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- anonymous

is the following set a subspace of v3
{x,y,z|x,y=0}

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- anonymous

is the following set a subspace of v3
{x,y,z|x,y=0}

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- anonymous

there is a test for subspace. make sure it has a zero element (0,0,0) and is closed under multiplication (we dont have to test all the axioms of a vector space)

- anonymous

so it has the zero element, set z = 0. and it is closed under multiplication
c(0,0,z) = (0,0,cz) and (0,0,cz) is an element of { x, y , z | x , y = 0 }

- anonymous

Oh i left out one more thing.
we have to make sure addition is closed.
so if (0,0,z1) + (0,0,z2) show that it is an element of {x,y,z|x,y=0}
which is easy, because the sum is (0,0,z1+z2) and thats an element of {x,y,z|x,y=0}.
Here are conditions for linear subspace.
Theorem: Let V be a vector space over the field K, and let W be a subset of V. Then W is a subspace if and only if it satisfies the following 3 conditions:
1. The zero vector, 0, is in W.
2. If u and v are elements of W, then any linear combination of u and v is an element of W;
3. If u is an element of W and c is a scalar from K, then the scalar product cu is an element of W;

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- anonymous

http://en.wikipedia.org/wiki/Linear_subspace

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