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Let $V$ and $W$ be vector spaces and let $T:V\to W$ be a linear transformation

(a)Prove that and only if $T$ carries linearly independent subsets of $V$ to linearly independent subsets of $W$.

(b) Suppose that $T$ is one-one and that $S$ is a subset of $V$. Prove that $S$ is linearly indepdent if and only if $T(S)$ is linearly indepdent.

(c)Suppose that $z=\{v_1,\ldots,v_k\}$ is a basis for $V$ and that $T$ is one-one and onto. Prove that $\{Tv_1,\ldots,Tv_k\}$ is a basis for $W$.

What i tried

(a) The statement '$T$ carries linearly independent subsets of $V$ to linearly independent subsets of $W$' means that $S$ is linearly independent implies $T(S)$ is linearly independent.

TO prove the forward implication means i have to prove:

$T$ is one-one if $S$ is linearly independent implies $T(S)$ is linearly independent.

I tried to prove by contradiction which means T(S) is linearly dependent.

T(S) is linearly dependent means $T(S)$ can be expressed as a linear combination of other vectors $T(S)$ and somehow i have to show that $S$ is also linear dependent. COuld anyone explain. Thanks

(b) I think this part is similar to part (a)

(c) Because T is both one-one and onto, $T$ is thus bijective. Then every vector $v_1$ can be match to a unique corresponding vector $Tv_1$ but im unsure of how to proceed from here. Could anyone please explain this whole question to me. Thanks

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Part (a): Hint: Assume that $S=\{v_i\}_{i=1}^n$ are linearly independent, but that $T(S)$ is not. Then this implies that $\sum_{i=1}^n a_iT(v_i)$ is linearly dependent, i.e. there exists $a_1,..., a_n$ such that $a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)=0$. Use the properties of linear transformations to make a conclusion about the linear independence of $S$.

Part (c): Assuming that (a) is true, then we know that the image of the basis will be linearly independent in $W$. Ask yourself, what do we know about the span of a linearly independent set with the same number of basis vectors as the dimension of the vector space?

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  • $\begingroup$ Oh Thanks, but is it possible to prove directly rather than to use contradiction. $\endgroup$ – ys wong Feb 17 '16 at 9:51

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