The matrix entries depend on the bases of the domain and range?

Question

Consider the basis $B=\{v_1,\dots,v_n\}$ of $\Bbb R^n$, and write $S=\{e_1,\dots,e_n\}$ to mean the standard basis.

Then this means (among other things) that $v_i\in \Bbb R^n$, and hence I can write each $v_i$ in the standard basis as: $$v_i=\sum_{j=1}^n a_{ji}e_j,$$ so then I suppose I am saying that I have a change of basis matrix: $$(a_{ij})_{ij}$$ where indeed (and similarly for the other $v_i$ written in basis $B$), $$\begin{bmatrix}a_{11}&a_{12}&\dots&a_{1n}\\&\ddots\\&&\ddots\\a_{n1}&a_{n2}&\dots&a_{nn}\end{bmatrix}\begin{bmatrix}1\\0\\\vdots\\0\end{bmatrix}_{B}=\begin{bmatrix}a_{11}\\a_{21}\\\vdots\\a_{n1}\end{bmatrix}_{S}$$ where I have used a subscript to denote the basis this vector is written with respect to.

I suppose the matrix entries are dependent on the basis of the vectors in the domain and image. Does one typically denote the matrix in a way that encodes the basis of the source and target?

I suppose I am asking for this matrix $A=(a_{ij})_{ij}:\Bbb R^n\to \Bbb R^n$ to be an isomorphism, and am saying that $A:V\to W$ where $V$ is freely generated by $\{v_1,\dots,v_n\}$ and $W$ is freely generated by $\{e_1,\dots,e_n\}$.

Answer 1

There is a difference between "matrix" and "linear operator between finite dimensional vector spaces".

A matrix is just a formal table of numbers, functions, letters, whatever.

A linear operator between two finite dimensional real vector spaces is a linear map between $V$ and $W$, where $V$ and $W$ are vector spaces with $\dim V = n$ and $\dim W = m$ (and so $V \cong \mathbb{R}^n$ and $W \cong \mathbb{R}^m$). As a linear operator $f: V \rightarrow W$ is completely determined by the image of the vectors of a basis, in order to find the action of $f$ it is enough to take a basis $\{e_1,\dots,e_n\}$ of $V$ and compute $f(e_1),\dots,f(e_n)$. These will be vectors in $W$, so can be expressed in terms of a basis of $W$, say $\{l_1,\dots,l_m\}$, that is $$f(e_i) = \sum_{k=1}^m a_i^kl_k$$ where $a_i^k, i = 1,\dots,n, k = 1,\dots,m$ are real numbers. These define a matrix $(a^i_k)$, which is the matrix associated to $f$. In this context, the entries of the matrix depend naturally on the bases chosen.

Answer 2

The columns of your matrix $A=(a_{ij})$ are exactly the coordinates of each new basis $B$ vector w.r.t the standart basis $\{e_1,\ldots,e_n\}$. For each $j$ the column $j$ of $A$ gives you the ordered coordinates of the new basis vector $v_{j}\in B$ with respect to the canonical basis $S=\{e_1,\ldots,e_n\}$. In equations:

$$ [v_i]_{S}=\left[\begin{array}{c} a_{1j}\\ a_{2j}\\ \vdots \\ a_{nj}\end{array}\right]= A\left[\begin{array}{c} 0\\ 0\\ \vdots\\ 1\\\vdots \\ 0\end{array}\right] $$ where $e_{j}=\left[\begin{array}{c} 0\\ 0\\ \vdots\\ 1\\\vdots \\ 0\end{array}\right]$ has $1$ at position $j$ and $0$ elsewhere.