Find $\operatorname{Ker} p$ with $p = \frac{1}{2}(f + \operatorname{Id}_E)$ $E$ is a finite dimension vector space over $K$, and $f$ is an endomorphism of $E$ such that $f^2 = Id_E$


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*Show that the spectre of $f$ is either {1}, {-1} of {1, -1}.

*a) Show that $p = \frac{1}{2}(f + \operatorname{Id}_E)$ is a projector, and $\operatorname{Ker} p$ and $\operatorname{Im}p$
b) Deduce that $f$ diagonalisable.
My work:
1. I did that question
2. $p(p(x)) = p(x)$ mean that $p$ is a projector.
 please help me for the rest.
 A: Hints:
$$x\in\ker p\implies 0=p(x)=\frac12\left(f(x)+x\right)\implies f(x)=-x\\
x\in \text{Im}(p)\implies \exists y\in E\;\;s.t.\;\;x=p(y)=\frac12(f(y)+y)\implies f(y)=-y+2x$$
Do you know what generalized eigenvalues are? You can also use that $\;f\;$ is diagonalizable iff its minimal polynomial is a product of different linear factors...
A: First of all, I'm going to assume $\text{char} \; K \ne 2$, since I think it's clear from the framing of the question that that is what is intended.  It is the multiple occurrances of $\frac{1}{2}$ which lead me this conclusion.
This being the case, we have, using the notation $I = \text{Id}_E$:
Since $f^2 = I$, the eigenvalues of $f$ must lie in the set $\{ 1, -1 \}$, for if $f x = \lambda x$ with $x \ne 0$, then $f^2 x = \lambda^2 x$, $0 = (\lambda^2 - I)x$, whence $\lambda^2 - 1 = 0$, so that we must have $\lambda = \pm 1$.  Furthermore, any possible (non-empty) subset of $\{ 1, -1 \}$ may occur as the collection possible eigenvalues for an endomorphism satisfying $f^2 = I$; for example, consider the following three matrices $F_1, F_2, F_3 \in M_2(K)$:
$F_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \; F_2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \; F_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}; \tag{1}$
each of these matrices satisfies $F_i^2 = I$; as pointed out by our OP Mohemez, any non-void subset of $\{ 1, -1 \}$ does in fact occur as the spectrum of some $f$ satisfying $f^2 = I$.  Thus item (1) is resolved.
Now as for item (2):  with 
$p = \dfrac{1}{2}(I + f), \tag{2}$
we have 
$p^2 = \dfrac{1}{4}(I + 2f + f^2) = \dfrac{1}{4}(I + 2f + I) = \dfrac{1}{4}(2I + 2f) = \dfrac{1}{2}(I + f) = p; \tag{3}$
the formula $p^2 = p$ shows, by definiton, $p$ is a projection operator.  We next find  $\ker p$ and $\text{Im} \; p$, apparently what the question means when it says, "Show . . . and $\ker p$ and $\text{Im} \; p$"; I'm guessing here, but I think this is the most likely interpretation, given the overall context of the question.  If $v \in \ker p$, then
$\dfrac{1}{2}(f + I)(v) = p(v) = 0, \tag{4}$
or
$f(v) = -v; \tag{5}$
i.e. $v$ is in the $-1$-eigenspace of $f$.  Furthermore, it is easy to see that (5) implies
(4) as well, so that
$\ker p = \{ v \in E \mid f(v) = -v \} = E_{-1}, \tag{6}$
The simplest way I know to find $\text{Im} \; p$ is to use the well-known relation, true for any projector, i.e. any operator satisfying $p^2 = p$, 
$\ker p = \text{Im} \; (I - p). \tag{7}$
To see (7), let $x \in \ker p$; then $p(x) = 0$ so $(I - p)(x) = x - p(x) = x$, whence $x \in \text{Im} \; (I - p)$, so $\ker p \subset \text{Im} (I - p)$; likewise if $x \in \text{Im} (I - p)$, then $(I - p)(y) = x$ for some $y \in E$, thus $p(x) = p(I - p)(y) = (p - p^2)(y) = 0$ by virtue of $p = p^2$, showing $\text{Im} \; (I - p) \subset \ker p$ as well.  We apply (7) to the discovery of $\text{Im} \; p$ by observing that $p^2 = p$ implies $(I - p)^2 = I - p$:  $(I - p)^2 = I - 2p + p^2 = I - 2p + p = I - p$.  Thus $I - p$ is a projector, and by (7) we have
$\ker (I - p) = \text{Im} \; (I - (I - p)) = \text{Im} \; p. \tag{8}$
From (2) we see that
$I - p = I - \dfrac{1}{2}(I + f) = \dfrac{1}{2}(I - f), \tag{9}$
and thus $\ker (I - p) = \ker \dfrac{1}{2}(I - f)$.  If $v \in \ker (I - p)$, then
$\dfrac{1}{2}(I - f)(v) = 0, \tag{10}$
or 
$f(v) = v, \tag{11}$
so that $v$ is in the $1$-eigenspace of $f$; thus we see that
$\text{Im} \; p = \ker (I - p) = \ker \dfrac{1}{2}(I - f) = \{ v \in E \mid f(v) = v \} = E_1: \tag{12}$
we have thus found $\ker p$ and $\text{Im} \; p$ in terms of the eigenspaces of $f$.
With the above properties of $f$ and $p$ at our disposal, it is easy to see that $f$ may be diagonalized.  For any $x \in E$, note that 
$x = x - p(x) + p(x) = (I - p)(x) + p(x), \tag{13}$
and also that any $x \in  \text{Im} \; p \cap \text{Im} \; (I - p)$ must vanish, for if $x = p(y) = (I - p)(z)$ then (13) yields 
$x = p(I - p)(z) + (I - p)p(y) = 0, \tag{14}$
by virtue of $p^2 = p$.  Thus we see that $E$ may be expressed as the direct sum of the $+1$ and $-1$ eigenspaces of $f$, $E_1$ and $E_{-1}$.  The diagonalization of $f$ may now be completed by choosing bases $v_1,  v_2, . . . , v_m$ of $E_1$ and $w_1, w_2, . . . , w_n$ of $E_{-1}$; then a matrix $Q$ whose columns are the $v_j$, $w_k$, for instance
$Q = [v_1, v_2, . . . , v_m, w_1, w_2, . . . , w_n], \tag{15}$
diagonalizes $f$ via $Q^{-1}fQ$, since the columns of $Q$ are all eigenvectors of $f$.
Hope this helos.  Cheerio,
and as always,
Fiat Lux!!!
