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Linear Algebra proofs

  1. Jun 14, 2006 #1
    Hello. I've been reading through Friebderg's Linear Algebra and doing some of the problem sets. I can do the problems with little problem, but I want to make sure my proofs are okay looking. These are pretty basic though. I'm pretty sure I got the first one, just want to make sure that's right. The second one I need a bit of suggetsions on.

    Thank you : )

    1: In any vector space V, show that [tex](a+b)(x+y) = ax + ay + bx + by[/tex], where x,y [tex] \in V[/tex] and a,b [tex] \in R [/tex] (a, b scalars.)

    Proof:
    Let c = a+b

    c(x+y) = cx + cy By Axiom 7 that states "a(x+y) = ax + ay, scalar a, and x,y, in V".

    Next, I said c = a+b, so by another Axiom that states (a+b)x = ax + bx

    (a+b)x + (a+b)y = ax + bx + ay + by
    That should complete the proof. I think that's right.

    Next one is one I'm not too sure if myproof is correct/clear. Any comments/suggestions would be appreciated.

    2: If W is a subspace of V and [tex] x_1, x_2, ...x_n \in W [/tex] Prove that [tex] a_1 x_1 + a_2 x_2 ..... a_n x_n \in W [/tex] ([tex] a_n [/tex] is a scalar real number.)

    Proof:

    By a theorem earlier in the chapter, I first noted that for W to be a vector space V, then

    A:[tex] x +y \in W | x,y \in W [/tex]
    and
    B:[tex] ax \in W [/tex] whenever a is a scalar and [tex] x \in W [/tex]

    So, here was how my poorly constructed argument went.

    I already know that W is a subspace of V, I just want to show that linear combination above is also in W.

    Let me arbitrarily choose a vector in that linear combination, I'll denote it [tex] a_i x_i [/tex], 1 <= i <= n. I know that this must be in W, as that is stated by B in the previous theorem mentioned.

    So, we know that each of those [tex] a_i x_i [/tex] in the linear combination themselves are vectors in W. This gets me a step closer to showing the sum itself is.

    Now, Let me denote any [tex] a_i x_i = y [/tex]. Now, I know that [tex] y_1 + y_2 \in W [/tex] as that is stated by the part A of the previously mentioned theorem. So, I can now say that the sum of any two of the [tex] a_i x_i [/tex] in that linear combination is in W. Now, If I know that the sum of two of those is in W, I can consider that a new element of W called Z then add another [tex] a_i x_i [/tex]. This will also be in W, as it is the sum of two vectors already in W. It is clear that continuing this adding process of the elemnts of W is producing elements within W. So, I can continue this repetively until all the elements in that linear combination are summed and I have a vector X which is in the element W that is that linear combination.

    This sounds rather wordy, and I'm concerned that it isn't rigorous enough. Anyone have any ideas? Does my argument even work? Thanks!
     
    Last edited: Jun 14, 2006
  2. jcsd
  3. Jun 14, 2006 #2

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    The first proof looks good. Two suggestions. First, you don't need all those intermediate variables. For example, instead of defining c=a+b and saying c(x+y)=cx+cy, just treat a+b as a scalar (which it is), so that you can say (a+b)(x+y) = (a+b)x+(a+b)y. You do this a few other places as well. Second, to make the second part more rigorous, you should use induction. That is, assume the sum of the first k elements belongs to W and show this means the sum of the first k+1 belong to W as well. Then show the initial case and you'll be done.
     
  4. Jun 14, 2006 #3
    Thank you for the reply. I like the idea of trying it by induction. I'll give htat a shot. I'll work on it. If I make any progress, I'll post it. Thanks for the suggestion : )
     
  5. Jun 15, 2006 #4
    You're first proof is incorrect, but you had the right idea. I'll quote the vector space axioms straight from the text for you(from the 4th edition).

    Here's a proof.

    Suppose we have some vectors x, y in V and scalars a, b in R, then
    (a + b)(x + y) = (a + b)x + (a + b)y by (VS 7)
    = (ax + bx) + (ay + by) by (VS 8)
    = ax + (bx +(ay + by)) by (VS 2)
    = ax + ((bx + ay) + by) by (VS 2)
    = ax + ((ay + bx) + by) by (VS 1)
    = ax + ay + bx + by.
    Since x, y in V and a, b in R were arbitrary, this holds for all x, y in V and all a, b in R.

    Try to follow this, the only tricky part might be where (VS 2) is used. Also the question in my text says any field F(maybe a different edition). Hope that helps.

    Edit yet again: Another problem(if you have the 4th edition) that is similiar to this one in the sense you are proving really basic properties about vector spaces, is #9 where it asks you to prove Corollaries 1 and 2 of Theorem 1.1 and Theorem 1.2(c). This is good practice.
     
    Last edited: Jun 15, 2006
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