TITLE: Rank of quaternionic matrix
QUESTION [7 upvotes]: Is there a method of founding of rank of quaternionic matrix by Dieudonne's determinants ?

REPLY [4 votes]: I think that the method is exactly the same as in the commutative case. The rank of a quaternionic matrix (or more generally, over any not necessarily commutative field) is equal to the maximal size of a minor with non-zero Dieudonne determinant. The proof of this fact is essentially the same as in the commutative case. The key point is: given a square quaternionic matrix, its rows (equivalently, columns) are linearly independent if and only if the Dieudonne determinant does not vanish. 
Unfortunaly I do not have a reference for the answer to your question. But non-commutative linear algebra and Dieudonne determinants are discussed in the book "Geometric algebra" by 
E. Artin.