Matrices and Linear applications Help me visualize it.
Example: i have the matrix 2×3
$$A= \begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & 3\end{bmatrix}$$
The linear application $L: \mathbb R^3\to \mathbb R^2$ represented by $A$ through the natural base $C=\{v_1,v_2,v_3\}$ of $\mathbb R^3$ and the natural base $C_1=\{e_1,e_2\}$ of $\mathbb R^2$ is:
$$L(v_1)=-e_2, L(v_2)= e_1, L(v_3)=2e_1+3e_2.$$
It follows $L(x,y,z)= (y+2z,3z-x)$.
Can you show me in general why the rows of the matrix, using natural basis, are the coefficients of the components of the image of the vector through $L$?
 A: That is just the definition of the matrix of a linear map.
Let $T : V \to W$ be a linear map. Let $v_1,. . .,v_n$ be a basis of $V$ and let $w_1,. . .,w_m$ be a basis of $W$. We can express $Tv_k$ as a linear combination of the basis $w_1,. . .,w_m$. Where $A_{1,k},. . .,A_{m,k}$ are just scalars.
$Tv_k $ = $A_{1,k}w_1+. . .+A_{m,k}w_m$
Let the matrix of $T$ be denoted by $A$. Then we define the entries in the $J^{th}$ row and the $k^{th}$ column to be the scalars needed to write $Tv_k$ as a linear combination of the basis $w_1,. . . ,w_n$.
Example: Define a linar map $T : R^2 \to R^2$ by $T(x, y) = (2x, 2y)$. We just need to compute the basis of $V$.
$T(1, 0) = (2, 0) = 2(1, 0) + 0(0, 1)$
$T(0, 1) = (0, 2) = 0(1, 0) + 2(0, 1)$.
Let $A$ denote the matrix of $T$. Then, $A$ = $\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}$
The columns of this matrix consist of the scalars needed to write the corresponding $Tv_k$ as a linear combination of the basis of $R^2$. Where $k$ denotes the column of the matrix.
From here you can see that given a matrix of a linear map $T$. The span of the columns of that matrix equals the range of $T$.
