# Quantum/linear algebra and vector spaces

1. Jan 20, 2009

### saraaaahhhhhh

I have never taken linear algebra, but we're doing some catch-up on it in my Quantum Mechanics class. Using teh Griffiths book, problem A.2 if you're curious.

Please explain how to solve this, if you help me. If you know of resources on how to think about this stuff, I'd greatly appreciate the assistance.
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Consider the collection of all polynomials (with complex coefficients) of degree less than N in x.
a.) Does this set constitutte a vector space (with the polynomials as vectors)? If so, suggest a convenient basis and give the dimension of the space. If not, which of the defining properties does it lack?
b.) What if we require that the polynomials be even functions?
c.) What if we require that the leading coefficient (i.e., the number multiplying x^(N-1)) be 1?
d.) What if we require that the polynomials have the value 0 at x=1?
e.) What if we require that the polynomials have the value 1 at x=0?

My attempt at a solution is:
a.) Yes, it doesw consitute a vector space. Any vector would be an ordered N-tuple (?) constructed from teh coefficients. How would I answer about the dimension of the space? Does it have N dimensions? I'm not sure if I understand what is being asked.
b.) Nothing changes?
c.) Then you'd have a pretty boring vector space? But I think all the rules would work.
d.) Still a vector space?
e.) Still a vector space? I don't see why that would change, I must be missing something.

2. Jan 21, 2009

### CompuChip

So first of all, what are the axioms for a vector space?

a) Yes it does. So they ask for a basis. That means, give a bunch of polynomials in x, so you can express every polynomial of degree less than N in x as a linear combination. Don't think too hard, it's really straightforward :P The dimension is the minimal number of such functions that you need (and it's easy to check once you have a set, just check if you can indeed express every polynomial as a linear combination in your basis, and if you take one out you can find an example where this no longer works).
Note that every vector is now a function of x, although you are right that it is isomorphic (i.e. bijectively mapped and equivalent as a vectorspace) with RN by writing down a vector of N coefficients, in a particular basis you have chosen. This is not unique though: just post your basis and I will give you another one, which is equally good (in terms of being able to express all the functions) but where the coefficients for the same function look entirely different in both.
Note how you have to let go of the idea of vector as a set of numbers, and think of it as an abstract "point" in a vector space... it's similar to having a vector in RN: although a given vector is a unique point in the space, the coordinates you write down between the brackets which you call "the vector" are actually dependent on the basis you have chosen for the vector space. [Perhaps this confuses you now, if you want to ever seriously do something like quantum physics, think about it ]

b) Why does nothing change? You need to show this: check the properties of a vector space (is 0 even? is the sum of two even polynomials even? ...)

c) Again, check the rules. I wouldn't call it a boring vector space... (boring is a subjective word, but vector space is not :P )

d) Show it! Check the rules!

e) Try it! Check the rules!

Last edited: Jan 21, 2009