How to prove that $X^{T}X$ is singular if $X$ is singular.

Question

Let's suppose $X$ is $n \times m$ not full rank matrix (ie any column vector in $X$ can be expressed as linear combination of others column vectors). If $$B = X^TX$$ why $B$ is singular?

The question is inspired from the statistics, wherein, if vector of independent variable is linearly dependent, then it is assumed that $X^TX$ is also dependent (or singular). It is difficult to grasp if $X$ is dependent then how $X^TX$ is dependent?

Answer 1

Exists $v\neq0$ such that

$$Xv=0$$

then

$$X^TXv=0$$