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Prove that if $T: V \to W$ is one to one and ${Tv_1, ... Tv_n}$ is a basis for W, then ${v_1,..., v_n}$ is also a basis for V.

My idea is to introduce a $T^{-1}$ and then do a proof that is similar to the converse which is proven in many textbook. However, for the converse, there is also a requirement for T to be onto. I am wondering, why for this case, onto is not needed.

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The proof of Nameless, as it stands now, only proves that the $v_{i}$are linearly independent. You must also prove that they span $V$.

Let $v \in V$, and consider $w = T(v)$, an element of the image of $T$. By assumption, $w$ can be written uniquely as $$ T(v) = w = c_{1} T(v_{1}) + \dots c_{n} T(v_{n}) = T(c_{1} v_{1} + \dots + c_{n} v_{n}). $$ Since $T$ is one-to-one, one obtains $$ v = c_{1} v_{1} + \dots + c_{n} v_{n}. $$


Also, the linear independence bit does not require a proof by contradiction. If $$ 0 = c_{1} v_{1} + \dots + c_{n} v_{n}, $$ apply $T$. BY a standard property of linear maps, we have $T(0) = 0$, thus $$ 0 = c_{1} T(v_{1}) + \dots c_{n} T(v_{n}), $$ and since the $T(v_{i})$ are linearly independent by assumption, we have that all $c_{i}$ are $0$, showing that the $v_{i}$ are also linearly independent.

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  • $\begingroup$ oh I see the question he asked me now. You are right I completely left that detail out. $\endgroup$ – IAmNoOne Sep 20 '14 at 18:52

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