For any matrix A elementof Rm times n prove that rankA rank

For any matrix A elementof R^m times n, prove that rank(A) = rank(A^A).

Solution

Let x Null(A) where Null(A) is the null space of A. Then, by the definition of Null(A), we have Ax = 0. Now, on multiplying on the left by A^T , we get A^TAx = 0 . Therefore, Null(A) Null(A^T A).

Now, let x Null (A^T A). Then, again by the definition of Null (A^TA), we have (A^TA)x = 0.Then, on, multiplying on the left by x^T , we get x^TA^T Ax = 0 or, (Ax)^T (Ax )= 0 ( as (Ax)^T = x^TA^T).This implies that Ax = 0. Hence x Null(A) so that Null(A^T A) Null(A). Therefore, Null(A) = Null(A^T A). Then dim(Null(A) = dim(Null(A^T A) i.e. nullity (A) = nullity(A^T A). Now, by the Dimension theorem, (Rank-Nullity theorem), Rank (A) = Rank(A^T