Let $u=(1,2,3,4)$ and $v=(0,1,1,1)$. Clearly, $(u,v)$ is an independent family of $\mathbb{R}^4$, and can be completed to a basis of $\mathbb{R}^4$ with two vectors. It's very easy to see that adding $e_1=(1,0,0,0)$ and $e_2=(0,1,0,0)$ to this family yields a basis of $\mathbb{R}^4$ (compute the rank of $(u,v,e_1,e_2)$).
Now a linear mapping is uniquely determined by the image of a basis of its domain. We already know the image of $u$ and $v$ (the nil vector of $\mathbb{R}^3$), hence we only need to determine the image of $e_1$ and $e_2$ by $F$. A bit later on, we'll need the coordinates of a vector of $\mathbb{R}^4$ in the basis $(u,v,e_1,e_2)$, so we might as well do this right now. A straightforward system solving yields:
$$\forall (x,y,z,t)\in\mathbb{R}^4,\quad (x,y,z,t)=(-z+t)u+(4z-3t)v+(x+z-t)e_1+(y-2z+t)e_2.$$
There's a caveat: we can't take just any vectors $a$ and $b$ of $\mathbb{R}^3$ for the images of $e_1$ and $e_2$, since we could create some other vectors in the kernel of $F$. The restriction is that the family $(a,b)$ should be independent (think of the Rank–Nullity Theorem).
We then have the general solution of your problem: a linear mapping $F:\mathbb{R}^4\longrightarrow\mathbb{R}^3$ satisfies all your requirements
if and only if there exists an independent family $(a,b)$ of vectors of $\mathbb{R}^3$ such that:
$$\forall(x,y,z,t)\in\mathbb{R}^4,\quad F(x,y,z,t)=(x+z-t)a+(y-2z+t)b.$$
For example, the linear mapping $F:\mathbb{R}^4\longrightarrow\mathbb{R}^3$ defined by:
$$\forall(x,y,z,t)\in\mathbb{R}^4,\quad F(x,y,z,t)=(x+z-t,y-2z+t,0)$$
fulfills all your requirements. The matrix of this $F$ in the standard bases of $\mathbb{R}^4$ and $\mathbb{R}^3$ is:
$$[F]_{\text{std}(\mathbb{R}^4),\text{std}(\mathbb{R^3})}=\begin{pmatrix}1&0&1&-1\\0&1&-2&1\\0&0&0&0\end{pmatrix}.$$
But there are lots more (I actually gave you all of them!).
The example given by YourAdHere corresponds to the case $a=(3,3,0)$ and $b=(1,2,0)$ (which is, in some respect, not the simplest).
Now, given a $3\times4$ matrix, how can you determine whether its the matrix (in the standard bases of $\mathbb{R}^4$ and $\mathbb{R}^3$) of a linear mapping satisfying your requirements? Easy:
- Check that the two first columns are not proportional,
- Check that $C_3=C_1-2C_2$ and $C_4=-C_1+C_2$.
This fact is easily seen, e.g., from the form of the matrix I gave above.
Best Answer
Let $\Bbb R^2$ have standard basis $\{e_1,e_2\}$ and $\Bbb R^3$ have standard basis $\{f_1,f_2,f_3\}$.
Then, $T(e_1) = 1f_1+2f_2+0f_3$ and $T(e_2) = 2f_1+(-5)f_2+7(f_3)$.
Therefore, the matrix representation for $T$ is $\begin{bmatrix}1&2\\2&-5\\0&7\end{bmatrix}$.
Let $\Bbb P^n$ have standard basis $\{1,t,t^2,\cdots,t^n\}$.
Then, $T(1) = 0 \cdot 1 + 0 \cdot t + 0 \cdot t^2 + \cdots + 0 \cdot t^{n-1} + 0 \cdot t^n$, and so on.
Also, $T(t^n) = 0 \cdot 1 + 0 \cdot t + 0 \cdot t^2 + \cdots + n \cdot t^{n-1} + 0 \cdot t^n$.
Therefore, the matrix representation is:
$$\begin{bmatrix} 0&1&0&\cdots&0\\ 0&0&2&\cdots&0\\ 0&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&n\\ 0&0&0&\cdots&0 \end{bmatrix}$$