Transformation linearity and eigenvalues Consider $E$ as a linear subspace in $\mathbb{R^n}$ with dimension $k$ and F as a linear subspace in $\mathbb{R^n}$ with dimension $n-k$, such that $E\cap F=\{0\}$
The transformation $p:\mathbb{R^n}\longrightarrow\mathbb{R^n}$ defined as followes:
for every $x\in\mathbb{R^n}$ consider the subspace $F^\prime$ that goes throgh $x$ and parallel to $F$ with the same dimension as $F$, since $E\cap F$ is a point, so $E\cap F^\prime$ is a point too, name this point $p(x)$
First I want to prove that this transformation is linear, so I should prove that for every $x\in\mathbb{R^n}$ we have:


*

*$p(x+x^\prime)=p(x)+p(x^\prime)$

*$p(cx)=cp(x)$
I know that $F^\prime$ is $F$ transferred by $x$, so these 2 conditions  seems true but I don't know how to express my prove mathematically.
Then I want to find a base for $\mathbb{R^n}$ like $\mathfrak{B}$ such that:
$M^{\mathfrak{B}}_{\mathfrak{B}^\prime}(p)=\begin{bmatrix}
I_k & O\\O&O
\end{bmatrix}$
I found this for $\mathfrak{B}$:
$\mathfrak{B}={\{e_1,e_2,\dots,e_k,f_{k+1},\dots,f_{n}\}}$
where $e_j$ are elements of $E$ and $f_j$ are elements of $F$
Is this correct? 
And how do I can find eigenvalues and characteristic polynomial of $p$?
 A: Let's work with the basis you suggested: Let $e_1,\ldots, e_k$ and $f_{k+1},\ldots, f_n$ be bases for $E$ and $F$ respectively. Then $\mathfrak B = \{e_1,\ldots, e_k, f_{k+1},\ldots, f_n\}$ is a basis for $\mathbb R^n$ (since $E$ and $F$ have trivial intersection). Let $x=(x_1,\ldots, x_n)$ in this basis, and let's compute $p(x)$. 
Edit: Remember that $p(x) \in E\cap F^\prime$. Because of the way we chose the basis, a vector in $E$ is of the form $(a_1,\ldots,a_k,0,\ldots,0)$ for some $a_i\in\mathbb R$. Similarly, a vector in $F$ is of the form $(0,\ldots,0,b_{k+1},\ldots,b_n)$ for some $b_i\in\mathbb R$, which means that a vector in $F' = x+F$ is of the form $x+(0,\ldots,0,b_{k+1},\ldots,b_n)$. Since $p(x)$ is both a vector in $E$ and in $F'$, we can now write $p(x)$ in two ways: 
$$
p(x) = \begin{pmatrix} a_1 \\ \vdots \\ a_k \\ 0 \\ \vdots \\ 0 \end{pmatrix}
\qquad {\rm and} \qquad
p(x) = x + \begin{pmatrix} 0 \\ \vdots \\ 0 \\ b_{k+1} \\ \vdots \\ b_n 
\end{pmatrix}
= \begin{pmatrix} x_1 \\ \vdots \\ x_k \\ x_{k+1}+b_{k+1} \\ \vdots \\ x_n+b_n 
\end{pmatrix}
$$
(Edit end)
We see that this forces $a_i=x_i$ for all $1 \le i \le k$, so that we have the unique solution:
$$
p(x) = \begin{pmatrix} x_1 \\ \vdots \\ x_k \\ 0 \\ \vdots \\ 0
 \end{pmatrix}
$$
We see that $p$ is the linear transformation given by the matrix you wrote. The characteristic polynomial is:
$$
{\rm det}(xI-M) = (x-1)^kx^{n-k}
$$
with eigenvalues 0 and 1 (assuming $0<k<n$). 
