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Math Help - subspace of a Linear Transformation?

  1. #1
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    subspace of a Linear Transformation?

    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.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: subspace of a Linear Transformation?

    3 Conditions must hold

    1) 0_{w} \text{ must be in } L(V_1)
    Well since its a linear transformation  L(x - x) = L(x) - L(x) = 0_{w} , so  L(0_{v}) = 0_{w}

    2) closed under addition. If  x_1, x_2 \in L(V_1) then  x_1 + x_2 \text{ must also be in } L(V_1)
    Pf: Take  x_1, x_2 \in L(V_1) , so  \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,  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  x_1 + x_2 \in L(V_1)

    3)closed under taking scalar product.
    Pf: Take  c \in \mathbb{F} , then for any  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 cv \in V ,  cx \in L(V_1)
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  3. #3
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    Re: subspace of a Linear Transformation?

    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.
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