So that shows that the subspace is closed under scalar


Please show how the subspace satisfies both addition & scalar multiplication!

In each of the following exercises 8-17, we will denote by S the set of all vectors x = (x1, x2, x3) E R3 whose coordinates satisfy the given condition. In each case determine whether the given set S is a linear subspace of R3 and if so, compute its dimension.

1. x1 = x2 = x3

2. x2 = 2x1 and x3 = 3x1

Partial solution to 15:

{x = (x1, x2, x3) E R3: x2 = 2x1, x3 = 3x1}

So x = (x1, 2x1, 3x1)
= x1(1, 2, 3)

So that shows that the subspace is closed under scalar multiplication. And it obviously contains the zero vector: (0, 0, 0). What about addition though?---

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Algebra: So that shows that the subspace is closed under scalar
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