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Homework Help: A question of 5 parts

  1. Feb 5, 2008 #1
    i have solved a full question
    (except the last part which i tried to come up with some formula but that didnt work out)

    i showed every step of the way

    unfortunetly i dont have the answers to this question
    so can you tell me if i solved wrong some subquestion
    and how to solve the 5th
    it a realy challenge for me

    1st page :


    2nd page:


    3rd page:

  2. jcsd
  3. Feb 5, 2008 #2


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    Science Advisor

    This looks good.

    Since V has dimension 1, and the intersection with U is a subspace of V, the intersection must be either the trivial subspace (consisting only of the 0 vector) or V itself.

    In problem 4, You have an orthogonal basis which is what they ask for. If they had asked for an orthonormal basis, you would need to divide each vector by its length.

    For problem 5, you are told that, for linear transformation, A, A2+ A= -I and are asked to find A-1 (so you may assume that A has an inverse).
    You have factored A2+ A= A(A+ I)= -I. I recommend you multiply both sides of that equation by A-1 and then solve for A-1.

    An equivalent way to do the same problem is to write the original equation as I= -A2- A and multiply both sides of that equation by A-1.
  4. Feb 6, 2008 #3
    so regarding the second page

    you say that if the the solution is one vector
    it always must be (0,0,0)

    so my answer is wrong

    where is my mistake?
  5. Feb 6, 2008 #4


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    I'm not sure where you made a mistake because it's not clear to me what you did. You have several matrices you have row reduced but you don't say where you got those matrices or what you are trying to do with them.

    You have, I think correctly, calculated that V has <4, -5, 2> as basis and you are told that U is spanned by <1, 1, 1> and <9, 0 7>. That means that any vector in their intersection, since it is in both U and V can be written both ways. If <4, -5, 2>a is a vector in the intersection, then it can also be written in the form <1, 1, 1>b+ <9, 0, 7>c. In other words, there must exist numbers, a, b, c, such that <4a, -5a, 2a>= <b+ 9c, b, b+ 7c>. We have the three equations 4a= b+ 9c, -5a= b, and 2a= b+ 7c. From the second equation, b= -5a. Putting that into the first equation, 4a= -5a+ 9c so 9a=9c and c= a. Putting that into the third equation, 2a= -5a+ 7a is true! Any vector of the form a<4, -5, 2> is a member of the intersection so <4, -5, 2> is a basis vector. (Of course, what we have shown is that U intersect V is just V itself.)
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