Proving U is a subspace of R3

[Mampac]

Question:
Determine whether U is a subspace of R³, if

U = {(x, y, z)| x, y, z >= 0}
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So I looked up on Internet, and my professor's lecture notes/videos, and it seems like to do this proof it's enough to prove that the sets are closed under addition and multiplication.

The first one is easily doable (if we take 2 vectors then sum of two non-negatives is also a non-negative, same about scalar which is always non-negative and non-negative times non-negative is non-negative).

However I referred to the axioms of the space and realized there's no opposite vectors for say (1, 1, 1) in case of x = y = z = 1. Does this mean U is not even a space, so the answer is NO?

Thank you in advance!

 

[HallsofIvy]

In order for U to be a subspace of vector space V, U must be a subset of V, must be "closed under vector addition" and "closed under scalar multiplication".

Here V is \(\displaystyle R^3\), the vector space of triples, (x, y, z), of real numbers with the standard definitions of addition and scalar multiplication.

U is {(x,y,z)|x, y, z> 0}. Yes, that is a subset of \(\displaystyle R^3\).

"Closed under vector addition" means that if we add two thing in U the result is in U.
If (x, y, z) and (a, b, c) are triples of positive numbers then (x+a, y+b, z+ c) is also a triples of positive numbers because the sum of any two positive numbers is positive. Yes, U is "closed under vector multiplication".

"Closed under scalar multiplication" means that if we multiply something is U by any scalar (number) the result is in U. For a any real number a(x, y, z)= (ax, ay, az). Given that x, y, and z are positive, are ax, ay, az positive? (What if a= -1?)

Your statement that "scalar is always non-negative" is false. A scalar can be any real number, 0 or positive or negative.