I have not seen anywhere written that it is impossible, but it seems impossible, so I want to check if I missed something.
According to a theorem, an $n\times n$ matrix is diagonalizable if it has $n$ independent eigenvectors.
Let's say, the matrix has $1$ row with only zeros (worst singular case). As it has one row with only zeros, it will zero out the corresponding row of any vector it is multiplied by. That means, the corresponding rows of its eigenvectors have to be zero. And if one of the rows is fixed, there cannot be n independent eigenvectors. Am I correct?
Did I miss anything?