I'm reading linear algebra done right by Axler and I've been confused with the basis. For instance,
Let's say $T \in \mathcal{L}(V)$ where $V$ is $\mathbb{R}^2$, $v_1 = (4,2), v_2=(5,3)$, and let \begin{equation*} \begin{array}{cc} & \begin{array}{ccc} & v_1 & v_2 \end{array}\\ \mathcal{M}(T, (v_1, v_2)) = & \begin{array}{c} v_1\\ v_2\end{array} \left(\begin{array}{cc} 4 & 5\\ 2 & 3\\ \end{array}\right) \end{array} \end{equation*} This implies that $Tv_1 = 4v_1 + 2v_2$ since $$ \begin{array}{cc} & \begin{array}{ccc} \end{array}\\ \begin{bmatrix} 4 & 5 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}= & \begin{array}{c} v_1\\ v_2\end{array} \left(\begin{array}{cc} 4\\ 2 \\ \end{array}\right) \end{array}$$
However, if $T \in \mathcal{L}(V, W)$, $v_1 = (4,2), v_2=(5,3), w_1=(1,0), w_2=(0,1)$, and let \begin{equation*} \begin{array}{cc} & \begin{array}{ccc} & v_1 & v_2 \end{array}\\ \mathcal{M}(T, (v_1, v_2), (w_1, w_2)) = & \begin{array}{c} w_1\\ w_2\end{array} \left(\begin{array}{cc} 4 & 5\\ 2 & 3\\ \end{array}\right) \end{array} \end{equation*} This implies that $Tv_1 = 4w_1 + 2w_2$ since $$ \begin{array}{cc} & \begin{array}{ccc} \end{array}\\ \begin{bmatrix} 4 & 5 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}= & \begin{array}{c} w_1\\ w_2\end{array} \left(\begin{array}{cc} 4\\ 2 \\ \end{array}\right) \end{array}$$
The two results are completely different geometrically, but if we omit the bases, the numerical results are the same. Often in statistics or in other matrix computations, bases are not given. My quetions are:
- In most matrix computations that we see in our other math textbooks, how are the bases defined (especially when the matrix is not a square matrix)? For instance, in statistics, we often have the data matrix $X$ and its size is $N \times d$ where $N$ is the number of samples and $d$ is the feature size. What can we say about the bases?
- Is it okay to ignore the bases despite different geometrical interpretations?