- #1

- 996

- 5

## Homework Statement

Use the below relation to prove that [itex]rank(A) = rank(A^T)[/itex]

## Homework Equations

[tex](Im A)^c = ker(A^T)[/tex] where c is the orthogonal compliment.

## The Attempt at a Solution

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

Is this a valid argument?