Prove that the rank of a matrix X is equal to the rank of th

Prove that the rank of a matrix X is equal to the rank of the matrix X^T X. Note that if X is of dimensions n x p then X^TX is of dimension p x p. We also know that for any n x p matrix A, r(A) + r(N(A)) =p.

Solution

Let us define Rank(X) as the column rank of X: col rank(X) = dim { Ax : x belongs to R }.
Let X^T be the transpose of A. Let us first show that X^TX x = 0 if and only if Xx = 0
X^TXx = 0
x^TX^TXx = 0 (multiplying with x^T)
  (Xx)^TXx = 0 => Xx = 0
   
Therefore columns of X^TX satisfies the same linear relationship as the colums of X. Number of rows doesn\'t matter, same number of columns => same column rank => same rank (as we have assumed initially)