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Prop: Let $v_1,...,v_n$ be a basis for the vector space $V$. And $T : V \rightarrow W$ is an isomorphism, then $T(v_1),..,T(v_n)$ is a basis for the vector space $W$.

To prove this proposition, I used the fact that if $v_1,...,v_n$ is a basis, then they must be linearly independent. So, $\sum \alpha_i v_i=0.$ Let $T(v_i) = w_i$, by definition of isomorphism, $T^{-1} : W \rightarrow V$ so $T^{-1}(w_i) = v_i$.

$$T(\sum \alpha_i v_i) = \sum \alpha_i T(v_i) = \sum \alpha _ i w_i \implies T(0) = \sum \alpha _ i w_i \implies \sum \alpha _ i w_i = 0 $$

Since we know this fact, we can now conclude that

$$\sum \alpha_i T(v_i) = \sum \alpha_i w_i \implies \sum \alpha_i T(v_i) = 0$$

And if $\sum \alpha_i T(v_i)=0$, then it is linearly independent so is a basis for $W \square$

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  • $\begingroup$ It looks convoluted but overall ok. Here is the gist $0=\sum a_i w_i=T(\sum a_i v_i),$ and since $T$ is injective, $\sum a_iv_i=0,$ so each $a_i=0,$ thus the $w_i$ are l.i. Q.E.D. $\endgroup$
    – William M.
    Nov 4, 2019 at 17:11
  • $\begingroup$ If you are trying to prove those vectors for a basis, you also have to show that they span the whole space. $\endgroup$
    – Aaron
    Nov 4, 2019 at 19:28

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Say we have $$\sum \alpha_i T(v_i) = 0$$ then we have$$\sum T(\alpha_i v_i) = 0\implies T(\sum \alpha_i v_i) = 0$$

Since $T$ is injective we have now $$\sum \alpha_i v_i = 0$$

and since $v_i$ are independent we have $\alpha _i=0$ for all $i$.

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  • $\begingroup$ Isn't this circular logic? You're starting with what you want to prove. $\endgroup$
    – Melz
    Nov 4, 2019 at 16:54
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    $\begingroup$ @Melz No, what you want to prove is that if the image sums to zero, then so does the argument, showing that the image can only be zero if all the coefficients are zero. $\endgroup$
    – Matt Samuel
    Nov 4, 2019 at 16:58
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    $\begingroup$ Being linearly independent isn’t “a linear combination is 0” it is “if a linear combination is 0, then the coefficients were all 0”. So the first step is to assume there is a linear combination that equals zero so that you can show the coefficients of that particular combination are all 0. $\endgroup$
    – Aaron
    Nov 4, 2019 at 18:33

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