Eigenvector of Matrix with Duplicate Columns I have this $5 \times 5$ matrix :
\begin{pmatrix}
1&1&1&1&1 \\ 2&2&2&2&2 \\ 3&3&3&3&3 \\
4&4&4&4&4 \\ 5&5&5&5&5
\end{pmatrix}
I need to find the eigenvalues and the eigenvectors. I found out that the eigenvalules are $15$ and $0$, $0$ with an algebraic multiplicity of $4$. 
I am now calculating the eigenvectors and I was wondering is there a simple direct way of knowing the eigenvector of the eigenvalue of $15$, not by the usual calculation? 
Is there a known eigenvector for an eigenvalue that is the sum of every column, as there is an eigenvector for an eigenvalue that is the sum of every row $[(1,1,1,1,...) ]$?
And what about cases in which the columns have the same sum, but are not identical?What is the eigenvector in that case?
For example: 
\begin{pmatrix}
3&5&4 \\ 2&2&1 \\ 2&0&2 
\end{pmatrix}

This is obviously a rank-one matrix, which you’ve verified by finding that the algebraic multiplicity of $0$ is four. Its column space (image) is spanned by $v=(1,2,3,4,5)^T$, so the only possibility for an eigenvector with a nonzero eigenvalue is a multiple of $v$.  
As for the second question, remember that there’s no such thing as “the” eigenvector: every nonzero scalar multiple of an eigenvector is also an eigenvector with the same eigenvalue. So, if the columns of a matrix are all multiples of some nonzero column, then the column space is again one-dimensional and every nonzero column is an eigenvector.  
Algebraically, such a matrix can be decomposed into the outer product $uv^T$ of a pair of vectors. We then have $(uv^T)u=u(v^Tu)=(v^Tu)u=\lambda u$, therefore $u$ is an eigenvector with eigenvalue $v^Tu$.
