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Math Help - Injective linear transformation

  1. #1
    MHF Contributor arbolis's Avatar
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    Injective linear transformation

    I got the following question wrong in the final exam : Tell whether the affirmation is true or false by justificating.
    T:\mathbb{R}^n \to \mathbb{R}^n is an injective linear transformation if and only if for all subspaces W of \mathbb{R}^n with dimension 2, T:W \to \mathbb{R}^n is injective.
    I said it was true but my explanation was wrong. I had a small doubt : I thought it might be false because n can be uneven but then this doubt disappeared.
    Can you help me with this exercise?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    If T is not injective, take 0\neq u \in \mbox{Ker T}, v \notin  \mbox{Ker T}; then u, u+v are linearly independent and T(u)=T(u+v). This implies that T(k_1u+k_2v)=k_1T(u)+k_2T(v) = k_2T(v); in other words T(\mbox{span }\{u,u+v\})=\mbox{span }\{T(v)\}, so that if T is not injective, there exists a two-dimensional subspace mapping to a one-dimensional subspace.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Bruno J. View Post
    If T is not injective, take 0\neq u \in \mbox{Ker T}, v \notin  \mbox{Ker T}; then u, u+v are linearly independent and T(u)=T(u+v). This implies that T(k_1u+k_2v)=k_1T(u)+k_2T(v) = k_2T(v); in other words T(\mbox{span }\{u,u+v\})=\mbox{span }\{T(v)\}, so that if T is not injective, there exists a two-dimensional subspace mapping to a one-dimensional subspace.
    Ok so if I understand well, T has to be injective in order to go from W \to \mathbb{R}^n.
    So you've proved \Leftarrow ?
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    MHF Contributor Bruno J.'s Avatar
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    Not sure what you mean by "T has to be injective in order to...", but yeah I proved \Leftarrow.

    If T is injective, then clearly the restriction of T to any W is also injective.

    If T is not injective, then there exists some 2-dimensional W whose image is 1-dimensional; hence surely if all 2-dimensional W have 2-dimensional images then T must be injective.

    This is a proof by the "contrapositive" - in order to prove that A implies B, we prove that (not B) implies (not A).
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