I have a matrix $A$ of which the SVD is $U \Sigma V^{\top}$. $A$ is of rank $r$, and therefore $A = U_r \Sigma_r V_r^{\top}$ where $U_r$ are the top left singular values, $V_r$ similarly for right singular values and $\Sigma_r$ is a square diagonal matrix with the singular values.
I also know that $A = XY^{\top}$ for some $X \in \mathbb{R}^{d \times M}$ and $Y \in \mathbb{R}^{d' \times M}$.
I want to show that $U_r U_r^{\top} X = X$. It doesn't HAVE to be the case because $U_r U_r^{\top}$, which is a $d \times d$ matrix does not have to be the identity in normal circumstances (because it is not the full SVD).
However, I get the sense that the columns of $X$ are in the space that is spanned by $U$, and therefore we don't lose any information by multiplying $U$ by $X$.
Is that so? How do I prove that?
$U_r$ is an orthogonal matrix that consitutes a basis for $\mathbb{R}^{d,d}$. The fact that $rank(A)=r$ does not stop $U_r$ from being a basis, you will still complete the existing left singular vectors with singular vectors from the kernel.
If $U_r$ is an orthogonal basis then $U_r U_r^T=I$ so the statement $U_r U_r^TX=X$ will always hold even if $A$ is not full-rank as far as I understand.