EbolaPox
- 99
- 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 (a+b)(x+y) = ax + ay + bx + by, where x,y \in V and a,b \in R (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 x_1, x_2, ...x_n \in W Prove that a_1 x_1 + a_2 x_2 ... a_n x_n \in W (a_n 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:x +y \in W | x,y \in W
and
B:ax \in W whenever a is a scalar and x \in W
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 a_i x_i, 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 a_i x_i 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 a_i x_i = y. Now, I know that y_1 + y_2 \in W 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 a_i x_i 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 a_i x_i. 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!
Thank you : )
1: In any vector space V, show that (a+b)(x+y) = ax + ay + bx + by, where x,y \in V and a,b \in R (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 x_1, x_2, ...x_n \in W Prove that a_1 x_1 + a_2 x_2 ... a_n x_n \in W (a_n 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:x +y \in W | x,y \in W
and
B:ax \in W whenever a is a scalar and x \in W
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 a_i x_i, 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 a_i x_i 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 a_i x_i = y. Now, I know that y_1 + y_2 \in W 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 a_i x_i 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 a_i x_i. 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: