Maximum Rank of a Matrix Maximum rank of a 4*6 matrix A is 4.... 
But we know that 
Rank(A) = Rank(A transpose)...
In that case 
Maximum Rank(A transpose) = 6
How can Max Rank(A) be 4 & 6 for the same matrix ? 
Or am I missing some rule or assumption here?
 A: Note that rank is the dimension of the space spanned by its rows (or columns, doesn't actually matter). Now of course you can't have more than 4 dimensions if spanned by four vectors. On the other hand if you transpose the matrix you get 6 vectors, but that doesn't help much as they are only in a four dimensional space - so they still cannot span more than four dimensions.
So the rank can never be larger than the number of rows, and it can never be larger than the number of columns either. This is what you missed.
On the other hand it's easy to construct a matrix with the rank equals the minimum of number of rows and number of columns - just make the diagonal elements $1$ and the rest of the elements $0$.
So the maximum rank therefore on a $4\times 6$ matrix is the smaller of $4$ and $6$, that is $4$. By the same the maximum rank for a $6\times 4$ matrix is the smaller of $6$ and $4$, that is $4$.
A: The maximum rank of a $4\times 6$ matrix is $4$.
The maximum rank of a $6\times 4$ matrix is also $4$.
So the mistake you made was in the sentence

In that case Maximum Rank(A transpose) = 6

which is both unfounded (i.e. there is no proof given for it) and, more importantly, false.
A: Hint: rank of rows= rank of columns 
