Matrices with columns which are eigenvectors In this question, the OP asks about finding the matrix exponential of the matrix $$M=\begin{bmatrix}
1 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1
\end{bmatrix}.$$ It works out quite nicely because $M^2 = 3M$ so $M^n = 3^{n-1}M$. The reason this occurs is that the vector $$v = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$$ is an eigenvector for $M$ (with eigenvalue $3$). The same is true of the $n\times n$ matrix consisting of all ones and the corresponding vector. With this in mind, I ask the following:

  
*
  
*Can we find a standard form for an $n\times n$ matrix with the property that each of its columns are eigenvectors? (Added later: This may be easier to answer if we allow columns to be zero vectors. Thanks Adam W for the comment.)
  
*What about the case when we require all the eigenvectors to correspond to the same eigenvalue?

The matrices in the second question are precisely the ones for which the calculation of the matrix exponential would be analogous to that for $M$ as above.

Added later: In Hurkyl's answer, he/she shows that an invertible matrix satisfying 1 is diagonal. For the case $n=2$, it is fairly easy to see that any non-invertible matrix satisfies 1 (which is generalised by the situation in Seirios's answer). 
However, for $n > 2$, not every non-invertible matrix satisfies this property (as one may expect). For example,
$$M = \begin{bmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 1
\end{bmatrix}.$$

Another partial answer: One can notice that matrices of rank 1 are such examples. 
Indeed, if $M$ is of rank 1, its columns are of the form $a_1C,a_2C,...,a_nC$ for a given vector $C \in \mathbb{R}^n$ ; if $L=(a_1 \ a_2 \ ... \ a_n)$, then $M=CL$. 
So $MC=pC$ with the inner product $p=LC$. Also, $M^n=p^{n-1}M$.
