One has given formula of linear map $T:\mathbb{R}^n \to \mathbb{R}^m$ and matrix of this linear map $ M(T)_\mathcal{A}^\mathcal{B} $ in (unknown) bases $\mathcal{A}$ of $\mathbb{R}^n$ and $\mathcal{B}$ of $\mathbb{R}^m$ and would like to find these bases. For example, get $ T:\mathbb{R}^3\to \mathbb{R}^2$ such that $T(x,y,z)=(x+y+z,2x-y+3z)$ with matrix $$M(T)_\mathcal{A}^\mathcal{B}=\begin{bmatrix} 1 & 3 & -1 \\ 4 & 0 & 7 \end{bmatrix}$$ and we would like to determine bases $\mathcal{A},\mathcal{B}.$
I would like to use a formula of base change: we have matrices $M(T)_\mathcal{A}^\mathcal{B}$ and $ M(T)_{st}^{st}.$ If I'll find $M(id)_\mathcal{A}^{st}$ and $M(id)_\mathcal{B}^{st}$, I have done. I tried by getting explicite form of these matrix in unknown elements but computations are so complicated, because we get a linear system with 13 coefficients… I would like to ask you for more satisfactory approach, more formal without such computations.
Best Answer
Let $A$ be the $m\times n$ matrix of $T\colon \mathbb R^n\to\mathbb R^m$ with respect to the standard basis and $B=M(T)_{\mathcal A}^{\mathcal B}$ be the given matrix with respect to unknown bases $\mathcal A$ and $\mathcal B$.
The two matrices $A$ and $B$ can describe the same linear map $T$ if and only if they have the same rank. In this case, there are invertible matrices $X\in\mathbb R^{n\times n}$ and $Y\in\mathbb R^{m\times m}$ such that $$ YAX = B. $$ Comparing this with $M(\operatorname{id})_{\text{st}}^{\mathcal B}\ A\ M(\operatorname{id})_{\mathcal A}^{\text{st}} = B$, we might interpret $X$ and $Y$ as change of basis matrices. Hence, we can take the columns of $X$ as the basis vectors in $\mathcal A$ and the columns of $Y^{-1}$ as the basis vectors in $\mathcal B$.
Note that $X$ and $Y$ and hence $\mathcal A$ and $\mathcal B$ are far from being unique for given matrices $A$ and $B$. For example the identity map will have the same matrix with respect to any basis (taken in domain and codomain).
One algorithmic way to find such $X$ and $Y$ is to use row and column transformations to transform $A$ and $B$ into the rank normal form $$ \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix} \in \mathbb R^{m\times n}, $$ where $r$ is the rank of both $A$ and $B$ and $I_r$ is the $r\times r$ identity matrix.
Keeping track of row and column operations seperately (by applying them simultaneously to $I_m$ and $I_n$, respectively) this yields $$ X_1 A Y_1 = \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix} = X_2 B Y_2, $$ and hence $$ \underbrace{X_2^{-1} X_1}_X\ A\ \underbrace{Y_1 Y_2^{-1}}_Y = B. $$