TITLE: Linear transformations which are not invertible in a vector space V of real valued sequences
QUESTION [1 upvotes]: Find a vector space $V$ and maps $F,G\in L(V,V)$ such that $F\cdot G=I$ but neither $F$ nor $G$ are invertible maps.
I tried to use the vector space of all real valued sequences as V. I need to find two linear transformations as above. I got one transformation, $G:V\rightarrow V$ such that $G(a_1,a_2,...)=(a_2,a_3,...)$. But I do not know how to get $F$ such that $F\cdot  G=I$. Please help.

REPLY [1 votes]: Suppose $\Bbb F$ is any field and $V$ is the space of sequences of elements of $F$, that is, of functions
$f: \Bbb N \to \Bbb F; \tag 1$
it is clear we may represent any such $f$ in the form
$f \equiv (f_1, f_2, f_3, \ldots), \; f_i \in \Bbb F, \; \forall i \in N; \tag 2$
it is easy to see that $V$ is a vector space over $\Bbb F$, with addition and scalar multiplication defined component-wise.
The function
$G:V \to V \tag 3$
given by
$G(a_1, a_2, a_3, \ldots) = (a_2, a_3, a_4, \ldots), \tag 4$
is easily seen to be $\Bbb F$-linear; it is sometimes known as the left shift operator on $V$.
It is easy to see that $G$ has no left inverse, that is, that there exists no
$F \in L(V, V) \tag 5$
with
$F \circ G = I; \tag 6$
for consider two sequences
$(a_1, a_2, a_3, \ldots), \; (a_1', a_2, a_3, \ldots), \; a_1' \ne a_1, \tag 7$
which differ only in the first position; if there existed $F$ as in (6), then
$(a_1, a_2, a_3, \ldots) = I(a_1, a_2, a_3, \ldots)$
$= F \circ G(a_1, a_2, a_3, \ldots) = F(a_2, a_3, a_4, \ldots), \tag 8$
and also
$(a_1', a_2, a_3, \ldots) = I(a_1', a_2, a_3, \ldots)$
$= F \circ G(a_1, a_2, a_3, \ldots) = F(a_2, a_3, a_4, \ldots); \tag 9$
together (8) and (9) yield
$(a_1, a_2, a_3, \ldots) = F(a_2, a_3, a_4, \ldots) = 
(a_1', a_2, a_3, \ldots) \Longrightarrow a_1 = a_1';  \tag{10}$
since $a_1$, $a_1'$ may be selected independently of one another, (10) is in fact a contradiction; hence there is no such $F$.
On the other hand, if we set
$G(a_1, a_2, a_3, \ldots) = (0, a_1, a_2, a_3, \ldots), \tag{11}$
and
$F(a_1, a_2, a_3, \ldots) = (a_2, a_3, a_4, \ldots), \tag{12}$
so that now $F$ is the left shift, then
$F \circ G(a_1, a_2, a_3, \ldots) = F(0, a_1, a_2, a_3, \ldots)$
$= (a_1, a_2, a_3, \ldots) = I(a_1, a_2, a_3, \ldots), \tag{13}$
showing that
$F \circ G = I; \tag{14}$
note that $F$ is not injective, since
$F(a_1, a_2, a_3, \ldots) = F(a_1', a_2, a_3, \ldots) \tag{15}$
even when
$a_1 \ne a_1', \tag{16}$
and $G$ is not surjective, since according to (11) no sequence whose first entry is non-zero is in $\text{Range}(G)$; so neither $F$ nor $G$ is invertible; we thus see that such operator pairs $F$, $G$ satisfying (14) do indeed exist.
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