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Math Help - question about linear transformation property

  1. #1
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    question about linear transformation property

    there is a property that states

    T:V\rightarrow W is a one-to-one linear transformation, and k={v_1,...,v_s} is a set of independent vectors in V. So the image T(K)={T(v_1),...,T(v_s)} is a set of independent vectors in W.

    My question is, does that necessarily mean that if T:V\rightarrow W is not a one-to-one linear transformation then the image of those vectors is a set of dependent vectors.

    If so, can someone please show an example against my claim.
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  2. #2
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    Sure. If T is not one-to-one, then two different (possibly linearly independent) vectors could be mapped by T to the same vector. That is, T(x) = T(y). It follows that T(x) - T(y) = 0, a linear combination of T(x) and T(y) that equals zero, and the coefficients are nonzero. Therefore, T(x) and T(y) are linearly dependent. QED.
    Last edited by Ackbeet; October 26th 2010 at 01:33 PM. Reason: different parenthesis placement.
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  3. #3
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    Thanks but that's not exactly what I asked (b/c I wasn't so clear...)

    T:V \rightarrow W is not a one-to-one linear transformation. k = (v_1,...,v_s) is a set of independent vectors in V.

    Can you show me a case where the image T(k) = (T(v_1),...,T(v_s)) is also independent
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  4. #4
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    Ok, but I don't think you're going to be able to have s=\dim(V). Try this:

    k=\left\{\begin{bmatrix}1\\0\\0\end{bmatrix},\begi  n{bmatrix}0\\1\\0\end{bmatrix}\right\}, and

    T=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}.

    T is not one-to-one, since any multiple of

    \hat{k}=\begin{bmatrix}0\\0\\1\end{bmatrix}

    gets mapped to the zero vector. Moreover, T(k)=k, which is a linearly independent set.

    This, I think, satisfies what you have stated, does it not?
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  5. #5
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    Yes! Thanks.

    Great example.
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  6. #6
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    You're welcome. Have a good one!
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