Eigenvector of a matrix of ones associated with $\lambda =0$ An $n\times n$ matrix consistent of all ones, will have two eigenvalues: $0$ and $n$. The eigenvector associated with $n$ will be $(1,1,...,1)$, but are there then infinite solutions for the eigenvector associated with $0$? Because, in order for the eigenvalue to be zero, all rows of the eigenvector must add to zero, there are surely many possible solutions.
How can I express infinite solutions for the eigenvectors?
Thanks in advance
 A: It's not that strange, there are an infinite number of solutions for the eigenvector associated with $n$ too! In fact every eigenvalue has an infinite amount of eigenvectors.
For a matrix $A$ with eigenvector $\mathbf{v}$ and associated eigenvalue $\lambda$, then for any scalar $k$ the vector $k\mathbf{v}$ is also an eigenvector with eigenvalue $\lambda$:
$$
A(k\mathbf{v}) = kA\mathbf{v} = k \lambda \mathbf{v} = \lambda(k\mathbf{v})
$$
A: First of all your matrix is positive semidefinite. Hence eigenvectors are orthogonal to each other. You have only one nonzero eigenvalue and the associated eigenvector is the vector of all ones $\boldsymbol{1}=[1~1~\dots~1]$.
As a result, the eigenvectors associated with $0$s are the ones that are orthogonal to the all ones vector $\boldsymbol{1}$.
This is a ${\bf{subspace}}$ of dimension $n-1$. So any vector in the subspace which is the orthogonal complement of $\boldsymbol{1}$ is an eigenvector with eigenvalue $0$. The converse is true as well.
A: An eigenvector with eigenvalue $\lambda$ is any nonzero vector solving $Av=\lambda v$, where the collection of all solutions of this equation is called an eigenspace, denoted $E_\lambda$.  You can verify that $E_\lambda$ is a non-trivial vector subspace.
Now, non-trivial vector subspaces always have infinitely many vectors in them (say we are working over $\mathbb{R}$ or $\mathbb{C}$).  How do we deal with infinities like this in linear algebra?  By determining a finite (and possibly minimal) spanning set.
In your example, the eigenspace $E_1$ is the span of $\{(1,\dots,1)\}$, and the eigenspace $E_0$ is the set of all vectors $x$ with $\sum_i x_i=0$, which happens to be the span of the $n$ vectors $\{(n-1,-1,\dots,-1),(-1,n-1,-1,\dots,-1),\dots,(-1,\dots,-1,n-1)\}$.  (These are linearly dependent, and any one of them may be removed.)
A: Usually people just write $\left\{ u \mid \sum_{i=1}^n u_i = 0 \right\}$ for the eigenspace corresponding to eigenvalue $0$. If you need to pick a basis of size $n - 1$ for this subspace, the simplest one I can think of is
$\{e_1 - e_2, e_2 - e_3, \ldots, e_{n-1} - e_n\}$. Of course there are infinitely many other choices.
