Construction of a Linear Transformation question Let there be a linear transformation going from $\mathbb R^4$ to $\mathbb R^3$, such that $(1,0,1,0)$ , $(2,1,3,0)$ span ker T , and $(0,1,2)$ , $(3,1,2)$ span Im T. If such a transformation possible , construct it. 
Attempt at a Solution: after Gaussian elimination on the span of the kernel, I got $(1,0,1,0)$ and $(0,1,1,0)$. given the fact these two span the kernel, after performing the transformation on them we get the zero vector. These two then need to be expanded to $\mathbb R^4$, and here is my question...I was told to choose the vectors $(0,0,1,0)$ and $(0,0,0,1)$ . Is it mandatory for these two vectors to be used in expansion or any vectors independent on each other and the kernel spanning vectors would do? 
After this, I assign the vectors spanning the Im as results of the transformation for the last two vectors discussed....from there I perform the whole transformation on a vector of $\mathbb R^4$ , using the bases, and reach the transformation matrix required.  
 A: To answer your bolded question: it seems you mean to take the vectors $(1,0,1,0)$ and $(0,1,1,0)$, and find two more vectors to this extend this set and form a basis on $\mathbb R^4$. 
Is it necessary to use the two standard basis vectors $(0,0,1,0)$ and $(0,0,0,1)$?  Of course not.  You could have also used $(1,0,0,0)$ and $(0,0,0,1)$, or if you wanted to go crazy, $(1,2,3,4)$ and $(4,3,2,1)$.  As long as you find four linearly independent vectors, you have a basis for $\mathbb R^4$.  However, the standard basis vectors are usually easier to work with. 
Furthermore, whenever you have a linearly independent set, you can always extend that set to form a basis using some right choice of the standard basis vectors.  You can do the same thing using any other basis for $\mathbb R^4$, but as I said before, this choice is a matter of convenience.  In your words, any vectors independent of each other and the kernel would do.
Now, with $v_1=(1,0,1,0),v_2=(0,1,1,0),v_3=(0,0,1,0),v_4=(0,0,0,1)$ (or even with $v_2=(2,1,3,0)$), you can define your transformation by setting
$$
T(v_1)=T(v_2)=\vec 0\\
T(v_3)=(0,1,2)\\
T(v_4)=(3,1,2)
$$
That is, with respect to the above basis, $T$ is given by the transformation matrix
$$
A_T=
\begin{bmatrix}
0&0&0&3\\
0&0&1&1\\
0&0&2&2
\end{bmatrix}
$$
A: I just wanted to mention the following:
1) You didn't necessarily have to use Gaussian elimination on the two given vectors that span Ker T; since they are not scalar multiples of each other, they are linearly independent and therefore form a basis for Ker T.
2) When you did use Gaussian elimination on these vectors, this actually gave you a way to extend them to a basis for $\mathbb{R}^4$:  
If $A$ is the matrix with these two vectors as its rows, then solving the equation $Ax=0$ (which is essentially what you did) and taking a basis for this solution space will give you two more vectors which, together with the two given vectors, form a basis for $\mathbb{R}^4$.   (Here this would give the vectors (1,1,-1,0) and (0,0,0,1), for example.)
