If $X$ is a singular matrix, then is $X'X$ singular? If $X$ is a singular matrix, then is $X'X$ singular? And why?
(' means transpose)
Thanks in advance!
 A: $Xv=0$ 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 $X v=0$.
A: Note that $rank(X'X)\leq\min\{rank(X'),rank(X)\}<n$, which implies $X'X$ is still singular.
A: 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.
A: Since $X$ is singular, there exists a non-zero vector $v$ such that $Xv=0$.
Now,
$$X^TXv=X^T(Xv)=X^T0=0,$$
which implies $v$ is also in the nullspace of $X^TX$. Thus, the matrix is singular.
A: $X \, \text{is singular} \Leftrightarrow \det(X) = 0, \tag{1}$
$\det (X) = 0 \Rightarrow \det (X^TX) = \det (X^T) \det (X) = 0, \tag{2}$
$\det (X^TX) = 0 \Rightarrow X^TX \text{is singular}; \tag{3}$
OR
one might observe that $X$ singular implies the existence of a vector $v \ne 0$
such that $Xv = 0$, whence $X^TXv = 0$, whence $X^TX$ is singular!  QED.
Hope this helps!  Cheerio, 
and as always,
Fiat Lux!!!
