Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$? Is $A$ diagonalizable? Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$ and a basis for each eigenspace? Is $A$ diagonalizable?
Having some trouble with this one. I tried using the fact that $\lambda$ is an eigenvalue of $A$ iff there exist non-zero solutions to $Ax=\lambda x$. Well, clearly the $1\times n$ non-zero vector $x$ with all entries equal is an eigenvector of $A$, with corresponding eigenvalue $\lambda=n$, since $Ax=nx$. But I can't seem to get much further than that in terms of finding eigenvalues for $A$, any hints/suggestions?

Consider the matrix $B = A - \lambda I$, where $ \lambda \in \text{Spec}(A)$.
Then $\lambda_{1} = n - \lambda$ is an eigenvalue of $B$ with an associated eigenvector $v_{1} = \begin{pmatrix} 1 \\ 1 \\ 1\\ \cdot \\ \cdot \\ \cdot \\ 1 \\1 \\1\\ \end{pmatrix}$.
The remaining eigenvalues of $B$ are $\lambda_{2} = \lambda_{3} =...=\lambda_{n} = -\lambda$ and the corresponding eigenvectors are:
$v_{2} = \begin{pmatrix} 1 \\ -1 \\ 0\\ \cdot \\ \cdot \\ \cdot \\ 0 \\0 \\0\\ \end{pmatrix}$, $v_{3} = \begin{pmatrix} 0 \\ 1 \\ -1\\ \cdot \\ \cdot \\ \cdot \\ 0 \\0 \\0\\ \end{pmatrix}$, ..., $v_{n} = \begin{pmatrix} 0 \\ 0 \\ 0\\ \cdot \\ \cdot \\ \cdot \\ 0 \\1 \\-1\\ \end{pmatrix}$.
Now, it is easier to see that $v_{1}, ..., v_{n}$ are linearly independent and they are eigenvectors of $A$ with $v_{1}$ associated with the eigenvalue $\lambda =n$ and $v_{j}$ for $j \geq 2$ associated with $\lambda = 0$. From the linearly independence of these eigenvectors we see $A$ is diagonalizable. 
