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:


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 $f(x)=0$, $x=0$?


  • $\begingroup$ 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

That is because $f$, being bijective, is injective, and injective linear maps are cheracterised by the condition $\ker f =\{0\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.