![]() Here is this expression is the same thing. V2, plus c3 times v3, all the way to ck times Product, what do I get? I get c1 times v1, plus c2 times This expression over here are completely identical. Vector c1, c2, all the way to ck, multiplied by this Another way to write thisĮxpression right there, is to say that a is equal to the So we're going to have n rows,Īnd we have k columns. If we assume that all of theseĪre a member of Rn, then each of these are going to have nĮntries, or it's going to an n by k matrix. So let me see I have some matrixĬ that looks like this, where its column vectors are If I had a matrix where theĬolumn vectors were the basis vectors of B- so let me write V1, plus c2 times v2, plus all the way, keep adding Linear combination of these guys, where these coordinatesĪre the weights. Literally means that I can represent my vector a as a ![]() All this means, by ourĭefinition of coordinates with respect to a basis, this Subspace, this is a k-dimensional subspace. So this is the coordinates ofĪ with respect to B are c1, c2, and I'm going to have kĬoordinates, because we have k basis vectors. ![]() Let's say I have some vectorĪ, and I know what a's coordinates are with (keeping in mind that a vector's coordinates in the second basis are its original coordinates in this basis) is applying the change of basis matrix to the coordinates of this vector in this same basis, we're getting the coordinates in the initial basis, not moving the vector, but getting it's coordinates knowing the change we did to the second base, so we were considering the second base the original base, but now we're applying the change on the coordinates.īasis B, and it's made up of k vectors. So the idea is, what we did to the initial basis to get the second basis, is change of basis,īut what we did to get the coordinates of a vector in the initial basis from its coordinates in the second basis, If we have a vector in the standard base, and we have its coordinates in this base, then to get its coordinates in a different base, we are not going to move it to that different base, but get its coordinates in that different base. SO what we're doing here exactly is not changing or moving the vectors from standard base to base B, but representing them in these bases. STransform.filename = input.I think the idea is that, C is the CHANGE OF BASIS matrix from standard base to base B. % Reorganize a single record of a location structure sTransform.line = input.location(1) In this case, the algorithm is the transformation of a single token location into a more user-friendly format. Note that tokenReorg is a function that focuses entirely on the algorithm. You have a function, tokenReorg, that takes a scalar structure as an input argument and separates out the location into three different fields, creating a new scalar structure. For presentation purposes, you may need a more user-friendly way of storing and displaying the data. SubStringData is a cell array of structure arrays, one cell for each file, and one structure for each token in the file. The location field contains the line number and the start and end columns of the token. The difference is essentially the difference between array(i) and array Applying arrayfun with a cell array as an input will perform the operation on each cell of the array as opposed to cellfun, which operates on the contents of each cell. ![]() Arrayfun is similar to cellfun but operates on one or more MATLAB arrays and on each element of an array.
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