# Math Help - Transition Matrix

1. ## Transition Matrix

Im a quite a bit behind in my studies and wanting someone to make sure im on the right track (unfortunately I study away from Uni so no direct contact with lecturer).

Let B be the standard basis for R3 and B' the basis {(1,0,0),(2,2,0),(3,3,3)}.

a) Find the transition matrix from B' to B.

From my readings at least in R2 space B is meant to be the identity matrix (unsure as why they call it the standard basis) as they are showing B to be e.g. {(1,0),(0,1)} in the examples. So the standard basis in R3 would be

B = {(1,0,0),(0,1,0),(0,0,1)}

From here we find u'1, u'2, and u'3 as like this. (u is from the standard basis)

u1 = 1(u1) + 0(u2) + 0(u3)
u2 = 2(u1) + 2(u2) + 0(u3)
u3 = 3(u1) + 3(u2) + 3(u3)

From here we can build the transition matrix. Am I on the right track or have I misunderstood the standard basis part of the question ?

(Apologies for not using latex, couldnt find out how to show the mutl single quote and number).

2. Originally Posted by gk99

3. Thread re-opened as per moderator discussion. I would echo the OP'er's comment, though: please no answers.

4. Originally Posted by gk99
Im a quite a bit behind in my studies and wanting someone to make sure im on the right track (unfortunately I study away from Uni so no direct contact with lecturer).

Let B be the standard basis for R3 and B' the basis {(1,0,0),(2,2,0),(3,3,3)}.

a) Find the transition matrix from B' to B.

From my readings at least in R2 space B is meant to be the identity matrix (unsure as why they call it the standard basis) as they are showing B to be e.g. {(1,0),(0,1)} in the examples.
You are using the word "matrix" incorrectly here. A basis is NOT a matrix and {(1,0), (0,1)} is not a matrix- it is a set of vectors. What I think you mean is that if you formed a matrix by using the basis vectors as columns, in the correct order, it would be the identity matrix. It is the "standard basis" because they correspond to the unit vectors along the coordinate axes in a Cartesian coordinate system.

So the standard basis in R3 would be

B = {(1,0,0),(0,1,0),(0,0,1)}
Yes, that is correct.

From here we find u'1, u'2, and u'3 as like this. (u is from the standard basis)

u1 = 1(u1) + 0(u2) + 0(u3)
u2 = 2(u1) + 2(u2) + 0(u3)
u3 = 3(u1) + 3(u2) + 3(u3)

From here we can build the transition matrix. Am I on the right track or have I misunderstood the standard basis part of the question ?
Yes, it looks like you are on the right track.

(Apologies for not using latex, couldnt find out how to show the mutl single quote and number).

5. in R^n, which is kind of the "proto-typical vector space", there are two ways of indicating which vector you mean:

given a basis, B = {b1,b2,...,bn}, and an arbitrary vector v, one can write:

v = v1b1 + v2b2 +... +vnbn (a linear combination of the basis vectors), or

v = (v1,v2,...,vn)[B] (a series of B-coordinates).

since the standard basis is {(1,0,...,0),(0,1,0,...,0),.....,(0,....,0,1)}, for the standard basis

these two ways of expressing a vector coincide, so that when someone writes

"let v = (v1,v2,...,vn) in R^n", it is easy to forget there is actually a choice of basis being implied.

physically (or geometrically), one can visualize the situation thus: space is just space, but

choosing an origin, and coordinate axes, is somewhat arbitrary. the usual choice of a perpendicular x-axis, y-axis and z-axis,

is a convention. it is perfectly possible to choose other axes, such as u = (1,0,0), v = (2,2,2), w = (3,3,3)

(where the right-hand sides are understood the be the "standard basis coordinates" of u,v, and w).

as such the transition matrix B'-->B maps u (which is (1,0,0) in B'-coordinates) to its value (1,0,0) in B-coordinates (the standard basis),

and similarly maps v (which is (0,1,0) in B'-coordinates) to its value (0,2,2) in B-coordinates,

ans so on with w.

6. Ackbeet thanks for reopening I'll remember for next time not to ask a direct question about my assignment.

Thanks for the replies, I know its only basic stuff but helped a lot.