So I've got a bijection. It clearly has an inverse, but how exactly do I prove that the inverse is a linear map as well?

Suppose that the linear map $T:U\to V$ is a bijection. So $T$ has an inverse map $T^{-1}:V\to U$. Prove that $T^{-1}$ is a linear map.


Let $x,y\in V$. Since $T$ is a bijection, there exist $u,v\in U$ such that $T(u)=x,T(v)=y$. It follows that $$T^{-1}(x+y)=T^{-1}(T(u)+T(v))=T^{-1}(T(u+v))=u+v=T^{-1}(x)+T^{-1}(y)$$ and $$T^{-1}(kx)=T^{-1}(kT(u))=T^{-1}(T(ku))=ku=kT^{-1}(x)$$ for all scalar $k$.


Let $T:U\rightarrow V$ be given. We must prove for all $v_1,v_2\in V$ and $\alpha\in \mathbb F$ we have:

  • $T^{-1}(u_1+u_2)=T^{-1}(u_1)+T^{-1}(u_2)$.
  • $T^{-1}(\alpha u_1)=\alpha T^{-1}(u_1)$.

which is equivalent to:

  • $u_1+u_2=T(T^{-1}(u_1)+T^{-1}(u_2))$.
  • $\alpha u_1=T(\alpha T^{-1}(u_1))$.

you should be able to show this using the lineality of $T$.


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