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?
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4$\begingroup$ Let $T$ be the injective map between two linear spaces of same dimension and If $x_1,x_2,\cdots,x_n$ is basis of domain, Is $T(x_1),T(x_2),\cdots,T(x_n)$ a basis of co-domain. $\endgroup$– SuhailApr 4, 2015 at 3:55
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$\begingroup$ @Suhail: you could flesh this out a bit and post it as an answer. It avoids the Rank-Nullity theorem. I don't know which order things are proved. $\endgroup$– Ross MillikanApr 4, 2015 at 4:25
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2 Answers
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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.
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By the Rank-Nullity Theorem, for a linear map $T: V \to W$,
$$ \textrm{dim(im(}T)) + \textrm{dim(ker(}T)) = \textrm{dim(}V) $$
and so if $V,W$ are of the same dimension and $T$ is injective,
$$ \textrm{dim(im(}T)) = \textrm{dim(}W). $$
