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 $rank(A^T)=rank(A)$