MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a known characterization of such spaces?

An example: the space of $n \times n$ matrices spanned by $I$ and $J$ (the identity and all-ones matrices, respectively) is inverse closed by the Sherman-Morrison formula.

A possible question of interest would be the maximum dimension of a non-trivial inverse-closed space.

I am reminded here of the work on spaces of matrices of bounded rank about which not a little is known but maybe it's a very false analogy.

EDIT: After reading Denis Serre's neat answer I started thinking what happens in the singular case. The Moore-Penrose inverse of $A$ in general is not polynomial in $A$. It does not even always commute with $A$. But in a paper by Edward Wong, Does the Generalized Inverse of A Commute with A? (Mathematics Magazine, Vol. 59, No. 4 (Oct., 1986), pp. 230-232) it was shown that $A^{\dagger}$ is a polynomial in $A$ if and only if $A$ and $adj(A)$ have the same row-reduced echelon form, so that's some answer to the generalized question.

share|cite|improve this question
up vote 3 down vote accepted

Such matrix subspaces have been characterized: They are Jordan algebras. A bit more generally, linear structure is preserved in inversion if a matrix subspaces is equivalent to a Jordan algebra.

(Jordan algebra is closed under the product MN+NM.) Equivalence means that you are allowed to multiply with an invertible matrix from the left and with an invertible matrix from the right.

Reference: (Marko Huhtanen, Differential geometry of matrix inversion. Math. Scand., 107, pp. 267-284, 2010.)

share|cite|improve this answer

Take any matrix $n\times n$ $A$, then $E$ the sub-algebra spanned by $A$ ,that is the set of $P(A)$ with $P$ polynomials. This space is inverse-closed (Caylet-Hamilton). Its dimension may be $n$.

share|cite|improve this answer
I blush and laugh at the same time. Thanks! – Felix Goldberg May 23 '12 at 11:07

The upper triangular $n \times n$ matrices are inverse-closed, and this subspace has dimension $n(n+1)/2$.

share|cite|improve this answer

Matrices for which a given vector is an eigenvector are inverse closed. This subspace has dimension $n^2-n+1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.