This makes sense out of Lay's statement, a few subsections later, that the imposition of a coordinate system makes a space of n vectors "behave like R^n". You must really think of vectors in a coordinate-free way to understand the difference between R^n and a space of n independent vectors
1)In that case the associated scalar Field's range will decide the range of vector space . 2) Please also explain how is Vector space is applicable to the relativistic Quantum Mechanics.
I have a question. I want to preface this with saying I think that this series of lectures is very helpful and I only am highlighting mistakes in order to help someone who might be confused by them. At aprox 15:30 in the video you say that p(x)=a sub 0 is called "the" zero polynomial. That makes it sound as if it should function as the identity element, but clearly it is not. So I am wondering what is the significance of the thing the you refer to as "the zero polynomial" ? Or did you mean to say that the "zero vector" is the"zero polynomial" where a sub 0 =0 ? Also at the 10:30 mark you state that matrix multiplication is not associative, which is not correct.
at 15:30 there are lots of mistakes. She never actually said what it means for polynomials to be a vector space, like what P_n is. Also the polynomials of degree 0 are the constant polynomials, not the zero polynomial.
Also, P_n denotes the set of polynomials of degree at most n, not the set of all polynomials. While I'm sure a lot of people find these videos helpful, there are moments where I don't think she actually knows what she's talking about.
Why would you use that book? Take Hubbard's, or Susan Colley's or Marsden's Book. Unless you are studying Biology or something like that, then Stewart's Book is fine.
I tried different values for the Practise exercise... and for u = [ - 2 and 7] and v = [3 and -1], u+v is a Vector space. Please elaborate. I'm confused
If you mean the practice example at 11:31 the thing is that even if for some u and v vectors they follow the rule and u + v still remains inside V with just one case of vectors u and v that when sumed dont belong to V then the axiom is broken. By just finding a case that doesnt follow the axiom(normaly called proof by counter example)you prove that the axiom isnt true, as for it to be true it has to be valid in ALL cases(no exceptions), else you cant generalize that property . Hope that helped.
Since vectors are essentially matrices, they can only be multiplied if the dimensions work out. If A is 2x3 and B is 3x4, then we can multiply AB but not BA.
