What will happen if we remove the hypothesis that $V$ is finite-dimensional in this problem Original problem:
Suppose $V$ is finite-dimensional and $S, T,U ∈L(V)$and $STU = I$. Show that T is invertible and that $inv(T) = US$.
I know that it is because of the hypothesis of finite-dimensional that we can take advantage of the basis or spanning list of vectors. But I'm not quite clear of the power of "Finite-Dimensional" in linear algebra.
Please help me with some examples and insights.
Really really appreciate it.
 A: This is definitely false in infinite dimensions.
Let $V$ be an vector space of countably infinite dimension, and pick a basis $\{v_n\}_{n \in \mathbb{N}}$ for $V$.
Let $T$ be the be the map which sends $v_1 \mapsto 0$ and $v_n \mapsto v_n$ for $n\geq 2$.
Let $U$ be the map which sends $v_n \mapsto v_{n+1}$.
Let $S$ be the map which sends $v_1 \mapsto 0$ and $v_n \mapsto v_{n-1}$ for $n \geq 2$.
Then $STU=I_V$ but $T$ is not invertible since nothing maps to $v_1$ under $T$.
A: It fails in general:
Let $V$ be the vector space - over the reals - of infinite sequences of reals (so an element of $V$ is of the form $a=(a_i)_{i\in\mathbb{N}}$).
Let $U$ be the transformation $V\rightarrow V$ which "shifts right": $$U(a_0, a_1, a_2, . . . )= (0, a_0, a_1, a_2, a_3, . . .).$$
Let $T$ be the transformation $V\rightarrow V$ which "kills the first coordinate": $$T(a_0, a_1, a_2, . . .)=(0, a_1, a_2, . . .).$$
Let $S$ be the transformation $V\rightarrow V$ which "shifts left": $$S(a_0, a_1, a_2, . . .)=(a_1, a_2, a_3, . . .).$$
Then $STU=I$ but $T$ is non-invertible.
