Show that W is a subspace of R^3.

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

2) For any two vectors, [tex]$ w_1 $[/tex] and [tex]$ w_2 $[/tex] in W, [tex]$ w_1 + w_2 $[/tex] should also be in W.

That is, it should be closed under addition.

3) For any vector [tex]$ w_1 $[/tex] in W and for any scalar, [tex]$ k $[/tex] in V, [tex]$ kw_1 $[/tex] should be in W.

That is it should be closed in scalar multiplication.

The conditions are mathematically represented as follows:

1) 0[tex]$ \in $[/tex] W.

2) If [tex]$ w_1 \in W; w_2 \in W $[/tex] then [tex]$ w_1 + w_2 \in W $[/tex].

3) [tex]$ \forall k \in V, and \hspace{2mm} \forall w_1 \in W \implies kw_1 \in W[/tex]

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

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 [tex]$ w_1 = (x_1, y_1, z_1) $[/tex] and [tex]$ w_2 = (x_2, y_2, z_2) $[/tex] are in W.

That means: [tex]$ x_1 - 2y_1 + 5z_1 = 0 $[/tex] and

[tex]$ x_2 - 2y_2 + 5z_2 = 0 $[/tex]

We should check if the vectors are closed under addition.

Adding the two vectors we get:

[tex]$ w_1 + w_2 = x_1 + x_2 - 2(y_1 + y_2) + 5(z_1 + z_2) $[/tex]

[tex]$ = x_1 + x_2 - 2y_1 - 2y_2 + 5z_1 + 5z_2 $[/tex]

Rearranging these terms we get:

[tex]$ x_1 - 2y_1 + 5z_1 + x_2 - 2y_2 + 5z_2 $[/tex]

So, the equation becomes, 0 + 0 = 0

So, it s closed under addition.

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

This means [tex]$ x - 2y + 5z = 0 $[/tex]

[tex]$ kw_1 = kx - 2ky + 5kz $[/tex]

Taking k common outside, we get:

[tex]$ kw_1 = k(x - 2y + 5z) = 0 $[/tex]

The equation becomes k(0) = 0.

So, it is closed under scalar multiplication.

Hence, W is a subspace of [tex]$ \math{R^3} $[/tex].