I know that for any endomorphism on a finite dimensional vector space, if it's injective then it's surjective, and vice versa. But how about any injective linear map between two finite dimensional vector spaces with same dimension? Since any two finite dimensional vector spaces with same dimension should be isomorphic, that should also true, correct?
[Math] Given any linear map between two finite dimensional vector spaces with same dimension, if it’s injective it’s also surjective
linear algebra
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Best Answer
Let $T:V\rightarrow W$ be an injective linear map between finite dimensional linear spaces $V$ and $W$ of dimension $n.$ Let $v_1,v_2,\cdots,v_n$ be the basis of $V.$ Let $\alpha_1T(v_1)+\alpha_2T(v_2)+\cdots+\alpha_nT(V_n)=0$ equivalently $T(\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_nv_n)=0$ as $T$ is injective, therefore $\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_nv_n=0$ $\Rightarrow$ $\alpha_j=0,\,j=1,2,\cdots,n.$ Hence $T(v_1),T(v_2),\cdots,T(v_n)$ forms basis of $W.$ Let $w\in W$ then there exist scalers $\beta_j,\,j=1,2,\cdots,n$ such that $\beta_1T(v_1)+\beta_2T(v_2)+\cdots+\beta_nT(V_n)=w\qquad $ or $ T(\beta_1v_1+\beta_2v_2+\cdots+\beta_nv_n)=w.$ Hence $T $ is surjective.