Eigenvalues of matrix with all $1$'s. Let $A$ be the $n \times n$ matrix over a field of characteristic 0, all of whose entries are 1. What are the eigenvalues of $A$, counted with their multiplicities?
 A: What is $Ax$, where $x =(x_1,\ldots,x_n)^\top$?  It is an $n\times 1$ column vector in which every entry is $x_1+\cdots+x_n$.  Thus $x$ is in the kernel, and hence $x$ is an eigenvector with eigenvalue $0$, if that sum is $0$.  So what is the dimension of the space of $n$-tuples in which the sum of the components is $0$?  For the mapping $(x_1,\ldots,x_n)\mapsto x_1+\cdots+x_n$, the dimension of the image is $1$ and the dimension of the domain is $n$; hence the dimension of the kernel is $n-1$.  So the geometric multiplicity of $0$ as an eigenvalue of $A$, i.e. the dimension of the eigenspace, is $n-1$.  That leaves room for a $1$-dimensional eigenspace with some other eigenvalue.  And $(1,\ldots,1)^\top$ is mapped to $(n,\ldots,n)$, so that gives you the other one.
A: Let $A$ be the $n$-by-$n$ all ones matrix. Since each row is a repeated copy of the first row, determinant of $A$ is $0$. This implies that $0$ is an eigenvalue of $A$. Also, observe that nullity of $A$ is $n-1$. This implies that the geometric multiplicity of $0$ is $n-1$. The remaning eigenvalue has geometric multiplicity $1$. Note that if $x=[1 \cdots 1]^T$, then $Ax=n x$. Thus, $n$ is an eigenvalue of $A$. 
