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Linear algebra

  1. Feb 28, 2007 #1
    I am in a problem seminar class and I have not taken Linear Algebra in over 4 years so I am having alot of problems with this. Please help....:eek:
    1. The problem statement, all variables and given/known data
    Let P be the set of all polynomials with real coefficients and of degree less than 3. Thus,
    P = {f:f(x)= a(sub0) +a(sub1)x +a(sub2)x^2, a(sub i ) is in the reals}
    P is a vector spaces over the field of Reals under the usual opperations of addition and scalar multiplication of polynomials.

    Let V = {f element of P : f(-2) =f(1)}

    Find a basis for V .

    2. Relevant equations

    Not sure

    I know that to be a basis, that the set must be linearly independent and span V.:confused:

    3. The attempt at a solution

    I proved in the first part of this problem that V is a subspace of P :smile:

    I also said that in order for f(-2) = f(1),
    a(sub0) - 2a(sub1) +4a(sub2) = a(sub0) + a(sub1) + a(sub2)
    -3a(sub1) + 3a(sub2) so
    a(sub1) = a(sub2)
    not sure if this is right or not....
    so I randomly chose 2 different elements of this set. and showed that they were linearly independent.
    the elements I chose were v(sub 1) (x)= 1 + 3x +3x^2 and v(sub 2 ) (x) = 0 +1x + 1x^2
    Now I found a theorem in my linear algebra book that says;
    "Let H be a subspace of a finite-dimentional vector space V. Any linearly independent set in H can be expanded, if necessary to a basis for H. Also H is finit dimentional and dim H < or = dim V "

    Now I know that dim P = 3
    Thus dim V < or = 3. and since the set of two numbers I used above were linearly independent, they should be able to be expanded to form the basis, but my text does not tell me how to do this....:frown:

    Am I going about this wrong? I dont really understand how to find span V either.

    Please point me in the right direction!!!
    Last edited: Feb 28, 2007
  2. jcsd
  3. Mar 1, 2007 #2

    Your answer is excellent.
    I think you should, as a further exercice, take an (any) element of V and see how you can express it as a linear combinaition of v(sub 1) and v(sub 2).

    In other words: find out a and b such that

    v(given) = a v(sub 1) + b v(sub 2) (for any x)

  4. Mar 1, 2007 #3
    I would start representing the polynomials by the coefficients

    So you are getting the space to be (represented by) the subspace of vectors

    (x,y,z) where x=y. Find a basis of this space.
    Last edited: Mar 1, 2007
  5. Mar 1, 2007 #4


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    The set of all polynomials of degree less that or equal to 3 is
    [tex]\{ax^3+ bx^2+ cx+ d\}[/tex]
    which has dimension 4. As gammamc said, you can think of that as simply {(a,b,c,d)}. Your subspace is defined by the single equation f(-2)= f(1) which should reduce the dimension by 1: your subspace has dimension 3 so you are looking for 3 basis vectors.
    f(-2)= -8a+ 4b-2c+ d and f(1)= a+ b+ c+ d. The requirement that f(-2)= f(1) is -8a+ 4b- 2c+ d= a+ b+ c+ d or -9a+ 3b- 3c= 0 which reduces to -3a+ b- c= 0 or c= -3a+ b. Here's a method I like to use: take each coefficient equal to 1 in turn, the other 0 if possible and use that to solve for the remaining coefficients. Here if we know a and b, we can solve for c:
    If a= 1, b= 0, d= 0, what is c? If a= 0, b= 1, d= 0, what is c? If a= 0, b= 0, d= 1, what is c? Those give you the three basis vectors.
  6. Mar 1, 2007 #5


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    It's the polynomials of degree less than 3.
  7. Mar 1, 2007 #6
    why (x,y,z) where x=y, wouldny it be y=z?

    and I am not really sure how to find a basis other than picking random elements like I did above...
  8. Mar 1, 2007 #7
    thanks for the encouragement, but I dont feel like I am getting anywhere closer to finding a basis...:frown:
    so pick another random element such as (4,5,5) and find out how to write it in terms of v(sub1) and v(sub2)?

