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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. – 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.

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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