Problem:
Let $X_k = \lbrace A^T A \mid \operatorname{rank} A \ge k, A \textrm{ is an $n \times n$ matrix} \rbrace $. Show that $X_k$ is convex.
I know from the classes that $\operatorname{rank} A^TA = \operatorname{rank} A $. So it suffices to show that for every $A, B \in X_k$ the following relation holds:
$$\lambda A^TA + (1-\lambda) B^TB = C^TC$$, where $\operatorname{rank} C \ge k$. But taking the scalar inside the matrix, we could show that for any $A,B \in X_k$ there exists $C \in X_k$ that: $$A^TA + B^TB = C^TC$$
I'm stuck here. Can you help me please?
Hint: Note that $A^TA$ is positive semi-definite. Now if $P$ and $Q$ are both positive semi-definite, then we have $$ \operatorname{rank}(P+Q)\geq \min\{ \operatorname{rank}(P), \operatorname{rank}(Q)\}.$$ A proof of this fact can be found in the following Math.SE question: Rank of the sum of two positive semi-definite matrices.