    4 + 5x + 25x^2 =a(1 + 3x +3x^2) + b(0 +1x + 1x^2)
    Now I see that a=4 since we need to have a constant of 4 so I get
    4 + 5x + 25x^2 =4(1 + 3x +3x^2) + b(0 +1x + 1x^2)
    4 + 5x + 25x^2 =(4 + 12x +12x^2) + b(0 +1x + 1x^2)
    this leaves me with
    -7x+13x^2 = b(0+x+x^2)
    If b=7, this will get rid of the 7x but it will leave me with 6x^2
    If b=13 this will get rid of the 13x^2 but will leave me with -20x...:grumpy:
    What did I do?
    Does this mean that this new number is also linearly independent? and can be added to the set?
  9. Mar 1, 2007 #8
    Thanks I was confused for a minute thinking that I was supposed to include degree 3 too.....
  10. Mar 1, 2007 #9


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    P has dimension 3. V is a subspace of P, so it's dimension is less than or equal to 3. (By the way, do you see why V really is a subspace, and not just some subset?) V is clearly a proper subspace, i.e. it is not all of P, because not every element of P has the same value at -2 and at 1, e.g. the polynomial f(x) = x. Hopefully your book also tells you that (for finite dimensional vector spaces) the dimension of V equals 3 iff V = P, otherwise the dimension of V is strictly less than 3. So V has dimension 0, 1, or 2. You've found a linearly independent subset of V containing 2 vectors. Your book should also tell you that this means that the dimension of V must be greater than or equal to 2. By this point, you should be able to deduce that the dimension of V must be exactly 2, and hence your linearly independent pair of vectors must also be a basis for V, i.e. it does span V.

    This is sort of an abstract way to show that you've found a basis. The more concrete way would be to show directly that the set you've found is a spanning set, i.e. show exactly how to express an arbitrary vector in V as a linear combination of v1 and v2. Now you've determined, I assume correctly, that V can be defined by the condition a1 = a2, so an arbitrary element of V looks like

    b + ax + ax2

    which we'll just write as (b,a,a). You have (1,3,3) and (0,1,1). Find c and d such that

    (b,a,a) = c(1,3,3) + d(0,1,1)

    You get equations:

    b = c
    a = 3c + d
    a = 3c + d

    right? Now it's really easy to solve for c and d in terms of a and b, so you're done.
    You made two mistakes here. One is that you started with (4,5,5) but then started working with 4 + 5x + 25x2. (4,5,5) is of course just 4 + 5x + 5x2. The main mistake you made though is that you don't want to find how to express one particular element of V as a linear combination of v1 and v2. You want to show how any arbitrary element can be expressed as a linear combination of your two basis vectors.
  11. Mar 1, 2007 #10

    I could not undersand clearly the meaning of your last post.
    However, maybe this reminder could help:

    two polynoms are equal for all values of x
    if and only if their coefficients are equal

    Your set V of polynoms is defines as {a0+a1x+x1x²}
    So, if you take

    P1 = 1
    P2 = x + x²

    you have a basis for V,
    and any polynomial in V is a combination of P1 and P2.
  12. Mar 1, 2007 #11
    thanks soooooo much!
    You are right the second way is sooooo much eisier!
  13. Mar 2, 2007 #12


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    :blushing: One of these days, I really have to learn to read!

    Of course, that makes it easier: the set is {ax2+ bx+ c}.
    That space has dimension 3 and one equation will reduce that by 1: we are seeking a basis for a subspace of dimension 2 and so 2 independent vectors in that subspace.

    f(-1)= f(2) means a- b+ c= 4a+ 2b+ c or 3a+ 3b= 0: b= -a.
    Taking a= 1, c= 0 gives, with b= -1, x2- x.
    Taking a= 0, c= 1, gives, with b= 0, 1.

    The two "vectors", x2- x and 1 form a basis for that subspace: any quadratic polynomial of the form f(x)= a(x2- x)+ c has the property that f(-1)= f(2) and, conversely, any quadratic polynomial having that property can be written in that form.
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