Standard Basis Math: Find Point X w/ B-Coord Vector X

[EvLer]

I think I sort of understand this but need confirmation/correction:

We have this formula in the book: X = P(b)X' (point = point-matrix * coord vector).

So the follow-up problem is this:

given basis X1 = [1, 0]t X2 = [1, 1]t what point X has b-coord vector X' = [-3, 2]t.
Well, it is straight-forward, by formula one gets X = [-1, 2]t.

So it looks to me that this point is with respect to the standard basis, am I correct?

 

[HallsofIvy]

Assuming that by "b-coord" means with respect to this given basis, then you are looking for X'= -3[1, 0]+ 2[1, 1]= [-3+2, 2]= [-1, 2] in the standard basis.

 

[M98Ranger]

Yes, you are correct. The standard basis refers to the standard unit vectors [1, 0] and [0, 1], which are used to represent points in the Cartesian coordinate system. In this case, the given basis X1 and X2 are equivalent to the standard basis, so the point X with b-coord vector X' = [-3, 2]t will also be in the standard basis. This means that the coordinates of point X are [-3, 2] with respect to the standard unit vectors [1, 0] and [0, 1].