Show that $\operatorname{rank} A = 3$

Question

Matrix $A \in R^{3, 2015}$ is given. It is known that matrix $AA^{T}$ is invertible. Show that $\operatorname{rank} A = 3$.

How to start this? What does the info that $AA^{T}$ is invertible gives us?

Answer 1

Since $A^\intercal A$ is an invertible matrix of order $3$, one has: $$\textrm{rank}(A^\intercal A)=3.$$ Besides, one has: $$\textrm{rank}(A^\intercal A)\leqslant\min\left(\textrm{rank}(A),\textrm{rank}\left(^\intercal A\right)\right).$$ But, one has: $$\textrm{rank}\left(^\intercal A\right)=\textrm{rank}(A).$$ Therefore, one gets: $$3\leqslant\textrm{rank}(A).$$ Hence the result, since the number of lines of a matrix is greater than its rank. (Edited)

Remark. Using $\textrm{rank}(AB)\leqslant\min(\textrm{rank}(A),\textrm{rank}(B))$ is overkill, instead you can just use $\textrm{rank}(AB)\leqslant\textrm{rank}(B)$, which is an easier result to prove.