Question

Show that W is a subspace of R^3.

Answer 1

Check the two conditions of Subspace.

Explanation:

If W is a Subspace of a vector space, V then it should satisft the following conditions.

1) The zero element should be in W.

Zero element can be different for different vector spaces. For examples, zero vector in
`$ \math{R^2} $` is (0, 0) whereas, zero element in
`$ \math{R^3} $` is (0, 0 ,0).

2) For any two vectors,
`$ w_1 $` and
`$ w_2 $` in W,
`$ w_1 + w_2 $` should also be in W.

That is, it should be closed under addition.

3) For any vector
`$ w_1 $` in W and for any scalar,
`$ k $` in V,
`$ kw_1 $` should be in W.

That is it should be closed in scalar multiplication.

The conditions are mathematically represented as follows:

1) 0
`$ \in $` W.

2) If
`$ w_1 \in W; w_2 \in W $` then
`$ w_1 + w_2 \in W $`.

3)
`$ \forall k \in V, and \hspace{2mm} \forall w_1 \in W \implies kw_1 \in W`

Here V =
`$ \math{R^3} $` and W = Set of all (x, y, z) such that
`$ x - 2y + 5z = 0 $`

We check for the conditions one by one.

1) The zero vector belongs to the subspace, W. Because (0, 0, 0) satisfies the given equation.

i.e., 0 - 2(0) + 5(0) = 0

2) Let us assume
`$ w_1 = (x_1, y_1, z_1) $` and
`$ w_2 = (x_2, y_2, z_2) $` are in W.

That means:
`$ x_1 - 2y_1 + 5z_1 = 0 $` and

`$ x_2 - 2y_2 + 5z_2 = 0 $`

We should check if the vectors are closed under addition.

Adding the two vectors we get:

`$ w_1 + w_2 = x_1 + x_2 - 2(y_1 + y_2) + 5(z_1 + z_2) $`

`$ = x_1 + x_2 - 2y_1 - 2y_2 + 5z_1 + 5z_2 $`

Rearranging these terms we get:

`$ x_1 - 2y_1 + 5z_1 + x_2 - 2y_2 + 5z_2 $`

So, the equation becomes, 0 + 0 = 0

So, it s closed under addition.

3) Let k be any scalar in V. And
`$ w_1 = (x, y, z) \in W $`

This means
`$ x - 2y + 5z = 0 $`

`$ kw_1 = kx - 2ky + 5kz $`

Taking k common outside, we get:

`$ kw_1 = k(x - 2y + 5z) = 0 $`

The equation becomes k(0) = 0.

So, it is closed under scalar multiplication.

Hence, W is a subspace of
`$ \math{R^3} $`.