Show that $im(L_{I_n+A}) = E_1(A)$ Hi I'm having trouble with this homework question:
"Let $F$ be a field and let $A ∈ M_{n×n}(F)$.
Suppose that $F = Q, R$ or $C$ and that $A^2 = I_n$. 
$i)$ Show that if $λ$ is an eigenvalue of $A$ then
$λ = 1$ or $λ = −1$. 
$ii)$ Show that $ker(L_{I_n+A}) = E_{−1}(A)$ and that $im(L_{I_n+A}) = E_1(A)$ where $E_\lambda$$(A)$ denotes the eigenspace of $A$ with eigenvalue $\lambda$ and $L_{I_n+A}=A+I_n$"
I have managed to do part $i)$ and show that $ker(L_{I_n+A}) = E_{−1}(A)$ fairly easily but I'm having trouble with the last part where it asks you to show $im(L_{I_n+A}) = E_1(A)$. I used this as the definition of the image: $im(L_{A+I_n})=${$y$ : $L_{A+I_n}(x)=y$} $=$ {$y$ : $(A+I_n)(x)=y$}.
Any help with this would be very much appreciated.
 A: OK, here we go!
I'll discuss this problem in the more general context of fields $F$ such that $char F \ne 2$.  Certainly Q, R, C and any other field of characteristic zero all satisfy this criterion.
Since the underlying vector space $V$ upon which $A$ operates has not been explicitly specified in the question text, I will take $V = F^n$.
Part (i):  Let $\lambda \in F$ be an eigenvalue of $A$; that is, there exists a non-zero vector $v \in F^n$ such that
$Av = \lambda v; \tag{1}$
then, since $A^2 = I$, we have
$\lambda^2 v = \lambda(\lambda v) = \lambda (Av) = A(\lambda V) = A(Av) = A^2v = Iv = v, \tag{2}$
whence
$(\lambda^2 - 1)v = 0; \tag{3}$
since $v \ne 0$ we conclude that
$\lambda^2 - 1 = 0.  \tag{4}$
The roots of (4) are $\lambda = 1$ and $\lambda = -1$; since $char F \ne 2$ they are distinct (note that if $1 = -1$ in $F$, $2 = 1 + 1 = -1 + 1 = 0$ in $F$, contradicting the hypothesis $char F \ne 2$).  This shows every eigenvalue of $A$ is in the set $\{-1, 1 \}$.
The preceding remarks answer Part (i.).
We observe in passing that, provided $A \ne \pm I$, both cases $\lambda = \pm 1$ actually occur as eigenvalues of $A$.  For if, say, $1$ does not occur as an eigenvalue of $A$,  then
$Av \ne v = Iv \tag{5}$
for any non-zero $v \in F^n$.  Thus $v \ne 0$ implies
$(A - I)v \ne 0 \tag{6}$
as well.  Now for any $w \in F^n$ we see that
$(A - I)(A + I)w = (A^2 - I)w = 0, \tag{7}$
whence by (6) we find
$(A + I)w = 0  \tag{8}$
or
$Aw = -Iw  ;\tag{9}$
thus 
$A = -I; \tag{10}$
a parallel argument shows that if $-1$ does not occur as an eigenvalue, $A = I$.
Now, as for Part (ii.):
Suppose $v \in \ker (A + I)$; then 
$Av + v = Av + Iv =(A +  I)v = 0, \tag{11}$
whence
$Av = -v, \tag{12}$
that is, 
$v \in E_{-1}.  \tag{13}$
Steps (11)-(13) may be easily reversed to shown that (13) implies $v \in \ker (A + I)$.  Thus we see that $\ker (A + I) = E_{-1}$.
To see that $im (I + A) = E_1(A)$, suppose that $y \in im (I + A)$; then
$y = (I + A)x \tag{14}$
for some $x \in F^n$.  Then
$Ay = A(I + A)x = (A + A^2)x = (A + I)x = y, \tag{15}$
i.e., 
$y \in E_1(A). \tag{16}$
Finally, if (16) binds, so that
$Ay = y, \tag{17}$
we have
$(A + I)y = Ay + y = y + y = 2y, \tag{18}$
or, since $char F \ne 2$ so that $2^{-1}$ exists, we may write
$y = (2^{-1}2)y = 2^{-1}(2y) = 2^{-1}(I + A)y = (I + A)(2^{-1}y), \tag{19}$
showing
$y \in im (I + A). \tag{20}$
And we are done.
