Let $V$ be a finite dimensional real vector space and let $A:V\to V$ be a linear map such that $A^2=A$ Let $V$ be a finite-dimensional real vector space and let $A:V\to V$ be a linear map such that $A^2=A$. Assume that $A$ not $0$ or $I$. Then show that

$1.$ $\ker(A)$ is not $\{0\}$.
$2.$ $V=\ker(A)\oplus R(A) $.
$3.$ The map $I+A$ invertible.

 A: $1.$
$$
A(I-A)x = (A-A^2)x = (A-A)x = 0 = (A-A^2)x = (I-A)Ax.
$$
Since $A\ne I$, there is some $x\ne0$ for which $Ax\ne x$. Thus $(I-A)x\ne 0$.  Hence $(I-A)x\ne0$ must be in the kernel of $A$.
$2.$
$$
x = Ax + (I-A)x = \text{something in $R(A)$} + \text{something just shown to be in $\ker(A)$},
$$
so $V=R(A)+\ker(A)$.
$3.$
Suppose $(I+A)x= 0$.  Then $Ax=-x$, so $A^2 x= -Ax$, so $Ax=-Ax$.  Thus $Ax=0$. So $0=(I+A)x = x+Ax = x+0$, and consequently $x=0$.  That means $0$ is the only member of the kernel of $I+A$, so $I+A$ is invertible.
A: For 1:
Let $v$ be such that $Tv\neq v$ (exists since $T\neq I$) then $T(Tv-v)=0$ and $0 \neq (Tv-v) \in Ker(T)$
For 2:
$v = (v-Tv)+Tv \in Ker(T)+Im(T)$
And if $v\in KerT,ImT$ then $Tv=0$ and $v=Tu$ applying $T$ and we get $0=v=Tu$.
For 3:
Assume $(T+I)v=0$ then applying $T$ we get $Tv = 0$ which means that $v = 0$ and $(T+I)$ has null kernel and is therefore invertible.
A: We know that the polynomial $x(x-1)$ annihilates $A$. So the minimal polynomial of $A$ is a monic polynomial which divides $x(x-1)$, so it must be a product of distinct linear factors, which implies that $A$ is diagonalizable.
Let $n$ be the dimension of $V$. Since $A$ is not the null matrix nor the identity, $A$ has $0$ and $1$ as characteristic values. Let $\alpha_1,\ldots,\alpha_n$ a basis such that $\alpha_1,\ldots,\alpha_k$ are proper vectors associated to $0$ and $\alpha_{k+1},\ldots,\alpha_n$ are proper vectors associated to $1$. Such a basis exists because as we saw, $A$ is diagonalizable. Then for any $\alpha\in V$ and some scalars $x_j$ we have
\begin{align*}
\alpha &= \sum_{j=1}^k x_j\alpha_j + \sum_{k+1}^n x_j\alpha_j\\
       &= \sum_{j=1}^k x_j\alpha_j + \sum_{k+1}^n x_jA \alpha_j\\
       &= \sum_{j=1}^k x_j\alpha_j + A\left(\sum_{k+1}^n x_j\alpha_j\right),
\end{align*}
that is, $V = \ker A + A(V)$. Now using that
$$
\dim V = \dim\ker A + \dim A(V)$$ and
$$\dim V = \dim (\ker A + A(V)) = \dim\ker A + \dim A(V) - \dim \ker A\cap A(V)
$$
we see that $\ker A\cap A(V)=\{0\}$ so $V$ is in fact the desired direct sum.
Finally to see that $I+A$ is invertible, notice that the matrix associated with $I+A$ in the above basis is a diagonal matrix without $0$s in its diagonal.
