Linear Transformation Basics (Constructing a matrix to perform a transformation)

Question

I want to be clear that I am understanding how to construct a matrix $A$ corresponding to specified linear transformation $T$.

I am working through "Introduction to Linear Algebra by Strang" 4th edition.

Here is what is mentioned:

Suppose $T$ transforms the space $V$ ($n$-dimensional) to the space $W$ ($m$-dimensional). We choose a basis for $v_1,\ldots,v_n$ for $V$ and a basis $w_1,\ldots, w_m$ for $W$.

To find the first column of $A$, apply $T$ to the first basis vector $v_1$.

Then $T(v_1) = a_{11}w_1 + \cdots + a_{m1}w_m$ and these numbers $a_{11},\ldots,a_{m1}$ go into the first column of $A$. Transforming $v_1$ to $T(v_1)$ matches multiplying $(1,0,\ldots,0)$ by $A$.

My questions/thoughts

This section opens up by saying this approach works for constructing a matrix for any linear transformation. However, it seems the basis in $n$-dimensional space is stated as being generic (i.e. $v_1,\ldots,v_n$) but if $T(v_1)$ matches multiplying by $(1,0,\ldots,0)$ am I supposed to assume $v_1,\ldots,v_n$ are a standard orthonormal basis in $n$-dimensional space?

I guess I don't see how this example generalizes, but perhaps I am missing something.

Answer 1

Consider that the components of $v_1$ are determined by $$v_1=1 v_1+0 v_2+\cdots+0 v_n.$$

Answer 2

I always explain it more simply:

You have a linear transformation $ L: \mathbb{R}^n \rightarrow \mathbb{R}^m $. You know that for any vector $ x = \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ \vdots \\ \chi_{n-1} \end{array} \right) = \chi_0 \left( \begin{array}{c} 1\\ 0 \\ \vdots \\ 0 \end{array} \right) + \chi_1 \left( \begin{array}{c} 0\\ 1 \\ \vdots \\ 0 \end{array} \right) = \chi_0 e_0 + \chi_1 e_1 + \cdots $ $$ L( x ) = L( \chi_0 e_0 + \chi_1 e_1 + \cdots ) = \chi_0 L( e_0 ) + \chi_1 L( e_1 ) + \cdots $$ Thus, the linear transformation is completely defined by how the standard basis vectors $ e_j $ are transformed. A matrix is merely a convenient way of then writing this information: the $ j$th column of the matrix equals the vector that results from evaluating $ L( e_j ) $.

What Strang is trying to do here is to generalize this to viewing vectors in different bases.