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:

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

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

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

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:

$$BA^{-1} = \begin{bmatrix} -5 & 3 \\ 2 & 0 \\ -2 & 1 \end{bmatrix}$$

  • $\begingroup$ Ah, I see, that's sort of the kind of reasoning I tried to do. Though I still don't quite understand what B is supposed to be. Would it be a change of basis matrix too, just as A is? Or is it just a transformation matrix for the range? Thanks! $\endgroup$ – Matt24 Sep 16 '16 at 20:48
  • $\begingroup$ It's probably best to picture $B$ as a function that takes vectors from $\mathbb{R}^2$, and maps them to $\mathbb{R}^3$. As you are mapping from one vector space to a different one, this is not a basis transformation. I'm not quite sure what you mean by a "transformation matrix for the range". :) $\endgroup$ – arriopolis Sep 16 '16 at 20:51
  • $\begingroup$ Heh, by that expression I meant to say what you ended up saying: some sort of function that gives you the transformation vector. One last thing, if you don't mind: why can't I use this function B as my transformation matrix? Is it because it's in a non-standard basis (i.e. generated by (1,2) and (1,0))? $\endgroup$ – Matt24 Sep 16 '16 at 21:16

HINT....You can write this as a matrix equation. Let the transformation matrix be $M$

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

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

Now you can finish this.

  • $\begingroup$ Unfortunately this doesn't seem to be right, the question talks about vectors in $\mathbb{R}^2$, this answer instead changes them to vectors in $\mathbb{R}^2$. $\endgroup$ – wilsnunn Sep 30 '18 at 14:05

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