How to find the transformation matrix given two vectors and their particular transformations? I know this is probably a pretty simple thing to do, but I can't really wrap my head around it: I have two vectors in R^2, (1,2) and (1,0), whose transformations are (1,2,0) and (3,0,1) respectively. How do I find out the transformation matrix from this information? 

I know I could manually write down a new system of equations using the elements of the matrix as my unknowns, but I supposed that's too tedious to be the right solution.

Also, do basis transformation matrices have anything to do with this? I thought I could use a new matrix consisting of my two column vectors in R2 (representing a change of basis from the standard basis vectors), but I'm not sure if doing the same thing to my transformed vectors in R^3 makes any sense (can I group them up too? Would that be a change of basis in the range?). I'm sorry if I'm mixing up two unrelated things, but I had a hunch the problem might be related to that.

Thanks for the help!
You could try the following. First map the two vectors in R^2 to the standard basis vectors in R^2. Then find a mapping that maps the standard basis vectors in R^2 to the ones in R^3.

In particular, in your example, this would yield something like this: The matrix $A$ that maps the standard basis vectors to (1,2) and (1,0) is:

$$A = 
        \begin{bmatrix}
        1 &amp; 0 \\ 2 &amp; 1
        \end{bmatrix}
$$

The matrix $B$ that maps (1,0) and (0,1) to (1,2,0) and (3,0,1) is the following one:

$$B = 
\begin{bmatrix}
1 &amp; 3 \\ 2 &amp; 0 \\ 0 &amp; 1
\end{bmatrix}
$$

So, you first want to do the inverse of $A$ to map (1,2) and (0,1) to the standard basis vectors, and then apply $B$ to get to (1,2,0) and (3,0,1). So, the resulting matrix becomes:

$$BA^{-1} = \begin{bmatrix}
-5 &amp; 3 \\ 2 &amp; 0 \\ -2 &amp; 1
\end{bmatrix}$$
HINT....You can write this as a matrix equation. Let the transformation matrix be $M$

Then $$M\left(\begin {matrix}1&amp;1\\2&amp;0\end{matrix}\right)=\left(\begin{matrix}1.2&amp;3\\0&amp;0.1\end{matrix}\right)$$

Now post multiply by the inverse and calculate $$M=\left(\begin{matrix}1.2&amp;3\\0&amp;0.1\end{matrix}\right)\left(\begin {matrix}1&amp;1\\2&amp;0\end{matrix}\right)^{-1}$$

Now you can finish this.