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Thread: 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.
    $\displaystyle T:\mathbb{R}^n \to \mathbb{R}^n$ is an injective linear transformation if and only if for all subspaces $\displaystyle W$ of $\displaystyle \mathbb{R}^n$ with dimension 2, $\displaystyle T:W \to \mathbb{R}^n$ is injective.
    I said it was true but my explanation was wrong. I had a small doubt : I thought it might be false because $\displaystyle n$ 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 $\displaystyle T$ is not injective, take $\displaystyle 0\neq u \in \mbox{Ker T}$, $\displaystyle v \notin \mbox{Ker T}$; then $\displaystyle u, u+v$ are linearly independent and $\displaystyle T(u)=T(u+v)$. This implies that $\displaystyle T(k_1u+k_2v)=k_1T(u)+k_2T(v) = k_2T(v)$; in other words $\displaystyle T(\mbox{span }\{u,u+v\})=\mbox{span }\{T(v)\}$, so that if $\displaystyle T$ 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 $\displaystyle T$ is not injective, take $\displaystyle 0\neq u \in \mbox{Ker T}$, $\displaystyle v \notin \mbox{Ker T}$; then $\displaystyle u, u+v$ are linearly independent and $\displaystyle T(u)=T(u+v)$. This implies that $\displaystyle T(k_1u+k_2v)=k_1T(u)+k_2T(v) = k_2T(v)$; in other words $\displaystyle T(\mbox{span }\{u,u+v\})=\mbox{span }\{T(v)\}$, so that if $\displaystyle T$ is not injective, there exists a two-dimensional subspace mapping to a one-dimensional subspace.
    Ok so if I understand well, $\displaystyle T$ has to be injective in order to go from $\displaystyle W \to \mathbb{R}^n$.
    So you've proved $\displaystyle \Leftarrow$ ?
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  4. #4
    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 $\displaystyle \Leftarrow$.

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

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