I'm trying to answer the following question (which I've found in a linear algebra textbook):
"Why does a matrix $A$ represent the linear application $L_A \colon \mathbb{R}^n \to\mathbb{R}^m , L_A (x)=Ax$ with respect to the standard basis?"
The fact is obvious looking at the definition of $L_A$ and that's exactly the reason I'm having a hard time giving a rigorous explanation of it. My answer:
"We know that the columns of the matrix which represents a linear transformation $T\colon V\to W$, $A^j \ (j=1,\dots , n) $ are given by the coordinates with respect to the basis of $W$ of $T(v_j)$, where $v_j \ (j=1,\dots , n)$ is the basis of $V$ so in this case $A^j $is the column vector $A^j= (L_A (e_j))=(Ae_j)$ which coincides with the j-th column of $A$ and the same holds for all the other columns of $A$."
I think mine is a very clumsy (if ever correct) answer so I would appreciate any hint about how to clarify/correct/improve it.
Best regards,
lorenzo
Go by the definition:
Let $\;E=\{e_1,...,e_n\}\;$ be the standard basis and suppose $\;A=(a_{ij})\;$, then
$$L_Ae_i=Ae_i=\begin{pmatrix}a_{1i}\\a_{2i}\\\ldots\\a_{ni}\end{pmatrix}=a_{1i}e_1+a_{2i}e_2+\ldots+a_{ni}e_n$$
and thus the matrix representation of $\;L_A\;$ wrt the standard basis has the $\;i\,-$ th column equal to $\;\begin{pmatrix}a_{1i}\\a_{2i}\\\ldots\\a_{ni}\end{pmatrix}\;$ ...and so it is $\;A\;$ itself!