Question

Show that W is a subspace of R^3.

Answer 1

Answer:

Check the two conditions of Subspace.

Step-by-step 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 $\math{R^3}$.