A generalised permutation matrix is a matrix where each row and column contains exactly one non-zero entry.
I am interested in knowing if an arbitrary generalised permutation matrix is diagonalisable over the complex field. If this is not true, are there any sufficient and necessary conditions for it to be diagonalisable?
It has been a while since I have studied linear algebra so my knowledge is a bit sketchy. I tried to calculate the characteristic polynomial however I wasn't able to get a nice expression for it. I also tried to use the fact that any permutation matrix is diagonalisable and any generalised permutation matrix can be written as $DP$ where $D$ is a diagonal matrix and $P$ is a permutation matrix. However, I wasn't able to make much progress with that.
Edit
I have added that I am working over the complex numbers.
Over a general (algebraically closed) field, the answer is negative. E.g. $\pmatrix{0&1\\ 1&0}$ is not diagonalisable over the algebraic closure of $GF(2)$.
However, the answer is affirmative if the underlying field is complex. Call your matrix $A$. View $A$ as an adjacency matrix of a directed graph. Since each column or row of the matrix has exactly one nonzero entry, the graph is a disjoint union of cycles. Therefore, the matrix is permutation-similar to a direct sum of a diagonal matrix and some submatrices of the form $$ C=\pmatrix{0&c_1\\ &0&c_2\\ &&\ddots&\ddots\\ &&&\ddots&c_{k-1}\\ c_k&&&&0} $$ where the $c_i$s on the circulant transversal are nonzero. By inspecting the locations occupied by the nonzero entries of $C,C^2,\ldots$, we see that $I,C,C^2,\ldots,C^{k-1}$ are linearly independent. Therefore the minimal polynomial of $C$ is degree-$k$. However, as $C^k=\left(\prod_{i=1}^kc_i\right)I$, the minimal polynomial of $C$ is $m(x)=x^k-\left(\prod_{i=1}^kc_i\right)$. Therefore $m(x)=0$ has $k$ distinct roots and $C$ is diagonalisable. In turn, $A$ is diagonalisable too.