Timeline for If $AB = I$ then $BA = I$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 10, 2022 at 3:05 | comment | added | Pietro Paparella | Yes. It is stated in the paper immediately preceding theorem. | |
| Feb 9, 2022 at 4:34 | comment | added | Morad | Did you use the assumption that $A$ and $B$ are square? | |
| Feb 3, 2022 at 22:06 | history | edited | Pietro Paparella | CC BY-SA 4.0 | deleted 8 characters in body |
| Nov 21, 2021 at 11:42 | comment | added | user920694 | Alternatively, rom the 2nd paragraph, since $\exists x;\, Bx = b,\,\forall b\Rightarrow \exists X;\,BX = I.$ Then, $AB = I\Rightarrow ABX = IX\Rightarrow A = X\Rightarrow BA = BX = I.$ | |
| Nov 13, 2021 at 19:13 | comment | added | Max Herrmann | Ping! Pinpoint. | |
| Sep 10, 2021 at 22:56 | history | edited | Pietro Paparella | CC BY-SA 4.0 | improved exposition |
| Dec 7, 2020 at 22:39 | history | edited | Pietro Paparella | CC BY-SA 4.0 | deleted 19 characters in body |
| Nov 22, 2017 at 17:21 | history | answered | Pietro Paparella | CC BY-SA 3.0 |