Some questions about the Eigenvalues of this $4\times 4$ matrix \begin{bmatrix}
1 & 1 & 1 & 1\\ 
1 & 1 & 1 & 1\\ 
1 & 1 & 1 & 1\\ 
1 & 1 & 1 & 1
\end{bmatrix}
The rank is $1$ as there is only $1$ linearly independent row.
The answer states that the four Eigenvalues are $0,0,0,4$. 
The $4$ comes from the matrix trace $1+1+1+1$ however I am clueless as to where the three zeros are from? The answer does not provide that information. I know the determinant of the matrix is $0$.


*

*So are there always as many Eigenvalues as there are rows($n$ many Eigenvalues)?

*Could someone shed light on where the three zeros come from.

 A: An $n$ by $n$ matrix always has $n$ eigenvalues (if we include repetition). If we know that the rank is $r$ then at least $n-r$ eigenvalues must be zero.
Since the rank is $1$, there must be at least $3$ zero eigenvalues.
A: Clearly $(1,1,1,1)^T$ is an eigenvector, and $4$ is its eigenvalue, not particularly because of the trace (the trace of a matrix doesn't have to be an eigenvalue), but because $(1,1,1,1)^T$ becomes $4\cdot(1,1,1,1)^T$ when multiplied by the matrix.
$0$ is also an eigenvalue, for example with eigenvector $(1,-1,0,0)^T$. Why are there three zeroes. There are two possible explanations, and without more, your text could mean either of them:
Algebraically, there are three zeroes because $\lambda=0$ is a triple root of the characteristic polynomial.
Geometrically, there are three zeroes because the eigenspace for eigenvalue $0$ is three-dimensional. For example, is is spanned by $(1,-1,0,0)^T$, $(1,0,-1,0)^T$ and $(1,0,0,-1)^T$.
In this case the algebraic and geometric view leads to the same count, but this is not always the case -- the geometric multiplicity of an eigenvalue can be smaller than the algebraic one.
