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Show that g1 and g2 form a basis for V

  1. Aug 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Let V be the space spanned by f1 = sinx and f2 = cosx.
    (a) Show that g1 = 2sinx + cosx g2 = 3cosx form a basis for V.
    (b) Find the transition matrix from B' = {g1, g2} to B = {f1, f2}.

    2. Relevant equations

    P = [[u'1]B [u'2]B...[u'n]B]

    3. The attempt at a solution

    I'm sorry but I don't know exactly how to start? Would the vectors f1 and f2 be represented in some form of coefficient matrix? Then how would that look? And then what would the vecors g1 and g2 look like in a coefficient matrix?

    Please. I know this is not much of a solution...but any help would go a long way.

    Thanks

    Derryck
     
  2. jcsd
  3. Aug 8, 2011 #2
    What is the definition of a space spanned by two vectors?

    If you find this difficult, try to answer to this question:
    Let's take [itex]R^2[/itex]. It is spanned by two vectors, for example [itex]\left[\begin{array}{c}1\\0\end{array}\right],\left[\begin{array}{c}0\\1\end{array}\right][/itex].
    How can you describe each element of [itex]R^2[/itex] in terms of these vectors?

    If you answer to this question, it is not difficult to understand HOW is built V.
     
  4. Aug 8, 2011 #3
    Ok, so to represent any element in R2 I would say (a, b) = t(1, 0) + v(0, 1)

    Are you then saying that (a, b) = cosx(1, 0) + sinx(0, 1)?
     
  5. Aug 8, 2011 #4
    You're almost there.

    Saying that [itex]R^2=span\{e_1=\left[\begin{array}{c}1\\0\end{array}\right],e_2=left[\begin{array}{c}0\\1\end{array}\right]\}[/itex] , means that, as you wrote,
    [item]\forall \mathbf{v}\in R^2,\,\, \mathbf{v}=\alpha\cdot e_1+\beta\cdot e_2,\,\,\alpha,\beta\in R[/item]

    In this case, the problem says that f_1=sin(x) and f_2=cos(x) span the space V. Hence, you can write that [item]\forall \mathbf{v}\in V,\,\, \mathbf{v}=\alpha\cdot f_1+\beta\cdot f_2=\alpha\cdot\sin(x)+\beta\cdot\cos(x)[/item]

    Now you should be on the right way...
     
  6. Aug 9, 2011 #5
    um? your R^2=SPAN/{E_1/LEFT[...is not turning into the nice representation you had previously accomplished? I'm not quite understanding:( I'm really sorry DiracRules...
     
  7. Aug 9, 2011 #6
    Sorry, maybe some problem with the forum.

    However, the vectors e_1=[1;0] and e_2=[0;1] spans R^2 means that every element of R^2 is a linear combination of e_1 and e_2, that is, you can write it as v=a e_1+b e_2.

    In you exercise, it says that the vectors f_1=sin(x) and f_2=cos(x) span the space V.
    What does this mean? How can you describe every element of your space?
     
  8. Aug 9, 2011 #7
    By V = asinx + bcosx...I knew/thought of this approach at the beginning. But I don't understand how I would prove that g1 and g2 form a basis for the same space?
     
  9. Aug 10, 2011 #8
    You have to show that every element of V can be written in terms of g_1 and g_2.
    If v is an element of V, you can write, as we said, v=a f_1+b f_2=a sin(x)+b cos(x).
    Try to do the same thing with g_1 and g_2 and see if you can get to the same form.

    To answer to the last question, try to think at f_1,f_2,g_1 and g_2 as two dimensional vectors: f_1=(sin(x), 0) f_2=(0, cos(x) ) g_1=... g_2=...
    This way you can find (through simple observations or trial-and-error :P ) the transition matrix.
     
  10. Aug 10, 2011 #9
    Ok. Now we getting somewhere:) I'm sorry for actually being so stupid! But I have no lecturer since this is a correspondence course I am doing...

    Yes...and I worked through it now and got the right answer in the textbook:)

    Thanks a lot Diracrules!

    Ciao
     
  11. Aug 10, 2011 #10
    Do not worry... just think easily and all will go well :D
     
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