linear Groups by Wulf Rossman].  Let $Z:\Bbb{R}\to M_n(\Bbb{C})$ be a differentiable matrix function.  Then,
  $$\frac{d}{d\tau}e^{Z(\tau)}=e^{Z(\tau)}\left(\frac{1-e^{-\operatorname{ad}Z(\tau)}}{\operatorname{ad}Z(\tau)}\right)\frac{d}{d\tau}Z(\tau).$$
  Here, $$\left(\frac{1-e^{-\operatorname{ad}Z(\tau)}}{\operatorname{ad}Z(\tau)}\right)=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)!}\big(\operatorname{ad}Z(\tau)\big)^k.$$

Let's look at your counter-example.  I will illustrate here how to correctly apply the theorem above.  For $L=\Bbb{C}X$, we set, WLOG, $Y=X$.  So, 
\begin{align}\frac{d}{dt}e^{-X}e^{X+tY}&=e^{-X}e^{X+tY}\left(\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)!}\big(\operatorname{ad}X+t\operatorname{ad}Y\big)^k\right)Y\\&=e^{-X}e^{(1+t)X}\left(\sum_{k=0}^\infty\frac{(-1)^k(1+t)^k}{(k+1)!}(\operatorname{ad}X)^k\right)X
\\&=e^{-X}e^{(1+t)X}X=Xe^{tX}.\end{align}
That is, for $L=\Bbb C X$ and $Y=X$,
$$e^{-X}e^{X+tY}=e^{tX}.$$
In particular, $e^{-X}e^{X+Y}=e^X=e^Y$ when $Y=X$.  This is not in the form $I+aX$ in general just as Kavi Rama Murthy's comment shows, unless we have a stronger condition on $L$ such as $L$ is closed under left multiplication by any element of $\tilde{L}=L+\mathbb{C}X$.
In the section below, I verify that the assumption that $L$ is a subspace of $M_n(\mathbb{C})$ with the properties that $[X,T]\in L$ for all $T\in L$ and that $L$ is closed under left multiplication by $X$ and elements of $L$ itself, then the claim holds.  Without this extra assumption, we already see that the problem is wrong.


Modified Problem Statement.  Fix $X\in M_n(\mathbb{C})$.  Let $L$ be a subspace of $M_n(\mathbb{C})$ with the properties that $[X,T]\in L$ for all $T\in L$, and that $L$ is closed under left multiplication by any element of $\tilde{L}=L+\mathbb{C}X$, then $$e^{-X}e^{X+Y}\in I+L$$ for all $Y\in L$.

From the theorem, you should take $Z(t)=X+tY$.  That is,
$$\frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}\frac{d}{dt}e^{Z(t)}=e^{-X}\Biggl(e^{Z(t)}\left(\frac{1-e^{-\operatorname{ad}Z(t)}}{\operatorname{ad}Z(t)}\right)\frac{d}{dt}Z(t)\Biggr).$$
Since $\frac{d}{dt}Z(t)=Y$, we get
$$\frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}\left(\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)!}\big(\operatorname{ad}X+t\operatorname{ad}Y\big)^k\right)Y.$$
Because $\operatorname{ad}X(Y)=[X,Y]\in L$ and $\operatorname{ad}Y(Y)=[Y,Y]=0\in L$, we obtain $$\big(\operatorname{ad}X+t\operatorname{ad}Y\big)^kY\in L.$$  This proves that
$$\left(\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)!}\big(\operatorname{ad}X+t\operatorname{ad}Y\big)^k\right)Y \in L\,.$$  Now, because $e^{X+tY}T\in L$ for all $T\in L$ because $L$ is closed under left multiplication by any element of $\tilde{L}=L+\mathbb{C}X$, we see also that $e^{-X}e^{X+tY}T\in L$ as well.  This means
$$\frac{d}{dt}e^{-X}e^{X+tY}=e^{-X}e^{X+tY}\left(\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)!}\big(\operatorname{ad}X+t\operatorname{ad}Y\big)^k\right)Y\in L.$$
This proves that
$$e^{-X}e^{X+\tau Y}-I=\int_0^\tau\frac{d}{dt}e^{-X}e^{X+tY}\ dt \in L,$$
and the conclusion follows.
