Let $A \in M_{nxn}$ such that $A_{ij}=1$ for all $i,j \in \{1,...,n\}$
Determain the eigenvalores of $A$, the eigenspacies and say if it´s diagonalizable or not.
Well, I think that the eigenvalues has to follow a certain rule in order to be able to generalize it to any $n$. So I made a try with a 3 by 3 matrix. I end up with $\lambda _1=0$ with multiplicity of 2 and $\lambda _2=3$.
Then we know that $A$ is not diagonalizable because has a multiplicity higher than 1.
I belive that while we increse the n the multiplicity of $\lambda _1=0$ increses in one.
How could I formalize this intuitions?
Your example seems to suggest that $0$ is an eigenvalue of multiplicity $n-1$, while $n$ is an eigenvalue of multiplicity $1$.
To check the first thing, note that $$ Ax=0\iff x_1+\dots+x_n=0 $$ The equation on the RHS defines a space of dimension $n-1$, so the eigenspace associated to $0$ has dimension $n-1$.
Also, if $v$ is the vector whose coordinates are all $1$, we get that $$ Av=nv $$ so that $n$ is an eigenvalue, and $v$ is an associated eigenvector.