Let L: V--> W be a Linear Transformation from a vector space V into W. The image of the subspace V1 of V is: L(V1) ={β ϵ W | Ǝ α ϵ V where β = L(α)}. Show that L (V1) is a subspace of W.
3 Conditions must hold
1) $\displaystyle 0_{w} \text{ must be in } L(V_1) $
Well since its a linear transformation $\displaystyle L(x - x) = L(x) - L(x) = 0_{w} $, so $\displaystyle L(0_{v}) = 0_{w} $
2) closed under addition. If $\displaystyle x_1, x_2 \in L(V_1) $ then $\displaystyle x_1 + x_2 \text{ must also be in } L(V_1) $
Pf: Take $\displaystyle x_1, x_2 \in L(V_1) $, so $\displaystyle \exists v_1, v_2 \in V \text{ such that } L(v_1) = x_1 \text{ and } L(v_2) = x_2 $. So again, linear transformation, $\displaystyle L(v_1) + L(v_2) = L(v_1 + v_2) = x_1 + x_2 \text{ since } v_1+ v_2 \in V $, and so $\displaystyle x_1 + x_2 \in L(V_1) $
3)closed under taking scalar product.
Pf: Take $\displaystyle c \in \mathbb{F} $, then for any $\displaystyle x \in L(V_1) \exists v \in V \text{ such that } L(v) = x \text{ and so } c * x = c * L(v) = L(cv) $, since $\displaystyle cv \in V $, $\displaystyle cx \in L(V_1) $
Note: the first of jakncoke's requirement is, in some books, change to "the set is non-empty. Obviously if it contains the 0 vector, as jakncoke requires, it is non-empty. Conversely, if a nonempty set satisfies the other two requirements, since it is non-empty, there exist some vector, v. Because it satisifies (3), it also contains (-1)v= -v. And because it satisfies (2), v+ (-v)= 0 is in the set.