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$f$ is an isomorphism if and only if $f$ carries a basis to a basis.

This was stated in my linear algebra textbook without justification. I was wondering if people could please take the time to clarify this by proving that it is true.

The Pointer
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2 Answers2

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Hints:

1) Prove that a linear transformation $\;f:V\to V\;$ is $\,1-1\,$ (injective) iff $\;\ker f=\{0\}\;$, and that it is onto (surjective) iff it carries a generating set to a generating set.

2) if $\;\dim V<\infty\;$ , then $\;f:V\to V\;$ is $\;1-1\;$ iff it is onto iff is bijective.

DonAntonio
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  • Just to clarify a question someone wrote in the comments. Finiteness assumption necessary. For instance https://math.stackexchange.com/questions/2091104/must-an-injective-or-surjective-map-in-an-infinite-dimensional-vector-space-be-a – AlvinL Nov 15 '17 at 11:27
  • @AlvinLepik Yes, of course...in (2), but not in (1) . As written, imo. It is certainly true that an operator is bijective iff it carries basis to basis, whether finite dimensional or not...if this is what you meant. – DonAntonio Nov 15 '17 at 11:28
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    It's worth pointing out that injective maps preserve linear independence and the surjective maps preserve spanning. In this way, you can prove surjectivity directly. – Theo Bendit Nov 15 '17 at 11:31
  • @TheoBendit Are you addressing my question (and thus addressing me), or advising the poster under my answer (and thus addressing him)? – DonAntonio Nov 15 '17 at 11:32
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If $f : V \rightarrow W$ is an isomorphism, then $V$ and $W$ has same dimension. By this its enough to show that the image of the basis for $V$ are linear independent. Conversely if $f$ carries basis to basis, then their dimensions are equal. So $f$ must be an isomorphism.

Kelvin Lois
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  • For my own reference: 1) $f:V \to W$ is injective, which means that each element in the domain maps to a unique element in the codomain. 2) $f:V \to W$ is surjective, which means that, for every element in the codomain, there is at least one element from the domain that maps to it. This is because it carries a spanning set to a spanning set. 3) $dim(V) = dim(W)$. Therefore, it is obvious that an isomorphism must map a basis to a basis. – The Pointer Nov 15 '17 at 11:39
  • Why the down-vote? – The Pointer Nov 15 '17 at 11:39
  • Am i arguing wrong ?. I quite new to math so please be generous. – Kelvin Lois Nov 15 '17 at 11:41
  • I don't see anything wrong with your reasoning. I would appreciate justification for the down-vote; otherwise, I will up-vote to compensate. – The Pointer Nov 15 '17 at 11:42
  • First implication: the fact that isomorphic spaces have the same dimension is a direct consequence of the fact we are trying to prove (and I don't think there is any other way to prove it). So the reasoning becomes circular. Second implication: how does it follow from equal dimensions that $f$ is an isomorphism? It doesn't look correct, certainly not complete. – Adayah Nov 15 '17 at 11:49
  • @Adayah I'm not sure what you're referring to. No one is trying to prove isomorphism -- it is assumed. Did you read the main post? The goal is to show that $f$ is an isomorphism if and only if $f$ carries a basis to a basis. – The Pointer Nov 15 '17 at 11:55
  • For first implication, i use the reasoning that if $V$ and $W$ are isomorphic, then they have same dimension. Which we can get that by Rank-Nullity theorem (Axler p.55). And for the second implication i think you are correct. Maybe a few step needed for the conclusion. – Kelvin Lois Nov 15 '17 at 12:03
  • @ThePointer Do you mean that in order to prove the $\impliedby$ implication you don't need to prove that $f$ is an isomorphism?... – Adayah Nov 15 '17 at 12:22
  • @Adayah No, I'm saying that isomorphism is already assumed, so there is no point in proving it again. – The Pointer Nov 15 '17 at 12:24
  • @Sou燈馬想 The proof on p.55 of Axler relies on Theorem 3.4 (p.45) which proves a more general result than that of the question. In particular, the first paragraph of the proof of 3.4 states that it is enough to prove that whenever $w_1, \ldots, w_n$ is an extension of a basis of the kernel to some basis of $V$, then $(Tw_1, \ldots, Tw_n)$ is a basis of $W$. The proof then continues to justify that, therefore the fact you're using basically relies on the fact you're trying to prove. – Adayah Nov 15 '17 at 12:29
  • @ThePointer The $\impliedby$ part is literally: if $f$ carries basis to a basis, then $f$ is an isomorphism. The "$f$ is an isomorphism" part is the claim, not an assumption. – Adayah Nov 15 '17 at 12:31
  • @Adayah Oh, my apologies. I was not understanding what you meant by $\impliedby$. This makes sense now. – The Pointer Nov 15 '17 at 12:34
  • Yeah. I think you are right. Thanks. – Kelvin Lois Nov 15 '17 at 12:44