# If $T$ is a linear map defined by both basis, why is $T$ injective?

I have been trying to understand this statement from Axler's Linear Algebra Done Right:

If $T:V \to W$ is a linear map with $v_1,...,v_n$ a basis of V and $w_1,...,w_n$ a basis of W, defined by $T(a_1v_1+...+a_nv_n)=a_1w_1+...+a_nw_n$, why is $T$ injective? Is it because $w_1,...,w_n$ is linearly independent?

What has the fact $w_1,...,w_n$ being linearly independent have to imply that T is injective? Is it because the nullspace is trivial.

Thanks everyone!

It is well known that $T$ is injective iff $\ker T=\{0\}$ (because $T x=T y$ iff $T(x-y)=0$).
So here, we just have to prove that if $Tu=0$ then $u=0$. But $Tu=0$ means that all coefficients $a_i$ are zero, since the $w_i$ are linearly independent. So the original vector $u$ was also zero, and we are done.