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!
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 $$B'P_{\mathcal B'\leftarrow\mathcal B}\begin{bmatrix}a\\b\end{bmatrix} = B'\begin{bmatrix}\frac{a+b}2\\\frac{a-b}2\end{bmatrix} = \begin{bmatrix}a+b\\a\\a+b\\a\end{bmatrix}.$$