Let $V$ be a vector subspace of $\mathbb{R}^p$ of dimension $m$, where $p < m$. I'm checking with you if there alawys exists a rotation $A \in O(p)$ so that $A(V) = \mathbb{R}^p \times \{0\}^{p-m}$, i.e. $V$ is the rotational image of a rotation in $\mathbb{R}^p$ .
I think the answer is yes, by the following argument, could you please check this?
Let $V$ be the span of the orthonormal basis $\{v_1,...v_n\} \subset \mathbb{R}^p$. It's enough to show that there exists $A \in SO(p)$ so that $< Av_i , e_j> = 0 \forall m+1 \leq j \leq p$. Solving this for $A$ is basically solving $m(p-m)$ equations. But since $m(p-m) \leq \frac{p^2}{4} \leq \frac{p(p-1)}{2} = dim(O(p))$ as a manifold. So we can always solve for $A \in O(p)$ so that the above equations hold, or equivalently, there exists an $A \in O(p)$ so that $A(V) = \mathbb{R}^p \times \{0\}^{p-m}$.
Is the above correct?