How many different linearly independent Eigen Vectors of A? In $A = \left [ a_{ij} \right ]_{nxn}$ where $a_{ij}=1$ for all $i,j$ then the number of linearly independent eigen vectors of $A$ are ?

I am getting $(n-1)$ linearly independent eigen vectors. Am I right here ?
 A: The number of linearly independent eigenvectors is bounded above by the rank of the matrix since the span of those eigenvectors is a subspace of the image.  Your matrix has rank 1 (dimension of the column space is 1), and so any linearly independent set of eigenvectors has cardinality at most 1.
On the other hand, the vector $v = (1, 1, ..., 1)^T$ is an eigenvector.  Therefore, any set of linearly independent eigenvectors has cardinality 1.
A: $A$ is a real symmetric matrix, so is orthogonally diagonalizable. Not only can $n$ linearly independent eigenvectors be found, but one can arrange for them to be orthonormal as well.  
If you didn’t know this about symmetric matrices, you could work the answer out directly. The rank of $A$ is obviously $1$, which means that $\dim\ker A=n-1$, so any basis of the kernel provides us with $n-1$ linearly independent eigenvectors of $0$. The other eigenvalue is $\operatorname{tr} A=n$, and since eigenvectors of different eigenvalues are linearly independent, this gives us another eigenvector that’s linearly independent of the other $n-1$, completing a basis of the space. All of the rows sum to $n$, so it shouldn’t be hard to see that the vector consisting of all $1$s is an eigenvector of $n$.
