Prove that rank(A) = rank(A^T)

[hotvette]

Homework Statement

Use the below relation to prove that rank(A) = rank(A^T)

Homework Equations

(Im A)^c = ker(A^T) where c is the orthogonal compliment.

The Attempt at a Solution

Let A be an n x m matrix with (n \ge m) and rank(A) = k. Then, rank(A^c) = n-k. From the given relation, rank(ker A^T) must also equal n-k. Thus, rank(A^T) = n - (n-k) = k = rank(A).

Is this a valid argument?

 

[andrewkirk]

It looks like it's probably correct, but the explanation is a bit confusing because it uses operators in what appear to be non-standard ways. For instance if c denotes orthogonal complement, that is a function whose domain is a vector space, so it cannot be applied to a linear operator, which is what the symbol $A^c$ suggests. Also, I have not seen the rank operator applied to a space rather than to a linear operator, which is what $rank(ker\ A^T)$ suggests. Perhaps it means 'dim'(ension)?

 

[hotvette]

Thanks, I agree that my explanation used sloppy terminology. Maybe this is better: Let A be an n x m matrix with n \ge m and rank=k. Then dim((Im A)^c) = m-k = dim(ker(A^T)). Also, dim(ker(A^T)^c)=m-(m-k)=k. But, (ker(A^T))^c = Im(A^T). Therefore, dim(Im(A^T))=k=dim(Im A), and thus rank(A^T)=rank(A)