Rank and Nullity of a linear transformation

Okay so i have been breaking my head over this for a couple of hours but don't really know if I'm even doing the right thing.

Okay so we are looking at a transformation defined in the following way:

T$\displaystyle \left(\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{2 1}&a_{22}&a_{23}\end{array}\right)$ =$\displaystyle \left(\begin{array}{cc}2a_{11}-a_{12}&a_{13}+2a_{12}\\0&0\end{array}\right)$

They want to know if this is a Linear transformation, which as per my calculations it is. And then they want to know the nullity and the rank of the transformation. So since the dimension of vector space "V" is 2*3=6 the nullity and rank need to add up to this. Now I'm a little confused as to how to proceed with finding the null space and range, It seems that the null space needs the top row to be zeros and the bottom row can take on any values, but then how many matrices does that make? if it was just the 0 matrix then the null space would have dimension 0 but there are obviously other vectors, so my first question is what does the null space look like?

Then for the range, the thing that I think we are supposed to do is, take the basis for all 2X3 matrices and then transform each element, so this gave me 6 matrices, $\displaystyle \begin{array}{cc}1&0\\0&0\end{array} , \begin{array}{cc}-1&1\\0&0\end{array} , \begin{array}{cc}0&1\\0&0\end{array} $ with the remaining three matrices being the zero matrix.

So then since only the first 3 are linearly independent the rank is 3? which means the nullity must be three? But I don't know what the null space looks like so I don't know how to verify this. I don't even know if I'm doing this stuff right(Crying)!

Thank you so much for your help!