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Cool. I haven't gotten to the change of basis lecture video yet, or I might have tried that :)

by fferen 14y ago

btilly 14y ago
If you do the linear function/notation thing right, the idea of a change of basis comes for free. The basic idea is this. Suppose that we have a linear function F between two vector spaces (which might be the same), and we have coordinate systems on the vector spaces (which might be the same). Then we can take the elements of the first basis, v1, v2, ..., vn, and write out a matrix whose columns are F(v1), F(v2), ... F(vn) in the second coordinate system. This matrix uniquely represents F. The function F can be applied to a vector by writing out the vector in the first coordinate system, putting that on the right of the matrix, then doing matrix multiplication. You get an answer in the second coordinate system. And the operation of matrix multiplication turns out (if the coordinate systems match up) to be exactly the same as composition of functions. (This is no coincidence, this relationship is the motivation for matrix multiplication being defined as it is.) OK, with all of that mess, what is a change of basis matrix? It is simply the matrix you get for the identity function going from one coordinate system to another. Now in this case the matrix A = [[1 0 0] [1 1 1] [1 2 4]] is the matrix to change from the basis where a+bx+cx^2 is represented by [a b c] to the basis where it is represented by [a a+b+c a+2b+4c] (ie [p(0) p(1) p(2)]). Its inverse is the change of basis the other way. And B = [[0 1 0] [0 0 2] [0 0 0]] is, of course, in the [a b c] basis the representation of the function d/dx. Then A B A^(-1) represents change from pointwise coordinates to coefficients, then differentiate in coefficient coordinates, then change from coefficient coordinates back to pointwise coordinates. (Remember, the matrices are applied right to left, so you do A^(-1) first.) If you can keep that straight, you now understand change of basis matrices.

Cool. I haven't gotten to the change of basis lecture video yet, or I might have tried that :)

by fferen 14y ago

Cool. I haven't gotten to the change of basis lecture video yet, or I might have tried that :)

If you do the linear function/notation thing right, the idea of a change of basis comes for free. The basic idea is this. Suppose that we have a linear function F between two vector spaces (which might be the same), and we have coordinate systems on the vector spaces (which might be the same). Then we can take the elements of the first basis, v1, v2, ..., vn, and write out a matrix whose columns are F(v1), F(v2), ... F(vn) in the second coordinate system. This matrix uniquely represents F. The function F can be applied to a vector by writing out the vector in the first coordinate system, putting that on the right of the matrix, then doing matrix multiplication. You get an answer in the second coordinate system. And the operation of matrix multiplication turns out (if the coordinate systems match up) to be exactly the same as composition of functions. (This is no coincidence, this relationship is the motivation for matrix multiplication being defined as it is.) OK, with all of that mess, what is a change of basis matrix? It is simply the matrix you get for the identity function going from one coordinate system to another. Now in this case the matrix A = [[1 0 0] [1 1 1] [1 2 4]] is the matrix to change from the basis where a+bx+cx^2 is represented by [a b c] to the basis where it is represented by [a a+b+c a+2b+4c] (ie [p(0) p(1) p(2)]). Its inverse is the change of basis the other way. And B = [[0 1 0] [0 0 2] [0 0 0]] is, of course, in the [a b c] basis the representation of the function d/dx. Then A B A^(-1) represents change from pointwise coordinates to coefficients, then differentiate in coefficient coordinates, then change from coefficient coordinates back to pointwise coordinates. (Remember, the matrices are applied right to left, so you do A^(-1) first.) If you can keep that straight, you now understand change of basis matrices.