1. ## question about linear transformation property

there is a property that states

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

My question is, does that necessarily mean that if $\displaystyle T:V\rightarrow W$ 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.

2. 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.

3. Thanks but that's not exactly what I asked (b/c I wasn't so clear...)

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

Can you show me a case where the image $\displaystyle T(k) = (T(v_1),...,T(v_s))$ is also independent

4. Ok, but I don't think you're going to be able to have $\displaystyle s=\dim(V).$ Try this:

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

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

$\displaystyle T$ is not one-to-one, since any multiple of

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

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

This, I think, satisfies what you have stated, does it not?

5. Yes! Thanks.

Great example.

6. You're welcome. Have a good one!