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I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:V\rightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by

Surjectivity of $T \leftrightarrow \forall w \in V, \exists v \in V$ s.t. $ Tv = w.$

Let $v = v^ie_i$ for some basis $\{e_i\}$ of $V$.

$w = v^i(T e_i) = v^ie'_i$.

Surjectivity of $T$ now implies that the $\{ e'_i\}$ are another (linearly independent) set of basis vectors.

Linear independence of $\{e'_i\}$ implies that $i\neq j \rightarrow e_i'-e'_j \neq 0$ or $ e_i'-e'_j = 0 \rightarrow i = j$ or $Te_i = Te_j \rightarrow e_i = e_j \leftrightarrow T$ is injective.

Firstly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?

J. W. Tanner
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user2582713
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  • Surjectivity of $T$ implies linear independence of the $\left{e_i^\prime\right}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible. – user39082 Jul 16 '16 at 12:26
  • For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$. – user39082 Jul 16 '16 at 12:27
  • Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed. – user2582713 Jul 16 '16 at 13:12
  • I've always thought of this as a rather immediate corollary of the standard result that if $T:V \rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case) – John Coleman Jul 16 '16 at 13:15
  • @John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case? – user39082 Jul 16 '16 at 13:18
  • @user2582713: you want to prove injectivity, so you have to show that $Tv=0$ implies $v=0$. You can decompose $v=v^ie_i$ and then $Tv=0$ means $v^ie_i^\prime=v^iTe_i=0$. But as you have already argued the $e_i^\prime$ are linearly independent, so all $v_i=0$ and hence $v=0$ which is what you wanted to show, – user39082 Jul 16 '16 at 13:21
  • @JohnColeman The textbook I'm following sets this exercise before introducing that result. I think I'm supposed to be able to complete the exercise without it. I imagine I'll get to that result eventually. – user2582713 Jul 16 '16 at 13:21
  • @user39082 There is of course nothing wrong with proving it as a special case. But -- if it is a homework problem in a linear algebra book which has just proved the more general theorem, then the intended solution of the exercise is probably to prove it as a corollary. Doing so would then be not simply less work but would actually be more insightful. – John Coleman Jul 16 '16 at 13:22
  • @user2582713 In that case, the author is probably building up to this result, so your approach is fine. I have seen books which use it as an exercise afterwards. – John Coleman Jul 16 '16 at 13:24
  • @user39082 Thanks for that. How about proving that injectivity of $T$ implies its surjectivity? – user2582713 Jul 16 '16 at 13:26
  • Could I say that $T$'s injectivity means $Tv = v^iTe_i = v^ie'_i = 0 \rightarrow v = 0$ and that the linear independence of ${e_i}$ further implies $v^i = 0$ which overall means $v^ie'_i = 0 \rightarrow v^i = 0$ i.e. linearly independence of ${e'_i}$, so that they span $V$ implying $T$ is surjective? – user2582713 Jul 16 '16 at 13:46
  • The computation in the first line yields linear indepence of the $Te_i$, so by dimension count they must span all of $V$. This is just what I had written in my second answer below. – user39082 Jul 16 '16 at 14:12
  • Does this answer your question? If $g:V \rightarrow V$ is an injective linear transformation. Prove if $V$ is finite dimensional then $g$ is surjective. –  Jul 27 '20 at 20:35

4 Answers4

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Do you have the Rank-Nullity theorem yet?

If so it's an easy application.

If $n=\operatorname {dim}V$, then $\operatorname {rank}T+\operatorname {nullity}T=n$, so $T$ has full rank (is surjective) precisely when the nullity is $0$ ($T$ is injective).

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No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $\left\{e_n\right\}_{n\in{\mathbb N}}$, then $$Te_i=e_{i+1}\ \forall i\in{\mathbb N}$$ defines an injective but not surjective operator, and $$Te_0=e_0, Te_i=e_{i-1}\ \forall i\ge 1$$ defines a surjective but not injective operator.

However the equivalence is true for finite-dimensional vector spaces.

user39082
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  • Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case? – user2582713 Jul 16 '16 at 11:58
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Assume T is surjective. Let $\{e_n\}_{n=1}^N$ be a basis for V. By assumption $\{Te_n\}_{n=1}^N$ covers V. Then $\{Te_n\}_{n=1}^N$ is a basis for V since it contains the same number of elements as $\{e_n\}_{n=1}^N$ which is a basis (hence is non-redundant). Then for $v_1,v_2\in V,\ Tv_1=Tv_2\Rightarrow v_1=v_2.$

Assume T is injective. Then $\{Te_n\}_{n=1}^N$ is linearly independent. Therefore it contains the same number of elements as a basis ($\{e_n\}_{n=1}^N$), hence $\{Te_n\}_{n=1}^N$ covers V.

Seth Briney
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Surjectivity ===> Injectivity:

Assume by contradiction $T$ is not injective, let $\left\{v_1,\ldots,v_m\right\}$ be a basis of $ker(T)$ and complete to a basis $\left\{v_1,\ldots,v_n\right\}$ of $V$. Here $n\ge m\ge 1$.

Surjectivity implies that $\left\{Tv_{m+1},\ldots,Tv_n\right\}$ is a basis of $V$. Indeed any $w\in V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+\ldots+a_nv_n$, so $w=a_1Tv_1+\ldots+a_nTv_n=a_{m+1}Tv_{m+1}+\ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.

But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.

Injectivity ===> Surjectivity

The argument is similar. For some basis $\left\{v_1,\ldots,v_n\right\}$ consider the images $\left\{Tv_1,\ldots,Tv_n\right\}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.

user39082
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