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Just wanted to quickly work out the following two facts, since I couldn't find simple proofs quickly on the internet: Suppose $\operatorname{dim}(V) = \operatorname{dim}(W) = n$.

  1. If $T:V \to W$ is surjective and linear, it is an isomorphism.
  2. IF $T:V \to W$ is injective and linear, it is an isomorphism.

Proof of 1.

Suppose $T(v_1) = T(v_2)$. Then for a basis $\{e_i\}_{i=1}^{n}$ of $V$, $$T(v_1) = \sum_{i=1}^{n} a_iT(e_i) = T(v_2) = \sum_{i=1}^{n} T(e_i).$$

Now since $T$ is surjective, $S:= \{T(e_i)\}_{i=1}^{n}$ spans $W$. Since $S$ has $n$ members and $\operatorname{dim}(W) = n$, it is a basis. Thus, since representations in a basis are unique, $a_i=b_i$ for all $i$. Moving backwards in linearity, this implies $v_1=v_2$.

Proof of 2.

Similarly, if $T$ is injective, $S$ is L.I. of size $n$ and thus forms a basis, so working backwards in the linearity again, we get a $v$ so that $T(v) = w$.

Anthony Peter
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1 Answers1

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This follows immediately from the rank nullity theorem, saying that $$ \dim (im (T))+\dim (\ker (T))=\dim V. $$ If $T$ is injective, then $\dim(\ker(T)=0$, so that $\dim im (T)=n=\dim W$, and $im(T)\subseteq W$, hence $im(T)=W$. Similarly for $T$ surjective.

Dietrich Burde
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