Change of basis in Linear Algebra
eli.thegreenplace.net
eli.thegreenplace.net
You can do that with a matrix operation (M*v = v'):
[0 0 1]___[0__ __[1
[0 1 0]___0__=__0
[-1 0 0]__1]__ __0]
In fact, this matrix rotates any vector 90 degrees about the y-axis.
Basis vectors can also be more abstract than that. For instance, they're useful in quantum mechanics for simplifying the schroedinger eq (sometimes from a second order diff eq to a first order one) by changing from a position basis to a momentum basis, in effect rewriting your derivatives from a different point of view.
Not necessary for every change of basis.
In that notation, a vector |a> which is basis independent has components <i|a> in some basis |i> , where (say) |i_1> , |i_2> are the basis vectors. Then to change to another basis <k| the "operator one" (which is the outer product of a basis, i.e. sum over i of |i><i|) is inserted to get <k|a> = sum over i <k|i><i|a> , which turns into the matrix mechanics.
(Each of the outer products e.g. |1><1| is a "projection operator" which projects a vector onto that basis vector. The sum of all of them projects onto the whole space spanned by the vectors, which is the same as doing nothing, which is therefore the identity operator.)
Once you get your head around the connections between coordinates (i.e. a_x = a_1) and the dot product with a basis vector (i.e. dot(i_1, a) = <i_1 | a> = <1|a>), this notation can make the whole thing intuitive and mechanical.
I have an explanation of this online at http://cs.marlboro.edu/talks/bra_ket.pdf .