How to think about the elements of $F^m$ in terms of a matrix representation? In sheldon axlers book, he says:
If T is a linear map from $F^n$ to $F^m$, then unless state otherwise you should assume that the bases in question are the standard ones(where the kth basis vector is 1 in the kth slot and 0 in all the other slots. If you think of elements of $F^m$ as columns of m numbers then you can think of the kth column of $M(T)$ as T applied to the kth basis vector. 
In the last sentence, could you help me with understanding how to think of or view how the matrix $M(T)$ would look like? Would it be columns of coefficients or would it be just 0 and 1's?
I would really really appreciate it if you would so happen to have a picture of the matrix. Thanks so much!
 A: Axler means the following: You are asked to write the matrix of a given linear transformation $T:F^n \to F^m$. I hope you got the sentence related to the basis. I feel the next sentence is best understood with an example in hand. 
Consider $T: P_4(\mathbb{R}) \to P_3(\mathbb{R})$ ($P_n(\mathbb{R})$ represents the vector space of polynomials with real coefficients of degree $\leq n$) ; $T(p(x)) = \frac{d}{dx}p(x)$. Lets get the matrix representaion of $T$.
As Axler says, consider $T$ acting on the first basis vector of $P_4(\mathbb{R})$ i.e. {$1$}. We know $\frac{d}{dx} 1 = 0$. So the first column of $M(T)$ will consist of $0$'s \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}. Take the second basis vector {$x$}. $\frac{d}{dx} x = 1$. So, the second column will have a $1$ as the first element. \begin{bmatrix} 1 \\ 0 \\ 0 \\0 \end{bmatrix}
Take the third basis vector {$x^2$}. $\frac{d}{dx} x^2 = 2x$ i.e $2$ times the basis vector {$x$} of $P_3(x)$. Hence, the third column of $M(T)$ will have $2$ as its second element. \begin{bmatrix} 0 \\ 2 \\0 \\0 \end{bmatrix}. Do similarly for {$x^3$} and {$x^4$} and you will finally get $M(T) = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4 \end{bmatrix}$
