Show that if $UT=I=VT$ for $T,U,V\in \mathcal B(X)$, then $U=V$.

Question

Problem: Let $X$ be a Banach space and $T\in \mathcal B(X)$. Assume that there are $U,V\in \mathcal B(X)$ such that, $UT=I=TV$. Prove that $U=V$.

Attempt: Take an arbitrary $x\in X$ and consider:

$$(UT)x=Ix=x$$

By our hypothesis. But,

$$x=Ix=(TV)x$$

Again by our hypothesis. But then we have that,

$$(UT)x=(TV)x,\,\forall\,x\in X$$

I was hoping to argue something along the lines of the operators $UT$ and $TV$ act exactly the same as each other on $X.$ Maybe using the linearity of the operators in play to reconsider $Tx=y\in X$ for each $x\in X$, but I think that this breaks down here.

I'm not sure if this is correct; I'm primarily concerned that it's a little too "hand wavy" with the points about "operators acting the same". Is there mileage in this approach? Or should I consider something completely different?

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Edited as formulation of the intended problem was incorrect. Thanks to those who pointed this out.

Answer 1

Since the composition of bounded operators is bounded, we can do $$ U = UI = U(TV) = (UT)V = IV = V$$

Answer 2

This only gives a counterexample to the question before the edit. The edited question is answered in the comments.

Let $T: \ell_2(\mathbb{N}) \to \ell_2(\mathbb{N})$ be defined by $T((x_{1}, x_2, \dots, )) = (0, x_{1}, x_2, \dots, )$.

Let $U : \ell_2(\mathbb{N}) \to \ell_2(\mathbb{N})$ be defined by $U((x_{1}, x_2, \dots, )) = (x_2, x_3, \dots, )$.

Let $V : \ell_2(\mathbb{N}) \to \ell_2(\mathbb{N})$ be defined by $V((x_{1}, x_2, \dots, )) = (x_2 + x_1, x_3, \dots, )$.

You have $UT = I = VT$ but $U \neq V$.