Change of basis of the kernel of a rectangular matrix Suppose that I have a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that $T$ can be written as an $m \times n$ matrix. Let $k = \dim\ker T$, where $k > 0$. Thus, $\ker T \cong \mathbb{R}^k$, and any basis of $\ker T$ is a set consisting of $k$ linearly-independent vectors of $\ker T$. Nevertheless, the vectors of $\ker T$ are written (in the standard basis of $\mathbb{R}^n$) as a column of $n$ scalars (elements of $\mathbb{R}$).
Given a vector $v \in \ker T$ written in terms of a basis $\mathcal{B}$ of $\ker T$, I want to write $v$ in terms of another basis $\mathcal{B}'$.
I know that this typically entails computing the change-of-basis matrix (and its inverse), but I am mostly finding discussions and examples of this involving scenarios where the number of basis vectors equals the number of scalars in the column that "constitutes" the vector $v$ (in the standard basis).
Am I seemingly running into an issue here because I am trying to think of $v$ in terms of the basis of $\mathbb{R}^n$ (that is, $[v]_{\mathcal{E_n}}$, where $\mathcal{E}_n$ is the standard basis of $\mathbb{R}^n$)? If I instead consider $[v]_{\mathcal{B}}$ and $[v]_{\mathcal{B}'}$, then $v$ can be expressed as columns of $k$ scalars, so that the typical square change-of-basis matrix formalism can be used, correct? Though it seems I would need to know how to express the basis vectors of $\mathcal{B}$ in terms of those of $\mathcal{B}'$ (or vice versa) in order to generate this change-of-basis matrix in the first place. Instead, I know how to express the elements of $\mathcal{B}$ and $\mathcal{B}'$ in terms of $\mathcal{E}_n$.
Perhaps I can generate two rectangular change-of-basis matrices, $P_{\mathcal{E}_n \leftarrow \mathcal{B}}$ and $P_{\mathcal{B'} \leftarrow \mathcal{E}_n}$ such that their product $P_{\mathcal{B'} \leftarrow \mathcal{B}} = P_{\mathcal{B'} \leftarrow \mathcal{E}_n} P_{\mathcal{E}_n \leftarrow \mathcal{B}}$ is a square matrix that can be inverted? I believe that I can obtain $P_{\mathcal{E}_n \leftarrow \mathcal{B}}$ and $P_{\mathcal{E}_n \leftarrow \mathcal{B}'}$ by writing down the vectors of $\mathcal{B}, \mathcal{B}'$ in the standard basis of $\mathbb{R}^n$, but how can I obtain $P_{\mathcal{B'} \leftarrow \mathcal{E}_n}$ without having to resort to computing an "inverse" for a rectangular matrix?
If someone could explicitly walk through the steps needed to do this change of basis, that would be greatly appreciated. Thank you!
 A: You’re on the right track. If you were working with $\mathbb R^n$, you might assemble the elements of $\mathcal B$ into the columns of the matrix $B = P_{\mathcal E_n\leftarrow\mathcal B}$ and the elements of $\mathcal B'$ into $B' = P_{\mathcal E_n\leftarrow\mathcal B'}$. The change-of-basis matrix $P_{\mathcal B'\leftarrow\mathcal B}$ is then the solution to the equation $B'X=B$, and since $B'$ is full rank, that’s just $X={B'}^{-1}B$. Conceptually, $P_{\mathcal B'\leftarrow\mathcal B}$ first maps from $\mathcal B$ to the standard basis, and then from that to $\mathcal B'$.  
What if we try the same thing with the two bases of $\ker T$, expressed as elements of $\mathbb R^n$? Well, the equation $B'X=B$ is still valid, but since ${B'}^{-1}$ doesn’t exist we can’t simply multiply both sides by it as we did above. However, $B'$ does have full column rank, so it has the left inverse $({B'}^TB')^{-1}{B'}^T$. (That this is reminiscent of the formula for the least-squares solution to a system of linear equations is no coincidence.) Therefore, our change-of-basis matrix is $$P_{\mathcal B'\leftarrow\mathcal B}=({B'}^TB')^{-1}{B'}^TB.$$ This is consistent with the calculation in the first paragraph: if $B'$ is invertible, this expression reduces to ${B'}^{-1}B$.  
Taking a concrete example, let $$B = \begin{bmatrix}1&1\\1&0\\1&1\\1&0\end{bmatrix}, B' = \begin{bmatrix}2&0\\1&1\\2&0\\1&1\end{bmatrix}.$$ (The elements of $\mathcal B'$ are the sum and difference of the elements of $\mathcal B$.) Applying the above formula, we get $$P_{\mathcal B'\leftarrow\mathcal B} = \begin{bmatrix}\frac12&\frac12\\\frac12&-\frac12\end{bmatrix}.$$ This looks plausible given how $\mathcal B'$ was constructed from $\mathcal B$. To check, convert to the standard basis of $\mathbb R^4$: $B[a,b]^T = [a+b,a,a+b,a]^T$ and $$B'P_{\mathcal B'\leftarrow\mathcal B}\begin{bmatrix}a\\b\end{bmatrix} = B'\begin{bmatrix}\frac{a+b}2\\\frac{a-b}2\end{bmatrix} = \begin{bmatrix}a+b\\a\\a+b\\a\end{bmatrix}.$$
