# Operators such that $T\circ S=I$ but $S\circ T\neq I$.

Suppose $S$ and $T$ are linear operators on a vector space $V$ and $T\circ S=I$ where $I$ is the identity map. It's easy to see that $S$ is one-to-one.

If $V$ is finite dimensional, rank-nullity implies $S$ is invertible, so a little manipulation shows that $S\circ T=I$.

Makes me curious, if $V$ is infinite dimensional, is there an example of $T$ and $S$ such that $T\circ S=I$ but $S\circ T\neq I$?

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Let $V=l_2$ (this is the space of sequences $(x_1,x_2,\ldots)$ of complex numbers such that $\Sigma_i |x_i|^2$ is finite), $S(x_1,x_2,\ldots) = (0,x_1,x_2,\ldots)$, and $T(x_1,x_2,\ldots) = (x_2,x_3,\ldots)$. Then $TS=I$, but $ST(x_1,x_2,\ldots) = (0,x_2,\ldots)$.
Thanks! ${}{}{}$ –  Kally Jun 9 '13 at 5:13
Let $V$ be the space of polynomials, $S(f)=\int_0^xf(t)dt$, $T(f)=\frac{d}{dx}f(x)$.
Thanks! ${}{}{}$ –  Kally Jun 9 '13 at 5:14