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Linear algebra vector space

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data
    I uploaded the problem because its easier to see.

    2. Relevant equations

    3. The attempt at a solution

    The solution is there from a practice exam. I don't understand the x,y,z part and how it fulfills the multiplication requirement of a vector space. Could someone please explain to me why it does?

    Attached Files:

  2. jcsd
  3. Aug 10, 2010 #2
    In the solution they are handling the addition and multiplication at the same time.

    They define u and v as vectors in the space, then show that u + dv (where d is any real) is also in the space.

    So after multiplying and adding the vectors, you have a new vector:

    [tex] \left(\begin{array}{cc} 4(a + da') + 3(b + db')\\0\\(c + dc') - 2\\(a + da') + (b + db')\end{array}\right) [/tex]

    So now all we have to do is show that all three (a+da'), (b+db'), (c+dc') are real numbers. Since a,b,c,d are all real, this is obviously the case, and hence the space is closed under addition and multiplication.

    In the solution, they converted these expressions to (x,y,z), but that's unnecessary.
  4. Aug 10, 2010 #3
    oh! That makes sense, i didn't see where they came from. Thank you
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