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Polynomial subspaces

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Let [itex]M[/itex] be a subspace of the vector space [itex]\mathbb{R}_2[t] [/itex] generated by [itex]p_1(T)=t^2+t+1[/itex] and [itex]p_2(T)=1-t^2[/itex], and [itex]N[/itex] be a subspace generated by [itex]q_1(T)=t^2+2t+3[/itex] and [itex]q_2(T)=t^2-t+1[/itex]. Show the dimension of the following subspaces:[itex] M+N[/itex], [itex]M \cap N[/itex], and give a basis for each.


    2. Relevant equations



    3. The attempt at a solution
    I have tried the following: if I take the linear combination of [itex]p_1[/itex] [itex]p_2[/itex] [itex]q_1[/itex] [itex]q_2[/itex], I get [itex](a+b+c+d)t^2 + (a+2c-d)t +(a+b+3c+d).[/itex] And a basis of this polynomial is [itex]\{1,t,t^2\}[/itex], which means the dimension of M+N is 3.

    And if M and N are finite dimension subspaces then [itex]dim(M+N)=dim M + dim N- dim(M \cap N)[/itex]. The diemnsion of the subspace generated by p1 and p2 is 2, and so is the dimension of the subspace generated by q1 and q2. Am I right? But then from this [itex]dim(M+N)=dim M + dim N- dim(M \cap N)[/itex] I get that [itex](M \cap N)[/itex] has a dimension of 1.

    Thank you!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 13, 2011 #2

    Deveno

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    Science Advisor

    1. how do you conclude that dim(M+N) = 3? clearly it's ≤ 3, but how do you know that we need all 3 basis elements to describe something in M+N? it's not immediately clear that span({p1,p2,q1,q2}) is all of R2[t].

    2. before you can conclude dim(M) and dim(N) = 2, you should show that {p1,p2} and {q1,q2} are actually linearly independent. this isn't hard to do.

    3. assuming you do (1) and (2) above, then you are in a position to use

    dim(M+N) = dim(M) + dim(N) - dim(M∩N).

    finding the dimensions isn't really the hard part. finding the basis elements is. what would an element of M∩N look like?
     
  4. Nov 13, 2011 #3
    1. If I take the linear combinations of p1 p2 q1 q2 as I have written and Im not wrong I think the basis {t^2 t 1} is ok.

    2. I found out that {p1,p2} and {q1,q2} are linearly independent, because a*p1+b*p2=0 s only solution is the trivial soulution, same for {q1,q2}. So dim(M)=2 and dim(N)=2.

    3. An element in [itex]M \cap N[/itex] is something which is in the span({p1,p2}) and in the span({q1,q2}) at the same time. But how do I find a basis ?
     
  5. Nov 13, 2011 #4

    Deveno

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    Science Advisor

    for span({p1,p2,q1,q2}) to equal R2[t], every polynomial must be in the span.

    if we have any abitrary polynomial At2 + Bt + C, in the span, this means that we have:

    a+b+c+d = A
    a+2c-d = B
    a+b+3c+d = C

    this is equivalent to:

    [tex]\begin{bmatrix}1&1&1&1\\1&0&2&-1\\1&1&3&1 \end{bmatrix}\begin{bmatrix}a\\b\\c\\d \end{bmatrix} = \begin{bmatrix}A\\B\\C \end{bmatrix}[/tex]

    now, it isn't obvious that the 4x3 matrix actually has rank 3. i think you might have made a lucky guess.

    as for M∩N, suppose we have a(1+t+t2) + b(1-t2) = c(1-t+t2) + d(3+2t+t2). what can we say about a,b,c and d?
     
  6. Nov 13, 2011 #5
    a(1+t+t2) + b(1-t2) = c(1-t+t2) + d(3+2t+t2) is true if

    a-b=c+d
    a=2d-c
    a+b=c+3d

    What do I do now?
     
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