If $X$ is a singular matrix, then is $X'X$ singular?
Solution 1:
$Xv=0$ implies $X^T X v = 0$.
$X^T X v = 0$ implies $v^T X^T X v = \|X v\|^2 = 0$, which in turn implies $X v=0$.
Solution 2:
Note that $rank(X'X)\leq\min\{rank(X'),rank(X)\}<n$, which implies $X'X$ is still singular.
Solution 3:
We know that a matrix is sigular is equivalent to the determinant being zero. Therefore from your statement we have $\det(X)=0$. We also know $$\det(X^{\prime})=\det(X),$$ and $$\det(AB)=\det(A)\det(B).$$ So we have $$\det(XX^{\prime})=\det(X)\det(X^{\prime})=0.$$ Which is equivalent to saying $XX^{\prime}$ is singular.