Determine $(n-1)$ eigenvectors of $f$ associated to $0$? 
$$A=\begin{pmatrix}1&1&\cdots&1\\1&1&\cdots&1\\\vdots&\vdots&\ddots&1\\1&1&\cdots&1\end{pmatrix}\in\mathcal M_n(\mathbb R).$$

*

*Prove that $0$ and $n$ are eigenvalue of $A$.

*Determine the characteristic polynomial. (Hint: You can use the basis $(e_1, e_2 - e_1, \cdots, e_n - e_1)$)

*Determine the eigenvalues of $A$.

*Prove that $f(e_1)$ is an eigenvector.

*Determine $(n-1)$ eigenvectors of $f$ associated to $0$.

*Prove that there exists a basis $\mathcal{B_2}$ in which the matrix is of the form:
$$D = \begin{pmatrix}
n & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0
\end{pmatrix}$$



*

*Taking the vector $ v = 
\begin{pmatrix}
1 \\
1 \\
\vdots \\
1 
\end{pmatrix}$
We get: $Av = nv$.
But I could not see how $0$ can be an eigenvalue.


*$$P_A(X) = \begin{vmatrix}
1 - X & 1 & \cdots & 1 \\
1 & 1 - X & \cdots & 1 \\ 
\vdots & \vdots & \ddots & 1 \\
1 & 1 & \cdots & 1 - X
\end{vmatrix}
=
(1 - X)\begin{vmatrix}
1  & 1 & \cdots & 1 \\
1 & 1 - X & \cdots & 1 \\ 
\vdots & \vdots & \ddots & 1 \\
1 & 1 & \cdots & 1 - X
\end{vmatrix}$$
I could not proceed to get an expression of $P_f(X)$ as I could not see how to use the hint.


*Computing $ Ae_1 = \begin{pmatrix}
1 \\
1 \\
\vdots \\
1 
\end{pmatrix}$ which is not related to the eigenvalues $0$ or $n$.


*I am stuck in this question too.
 A: $(1)$ For $\lambda=0$, try $v=\begin{pmatrix}1\\1\\\vdots\\1\\1-n\end{pmatrix}$.  
$(2)$ For instance,  use the change of basis matrix (whose columns are the $-e_1+e_i$).  
$(4)$ $f(e_1)$ is an eigenvector for the eigenvalue $n$.
$(5)$  Take the eigenvector in $(1)$ and move the $1-n$ around to different coordinates, i.e. put $1-n$ in the $i$th coordinate,  and $1$'s everywhere else.  $n-1$ of these are linearly independent. 
$(6)$  Since it has a $n-1$ linearly independent eigenvectors for $\lambda =0$, and $1$ for $\lambda =n$, there is such a basis (consisting of eigenvectors) . 
A: Let $f_1 = e_1$, $f_2 = e_2 - e_1$, ..., $f_n = e_n - e_1$ as in the hint. Note that $\sum f_i = (\sum e_i) - (n-1) e_1$.
Then $A$ annihilates $f_2$, $f_3$, ..., and $f_n$. Meanwhile, 
\begin{align*}
Af_1 &= \sum_i {e_i} \\
&= \sum_i f_i + (n-1)  e_1 \\
&= nf_1 + \sum_{i \neq 1} f_i, 
\end{align*}
so the matrix for $A$ in the new basis is
$$
\begin{pmatrix}
n & 0 & 0 & \ldots & 0 \\
1 & 0 & 0 & \ldots & 0 \\
1 & 0 & 0 & \ddots & 0 \\
1 & 0 & 0 & \ldots & 0 \\
\end{pmatrix}.
$$
