Showing a vector space is linearly independent

It's probably too late, but here's what you have to do:

Let a(u-v) + b(u-2v+w) + c(v+w) = 0, for real numbers a, b, and c.
You have to show that a,b and c are necessarily all zero.
That would prove the independence of your set.

Here's the full answer:

Suppose a(u−v) + b(u−2v + w) + c(v + w) = 0, for some real a,b and c.

Grouping the coefficients of u, v, and w, gives us:
(a+b) u + (-a-2b+c) v + (b+c) w = 0.

The fact that {u,v,w} are independent means the three coefficients in the the red equation are zero, so we get:

a+b = 0 (Eq. 1)
-a-2b+c = 0 (Eq. 2)
b+c = 0 (Eq. 3)

Add all 3 equations and you get 2c = 0, which means c=0.
So Eq. 3 simplifies to b=0.
So Eq. 1 simplifies to a=0.
And we've done it. We've shown a,b and c are necessarily zero.
And that means the second set of vectors are independent.

***

Two big takeaways from this problem:
(1) The fact that this vector space was R^50 was a red herring.
(2) The independence of the second set is built on the independence of u,v and w.