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Let $T$ be a linear operator on a vector space $V$ over a field $F$. Suppose there is a linear operator S on $V(F)$ such that TS = I where I is identity operator on $V(F)$.

Now I am looking for an example where neither of $T$ or $S$ is invertible.

As $TS$ is one-one and onto, S is one-one and T is onto. And for finding an example to satisfy the above condition, $V$ must be infinite-dimensional.

But I am not able to find such an example.

Gitika
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  • I'd suggest taking $V$ to be a "sequence space", the set of infinite sequences $(a_1,a_2,\ldots)$ of elements of $F$. – Angina Seng Jun 27 '20 at 09:28
  • Consider $V$ to be the vector space of all $\mathbb{R}$-valued sequences. Let $S$ be the shift to the right (i.e. $(a_n)_n \mapsto (0, (a_n)_n)$) and $T$ be the shift to the left. We have $TS = I$ but $ST \neq I$. – NiklasvMoers Jun 27 '20 at 09:30

2 Answers2

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For example let $V$ be the space of infinite sequences of elements of $F$. Define $T,S$ like this:

$T(x_1,x_2,...)=(x_2,x_3,...)$

$S(x_1,x_2,...)=(0,x_1,x_2,x_3,...)$

Note that indeed $TS=I$, but not the other way around. So they can't be invertible.

Mark
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On the space of all sequences of real numbers define $S(a_n)=(0,a_1,a_2,...)$ and $T(a_n)=(a_2,a_3,...)$. Then $TS=I$ but neither $T$ not $S$ is invertible. $T$ is not injective and $S$ is not surjective.

Kavi Rama Murthy
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