TITLE: Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$
QUESTION [113 upvotes]: How can I prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$?

This is an exercise in my textbook associated with orthogonal projections and Gram-Schmidt process, but I am unsure how they are relevant.

REPLY [4 votes]: Since elementary operations do not change the rank of a matrix we have $\text{rank}(A^TA) = \text{rank}(E^TA^TAE)$, where $E$ is a multiplication of several elementary operations which make $AE = [A_1, A_2]$, where $A_1$ is a column full rank matrix with $\text{rank}(A_1) = \text{rank}(A)$.
Thus we can find a matrix $P$ such that $A_1P= A_2$ and $AE = [A_1, A_1P] = A_1[I, P]$.
Thus $\text{rank}(E^TA^TAE) = \text{rank}(A_1[I, P])^T(A_1[I, P])$. In this equation,  the matrices are all of full rank and the rank equals $\text{rank}(A)$, so on a real space $\text{rank}(A^TA) = \text{rank}(A)$, completing the proof.