how can i find a transformation matrix provided i have vectors in both frames?

Answer 1

Is that the one Keanu Reeves was after when he took on Darth Vader, while 20,000 leagues under the sea?

Addendum for Doug:
Nah, good effort, but I'm sure my one was right.

Second time for Dougie boy:
Ooh, I just love it when they put up a fight... :-)

Average of £100,000 a year, baby. Degree in economics and all around world's biggest ego... erm, I mean, 'brain'! You may whimper your apology while kissing the ring for forgiveness. That's an allusion to the Godfather movie by the way, not asking you to kiss my a.ss.... although it's funny how both seem applicable! :-)

Answer 2

Assume the vectors are x and X and are given as column vectors (with x in the domain space and X in the range space) consisting of x1, x2, ...xn, and X1, X2, ...Xn. Then you want a transform matrix A such that
A∙x = X Just let all of the 'off-diagonal' terms of A be zero and solve
A(i,i)*xi = Xi for the A(i,i) element and you're done.
Note that this is *exactly* the n-space equivalent of asking for the equations of all lines passing through a given point in 2-space. If you have more than one set of x and X, then repeat the above process using the next 'upper diagonal' of A to get the next set of coefficients (and 'wrapping around' the column index (mod n) to 'fill in' the lower triangular portion of A).
And *don't* forget to keep the previous coefficients of A in place while you solve for the next set of coefficients. Repeat as needed until you have all n given vector pairs mapped.

Yeah, it's a sh|tload and a half of arithmetic ☺

Again, note that in n-space you need n 'points' (vectors) to determine a unique transform, just like you need 2 points in 2-space to determine a unique line (which can be thought of as a linear transform of order 1).

HTH ☺

Doug

Addendum for Oliver T: You don't need linear algebra to say, "Would you like an order of fries with that." ☺