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Math Help - Prove that the inverse image of V is a subspace in X.

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    Prove that the inverse image of V is a subspace in X.

    Prove that if f is a linear transformation from a vector space X to a vector space Y, then for any subspace V in Y, f^{-1}[V] is a subspace in X.
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    Quote Originally Posted by mathwizard View Post
    Prove that if f is a linear transformation from a vector space X to a vector space Y, then for any subspace V in Y, f^{-1}[V] is a subspace in X.
    Let \bold{x},\bold{y}\in f^{-1}[V] and a\in F, the base field for the vector spaces. It means by definition that f(\bold{x}),f(\bold{y}) \in V. Since V is a subspace it means f(\bold{x})+f(\bold{y}) \in V. Therefore, since f is a linear transformation it means f(\bold{x}+\bold{y}) \in V which means \bold{x}+\bold{y} \in f^{-1}[V]. Thus, f^{-1}[V] is closed under vector addition. Likewise, kf(\bold{x}) \in V which means k\bold{x}\in f^{-1}[V]. Thus, f^{-1}[V] is closed under scalar multiplications. All the other properties for being a vector space are satisfied because f^{-1}[V]\subseteq X. Thus, f^{-1}[V] is a vector space over F.
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    Quote Originally Posted by ThePerfectHacker View Post
    Let \bold{x},\bold{y}\in f^{-1}[V] and a\in F, the base field for the vector spaces. It means by definition that f(\bold{x}),f(\bold{y}) \in V. Since V is a subspace it means f(\bold{x})+f(\bold{y}) \in V. Therefore, since f is a linear transformation it means f(\bold{x}+\bold{y}) \in V which means \bold{x}+\bold{y} \in f^{-1}[V]. Thus, f^{-1}[V] is closed under vector addition. Likewise, kf(\bold{x}) \in V which means k\bold{x}\in f^{-1}[V]. Thus, f^{-1}[V] is closed under scalar multiplications. All the other properties for being a vector space are satisfied because f^{-1}[V]\subseteq X. Thus, f^{-1}[V] is a vector space over F.
    Thanks for your input; your proof confirms my belief that the same proof in my linear algebra textbook is incorrect; it messed up on x & y and f(x) & f(y) (and it wasted me a long time to try to make sense out of that flawed proof ).

    PS: I think there's one part that your proof is incomplete; it hasn't showed that the zero vector is in f^{-1}[V].
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    Quote Originally Posted by mathwizard View Post
    PS: I think there's one part that your proof is incomplete; it hasn't showed that the zero vector is in f^{-1}[V].
    Yes it has. If \bold{x}\in f^{-1}[V] we proved k\bold{x} \in f^{-1}[V]. Now let k=0.
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    Quote Originally Posted by ThePerfectHacker View Post
    Yes it has. If \bold{x}\in f^{-1}[V] we proved k\bold{x} \in f^{-1}[V]. Now let k=0.
    Using your logic, why must there be a condition that a subspace must contain a zero vector, since one of the other two conditions (i.e., closed under scalar multiplication) has already taken care of the first condition (by letting k=0 as you said)?
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    Quote Originally Posted by mathwizard View Post
    why must there be a condition that a subspace must contain a zero vector, since one of the other two conditions (i.e., closed under scalar multiplication) has already taken care of the first condition (by letting k=0 as you said)?
    The condition is necessary, in order to exclude the possibility of the set being empty (the empty set is by convention not considered to be a subspace). If you're trying to show that X is a subspace then you can only make use of the implication \bold{x}\in X\Rightarrow 0\bold{x}\in X\Rightarrow \bold{0}\in X if you know that there exists a vector x in X.
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