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.(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# Linear Algebra proofs

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