Invertible operator

Builds on Normed vector space Continuous linear map Inverse element Banach space Bounded inverse theorem

Definition

Let $X$ and $Y$ be normed spaces over $\R$ or $\C$ , and let $T \colon X \to Y$ be a continuous linear map. The operator $T$ is invertible if there is a continuous linear map $S \colon Y \to X$ such that $S T = \id_X \quad \text{and} \quad T S = \id_Y.$ The map $S$ is then unique; it is the inverse of $T$ , written $T^{-1}$ .

Remarks

So $T$ is an inverse element for composition, with the inverse required to be bounded.

Example: on $\ell^2$ , the diagonal operator $D(x_n) = ((1 + 1/n) x_n)$ is invertible. Its inverse is $D^{-1}(x_n) = (\tfrac{n}{n+1} x_n)$ , which is bounded with norm at most $1$ , and both compositions are the identity.

Non-example: the diagonal operator $E(x_n) = (x_n / n)$ on $\ell^2$ is injective, but it is not invertible. It is not surjective, since $(1/n)$ lies in $\ell^2$ but $(1)$ does not, so $(1/n)$ is not in the image of $E$ .

If $X$ and $Y$ are Banach spaces, every bijective bounded operator is invertible, by the bounded inverse theorem. For an operator $T$ on one space $X$ , invertibility means invertibility in the algebra of bounded operators on $X$ .

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Tag	`Y0V7AP2E09`
URI	`m2://Y0V7AP2E09`
Kind	definition
Writing	W0 stub
MSC	47A05
Defines	Invertible operator
Uses	Normed vector space, Continuous linear map, Inverse element, Banach space, Bounded inverse theorem
Changed	2 November 2026

Invertible operator

Definition

Remarks

Reading path

Further reading

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