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I am looking at my notes for linear algebra. My professor was trying to prove the linear independence of the bases for $W$, a vector space. This is what I wrote:

Claim:

T: $V \rightarrow W$. $V$ has the dimension of $n$. Want to show $dim(W)=n$. Let $V_1, ..., V_n$ be a basis for V. $T_{V_1},...,T_{V_n}$ form a basis for W.

$\alpha_1T_{V_1}+ ... + \alpha_nT_{V_n}=0$.

$T(\alpha_1V_1+...+\alpha_nV_n)=0$. Since T is one-to-one, $\alpha_1V_1+...+\alpha_nV_n=0$.

I don't understand the last part. Why is it that if $f(x)=0$, $x=0$?

Thanks!

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  • $\begingroup$ One-to-one means that $f(x)=f(y) \implies x=y$. In this case, we know $T(0)=0$ because that's a common property of all linear transformations. So $$T(a_1V_1 + \cdots + a_nV_n) = T(0) \implies a_1V_1 + \cdots + a_nV_n = 0$$ $\endgroup$ – user137731 Sep 30 '16 at 22:46
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That is because $f$, being bijective, is injective, and injective linear maps are cheracterised by the condition $\ker f =\{0\}$.

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