2:17 PM
The algebraic topology room has so many folks. Maybe we can have some too?
Or maybe I should make one called Group, Ring, Field and Galois theory

9 hours later…
11:13 PM
Say I have two sets of basis vectors \$E = {e_1, e_2, e_3}\$ and \$E' = {e_1', e_2', e_3'}\$ for a vector space \$V\$. If I express the \$e_i'\$ as linear combinations of the \$e_i\$ and write the coefficients as columns of a matrix, does that matrix represent the change of basis matrix for the linear map \$e_i \mapsto e_i'\$ and if we multiply it by a vector's (say \$v = a_1 e_1 + a_2 e_2 + a_3 e_3\$) coordinate \$(a_1,a_2,a_3)^{T}\$, do I then get the co-ordinate of \$v\$ in the basis \$E'\$?

Why don't you just test it?

I've tried it out but it seems that it's not the case.
I found the matrix of \$e_i \mapsto e_i'\$ and multiplied it by the coordinate of \$v\$ but it seems that the inverse of that matrix multiplied by the coordinate gives \$v\$ as a linear combination of the \$e_i'\$, @Bye_World.

If you construct your matrix with columns of \$e_i'\$ in terms of \$e_i\$ then you'll get the change of basis matrix from E' to E, not the other way around.

11:29 PM
Would that then be the matrix corresponding to the linear map \$e_i \mapsto e_i'\$?

@Khallil
No. It's from the primed basis to the unprimed basis. The inverse of that matrix would take \$e_i \mapsto e_i'\$.