I tagged a question of mine , but it got removed. Since a pseudoinverse is about matrices/operators that are not necessarily invertible, is not exactly adequate, so I'd prefer adding this new tag. But I don't want to start an edit war (especially across at least 79 questions that might deserve it) so I'd rather hear community consensus first:

Is it ok to tag questions about pseudoinverses ?

Proposed tag-excerpt and tag-wiki can be found in a Community Wiki answer below.

  • 2
    $\begingroup$ I'll point out that some users do not like the (inverse) tag either and it was suggested to delete that tag. $\endgroup$ Nov 27, 2014 at 9:51
  • 1
    $\begingroup$ It is recommended that a person who creates a new tag also creates tag-wiki or at least tag-excerpt explaining what the tag is intended for. (See, for example, this answer.) Could you add to your post proposed tag-excerpt to your post here on meta? (Or if you created the tag-info back then, you could at least add a link to your suggested edit, see here.) $\endgroup$ Nov 27, 2014 at 9:55
  • $\begingroup$ @MartinSleziak Ok, added. I wouldn't call the opposition to inverse that overwhelming judging from the little attention that answer got, and deleting that tag would really be counter-productive IMHO. $\endgroup$ Nov 27, 2014 at 10:10
  • $\begingroup$ I do believe that is too long (an "formatty") for a tag-excerpt, but might make a reasonable tag-wiki. Since the Moore-Penrose pseudoinverse is only one kind of pseudoinverse, you might consider using instead either (matrix-pseudoinverses) (to be more general) or (moore-penrose-pseudoinverse) (to be more precise). Also important information for the excerpt/wiki includes under which circumstances the tag should be used, and edge cases where it should not. $\endgroup$
    – user642796
    Nov 27, 2014 at 10:25
  • $\begingroup$ @ArthurFischer Thanks, I added an excerpt. I'd rather leave the tag as pseudoinverse and include other potential meanings in its wiki. If a matrix is concerned, the tag matrices should be added anyway, and the other pseudoinverses occur so rarely that a tag-diversification would IMHO be overkill. $\endgroup$ Nov 27, 2014 at 10:34
  • $\begingroup$ I've taken the liberty to move the proposed tag-excerpt/-wiki to a CW answer, so as to not clutter the question itself. $\endgroup$
    – user642796
    Nov 27, 2014 at 12:50
  • $\begingroup$ @ArthurFischer good point, thanks $\endgroup$ Nov 27, 2014 at 17:53
  • 1
    $\begingroup$ Noticing your yesterday's activity, I would point out that it is better not to do too many retags of old questions at the same time. See older discussion here or here. (But since nobody complained, the number of your edits was probably ok.) $\endgroup$ Dec 5, 2014 at 13:31
  • $\begingroup$ @MartinSleziak Actually I got a polite mod-message regarding that... I guess a "You've edited 20 posts in the last 5 minutes, please don't overdo it" warning might be useful... $\endgroup$ Dec 5, 2014 at 13:32

1 Answer 1


Proposed tag-excerpt:

The operator $A^+$ which best approximates a solution to linear equations $Ax+b=0$ even if $A$ is singular.

Proposed tag-wiki:

(Based on the Wikipedia entry)


(also "Generalized Inverse")

For linear operators and matrices that are not invertible there still exists a unique Moore-Penrose Pseudoinverse $A^+$ which fulfils the following conditions:

  • $A A^+A = A$
    ($AA^+$ need not be the general identity matrix, but it maps all column vectors of $A$ to themselves);
  • $A^+A A^+ = A^+$
    ($A^+$ is a weak inverse for the multiplicative semigroup);
  • $(AA^+)^* = AA^+$
    ($AA^+$ is Hermitian); and
  • $(A^+A)^* = A^+A$
    ($A^+A$ is also Hermitian).

Use it: when the Matrix/Operator involved is (probably) singular

Don't use it: when the Matrix/Operator is definitely invertible or its state is unknown

  • $\begingroup$ Lacking objections I guess it's ok... $\endgroup$ Dec 3, 2014 at 18:44
  • $\begingroup$ The tag excerpt should specify that the "best approximation" is in the $\|Ax-b\|_2$-sense. Also I've seen (and use myself) the notation $A^\dagger$ (\dagger). This might be worth mentioning in the tag wiki. $\endgroup$
    – AlexR
    Dec 3, 2014 at 23:26
  • 2
    $\begingroup$ My only objection is that it seems to conflate the Moore-Penrose pseudoinverse with the concept of a generalized inverse, which needs only to satisfy the first of the Penrose conditions. $\endgroup$ Dec 4, 2014 at 7:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .