Let me know if my problem is asked earlier. I will remove my question immediately.
Question: Let X be a $n\times p $ matrix, and let $X^{\prime} X$ is the corresponding Gram matrix. I have learnt that the null space of $X$ is equal to null space of $X^{\prime} X$, i.e., $\mathcal{N}(X)=\mathcal{N}(X^{\prime} X)$. Does this imply that row space of $X$ is equal to the space spanned by $X^{\prime}X$?
More specifically, suppose the Gram matrix can be written as $X^{\prime} X=V D V^{\prime}$ where where $V=({\bf v}_1, \ldots, {\bf v}_r)$ is a $p \times r$ matrix such that $V^{\prime}V=I$, and $D$ is a $r\times r$ diagonal matrix. In that case can I write $${\bf x}_1=a_1 {\bf v}_1+ \ldots + a_r {\bf v}_r, $$ where ${\bf x}_1$ is a particular row of $X$?
The answer is yes and the reason is quite simple. Indeed if $X =\left( \begin{array}{c} x_1 \\ x_2 \\ . \\ .\\ .\\ x_p\end{array}\right)$
since for all $y\in \mathcal{N}(X)$, $y'x_i = 0$, then $\mbox{Vect}(x_1, x_2,\cdots, x_p) \subset \mathcal{N}(X)^{\perp}$.
On other hand the Gram matrix $X'X$ is symmetric (hermitian in complex) so
$\mathcal{N}(X'X)^{\perp} = \mathcal{I}(X'X)$. Then $\mbox{Vect}(x_1, x_2,\cdots, x_p) \subset \mathcal{I}(X'X)$. Or the two spaces have the same dimension if becomes then $\mbox{Vect}(x_1, x_2,\cdots, x_p) = \mathcal{I}(X'X)$