---
tag: Y0V7AP2E09
kind: definition
name: Invertible operator
msc: [47A05]
defines: CPAD5W1CGW
uses: [03J5RAKB30, 03TM135G0N, 0ZNZH0EXFX, 06GZ956657, MR5FJXR348]
sources:
  - url: https://en.wikipedia.org/wiki/Inverse_element
---

## Definition

Let $X$ and $Y$ be [[normed spaces|03J5RAKB30]] over $\R$ or $\C$, and let $T \colon X \to Y$ be a [[continuous linear map|03TM135G0N]].
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|0ZNZH0EXFX]] 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|06GZ956657]], every bijective bounded operator is invertible, by the [[bounded inverse theorem|MR5FJXR348]].
For an operator $T$ on one space $X$, invertibility means invertibility in the algebra of bounded operators on $X$.