Suppose I have two $n \times d, n>d$ matrices $B$ and $B'$, both full column-rank $d$. Suppose moreover that another matrix $A$ can be expressed as $A = BC = B^\prime C^\prime$, with $C$ and $C^\prime$ two full row-rank matrices.
All in all, matrix $A$ is full column-rank $d$ and $\cal{R}(B)=\cal{R}(B^\prime)= \cal{R}(A)$. Since both $B$ and $B^\prime$ are (not necessarily orthogonal) basis for $\cal{R}(A)$, there exists an orthogonal matrix $Q$ such that $B = B^\prime Q$, i.e., mapping the two bases. Correct? How can I show it?
Is such matrix $Q$ equal to $Q=C^\prime C^\dagger$?
What does the orthogonality of $B$ and $B^\prime$ would change?
I will restate in my words, shall I?
Let $T_0$ and $T_1$ be a transformation matrix from $U$ to $V_0$ and $V_1$ respectively. Do they span equivalent vector space? If there is such, there must be vector $w \in W$ which maps some $v_0 \in V_0$ and $v_1 \in V_1$.
For transformation matrix $T_{20}: V_0 \to E$ and $T_{21}: V_1 \to E$ on a common basis $A$, there is always $v_0 = T_0 u$ and $v_1 = T_1 u$ and necessarily $T_{20} T_0 u = T_{21} T_1 u = w$, since we consider he case which the vector are the same but in different basis. Therefore, we obtain the tranformation matrix relation $T_0 = T_{20}^{-1} T_{21} T_1 = T_{02} T_{21} T_1$.