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Linear algebra, vector spaces (for quantum)

  1. Jan 20, 2009 #1
    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.
    ***
    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. jcsd
  3. Jan 20, 2009 #2

    Dick

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    I don't know where to start. If Griffith doesn't have the introductory material then start with the usual on-line reference. http://en.wikipedia.org/wiki/Vector_space The key point is that vector spaces are more then just n-tuples of numbers. They are sets of objects that combine 'like' n-tuples of numbers. To start with the first one, all of your polynomials look like p(x)=c0+c1*x+c2*x^2+...+cn*x^(N-1), right? Can you suggest a set of simple functions of x that you can combine with constant coefficients that are linearly independent (study that concept) with which you can make any polynomial of degree less than N? Yes, the dimension is N. For the second one, something does change. p(x) is even means p(x)=p(-x). Not all polynomials satisfy that. The dimension is lower. How much lower? Like I said, this is a big subject. Post these questions one at a time and you'll probably get more help.
     
  4. Jan 20, 2009 #3
    Thanks for the tip.

    I must be confused from what my teacher's notes are saying. He basically said the vectors would be defined as polynomials, like your p(x) above.
    And this sentence:
    "Can you suggest a set of simple functions of x that you can combine with constant coefficients that are linearly independent (study that concept) with which you can make any polynomial of degree less than N?"
    is nearly meaningless to me. I understand linear independence, but why would I use simple functions of x to make a polynomial?

    What do you mean that vector spaces are sets of objects that combine "like" n-tuples of numbers? A vector space is a set of vectors, right? Vector objects.

    Okay, so b.) has N/2 dimensions, then.
     
  5. Jan 20, 2009 #4

    Dick

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    You do use simple functions to make polynomials. You combine {1,x,x^2,...x^(N-1)} to make a general polynomial. And if you understand linear independence, great! They are linearly independent. So they are a basis for all polynomials. And they span a space that is a lot like the space of n-tuples. But they aren't "n-tuples", they are functions. Sort of right for b). If you have polynomials of degree less than 3, is the dimension 3/2?
     
  6. Jan 21, 2009 #5
    I see your point, but I don't know how to generalize the value for the dimension.

    From what you're saying: in part a.), the basis is simply the set of {1, x, x^2....x^(N-1)} and teh vectors are represented by the coefficients {c_0, c_1...c_(N-1)}. Okay, that makes sense.

    For part b.), the dimension is not N/2 but something like it...N/2 if N is even, and N-1/2 if N is odd? There may be a better way to say that. And the basis is now the set {1, x^2, x^4...x^(N...?)}, or something close? With coefficients {c_o, c_2, c_4...etc}? This would happen because only the even-exponent values in teh polynomial are even functions.

    For part c.), wouldn't the vectors span the space, because they are linearly dependent, because they are all just {1, 1, 1,...}? Then would the basis be 0? Would the dimension be 1?

    For part d/e I am still at a loss.
     
  7. Jan 21, 2009 #6

    Dick

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    Saying N/2 for N even and (N-1)/2 for n odd is just fine. For part c), review the definition of a vector space. If O is the set of polynominals with the coefficient of x^(N-1) equal to 1, is it closed under addition? If you add to elements of O, do you get an element of O?
     
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