verify that S=the x "axis" =((x,y):y=0) is a subspace of V=R^2

Answer 1

Clearly, (0, 0) is an element of S.

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Show that S is closed under addition:

Let (x₁ , 0) and (x₂ , 0) be elements of S

(x₁ , 0) + (x₂ , 0) = (x₁ + x₂ , 0) is also in S.

So, S is closed under addition.

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Show that S is closed under scalar multiplication:

Let c be a scalar (that is, a real number) and let (x₁ , 0) be an element of S.

c(x₁ , 0) = (cx₁ , 0) which is in S.

So, S is closed under scalar multiplication.

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Since S is closed under addition and scalar multiplication and since (0,0) is in S, S is a subspace of R².