Find the rank and nullity of the following matrix Find the rank and the nullity of the following linear map from $\mathbb{R}^{4}$ to $\mathbb{R}^{3}$ 
$\left(x,y,z,t\right)\rightarrow(x-t,z-y,x-2y+2z-t)$
I understand how to find the rank and nullity, since the nullity is the dimension of the null space and the rank is the dimension of the column space. But I am having difficulty putting it into a matrix to solve. 
And can someone also explain why we can take a linear map and essentially say that it is the "same" structurally as a the corresponding matrix. How do isomorphisms link into this idea ? 
From this how would you know if it is onto ?

There is a systematic way to construct a matrix representation of a linear transformation once ordered bases for the domain and range are chosen, however, for maps between $\mathbb{R}^n$ and $\mathbb{R}^m$ it is easy to figure out the matrix by inspection, if you understand how matrix multiplication works. 
You want a matrix that will multiply on the left times a $4\times 1$ column vector and will give back a $3 \times 1$ column vector, so you know that the matrix must be a $3 \times 4$ matrix.
The first row of the matrix times the vector $\begin{bmatrix}x\\y\\z\\t\end{bmatrix}$ must equal the first component of the output vector, i.e. $x-t$. Similarly for rows 2 and 3. Thus it is easy to construct the desired matrix as
$$
\begin{bmatrix}
1&0&0&-1\\0&-1&1&0\\1&-2&2&-1\end{bmatrix}
$$
