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Definition :

A linear transformation is a mapping $V \to W$ between two vector spaces that preserves the operations of addition and scalar multiplication.

Question :

Assume that $T,S:V\to V$ are two linear-transformations and $\dim(V) \lt \infty$.

(i) Prove that if $T \circ S=id_v$ then $S \circ T=id_v$.

(ii) Is this true when $\dim(V)$ is not finite?

Note 1 : From $T \circ S=id_v$, I concluded that $T=S^{-1}$ Immediately ! It seems obvious to me ... is this true ? If yes, then the first part is proved ... But, I think i might be wrong ... Even if i'm right, its not a formal proof ...

Note 2 : I have nothing in my mind about an infinite linear transformation which doesn't hold the property of part (ii). But, one of my friends told me that a contradiction exists.

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Arman Malekzadeh
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  • Here's a hint: if $ii$ is false, then to prove $i$ you must at some point use the finiteness of dim(V) in your solution. – guest Dec 08 '16 at 21:13
  • Part $(i)$ is [this duplicate](http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i). – Dietrich Burde Dec 08 '16 at 21:16
  • @DietrichBurde its not ... my question is about combination not multiplication – Arman Malekzadeh Dec 08 '16 at 21:17
  • Is the same: combination is for linear maps, which corresponds to product of matrices. – Dietrich Burde Dec 08 '16 at 21:18
  • It should be noted that the "immediate conclusion" is faulty: of course you're right that in finite dimension if $T\circ S = id$ then $T = S^{-1}$, but **defining** what $S^{-1}$ (or at least, saying that its existence follows from that equation) is **requires a theorem just like this**. Prior to that point there's just "left inverse" and "right inverse" and one needs a proof that they're the same thing. Once you're there and have an inverse then this *retrospectively* becomes trivial, but that would be using the knowledge of the theorem in its *own* proof. – The Vee Dec 09 '16 at 11:21

1 Answers1

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(i) $TS=id$ implies that $S$ is injective. Then the rank–nullity theorem implies that $S$ is invertible. Let $U$ be the inverse of $S$. Then $U=id\circ U = (TS)U=T(SU)=T\circ id=T$ and so $ST=id$.

(ii) Take $V=\mathbb R^\infty$, the space of all real sequences, and $S$ as the right shift: $S(x_1, x_2, \dots) = (0,x_1, x_2, \dots)$. Then $S$ is injective and $TS=id$ for $T$ the left shift, but $S$ is not surjective and so not invertible.

lhf
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  • what is rank nullity theorem? ... what is $R^ \infty $ ?!?! What shift ? sir , i know you're answer is right ... but my knowledge is not enough to understand what you wrote ... would you please explain it more ? – Arman Malekzadeh Dec 08 '16 at 21:16
  • @IStillHaveHope Rank nullity theorem: If $T:V\to W$ is linear between finite dim spaces $\dim V = \dim \ker T + \dim T(V)$. $\mathbb{R}^{\infty}$ is space of all sequences of reals – user160738 Dec 08 '16 at 21:29
  • I truely appreciate your help :) – Arman Malekzadeh Dec 08 '16 at 21:30