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I recall this claim that I have read in some book a long time ago, but now I do not remember and unfortunately I could not find anything on google about it. I was wondering if someone could help me with some reference about this.

For $n>8$ there is no a $n$-dimensional vector subspace of $M_n(\mathbb{R})$ which all non zero elements are invertible matrix.

I was also wondering if we can say something for $n \leq 8$. Thank you.

Remark: I think this should be related to Hurwitz's Theorem (https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(composition_algebras)). For example, for $n=1,2,4,8$ these $n$-dimensional vector subspaces are the ones isomorphic to $\mathbb{R},\mathbb{C},\mathbb{H}$ (quaternions) and $\mathbb{O}$ (octonions) respectively.

Remark 2: I think that the fact these matrices are real it is very important, but I don't know why.

Frank Zermelo
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  • I think that claim, as it appears, is true for any $;n;$, since the zero matrix (=the zero vector) is never invertible and thus no such subspace can exist of any dimension. – DonAntonio Mar 14 '16 at 05:08
  • There is a 'nonzero' requirement in the question – Eric Stucky Mar 14 '16 at 05:08
  • Invertability is a property that cannot be disscused in vector spaces. There is no product in vector space. We need Algebra to say about invertibility. But I know that in any Banach algebra the set of invertible elements is a group and also an opeb set, – Ali Mar 14 '16 at 05:44
  • Any chance you're actually thinking of the octonions? – Eric Towers Mar 14 '16 at 05:46
  • @Ali Yes, of course we cannot talk about product in a vector space, I never said that. I just look at the elements of this vector subspace and see if they are invertible or not. If these vector spaces turn out to be algebras, it is possible, I already said I don't remember exactly. – Frank Zermelo Mar 14 '16 at 06:56
  • @user1551 Yes, it means "vector subspace", corrected. But surely, with your definition, $\rho(8) \neq 8$. Are you sure about your answer? – Frank Zermelo Mar 14 '16 at 07:05
  • @user1551 No, that is happening if the eigenvalue of $A^{-1}B$ is real, since $t \in \mathbb{R}$. This is not necessarily true. – Frank Zermelo Mar 14 '16 at 07:08
  • I looked at your link and I saw that the Hurwitz-Radon number $\rho(n)$ is the maximum number of linearly independent vector fields on $\mathbb{S}^n$. Could you detail more on why this is related to my problem? Thank you. – Frank Zermelo Mar 14 '16 at 07:22

2 Answers2

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The maximal dimension of subspace of real invertible $n\times n$ matrices is given by the Hurwitz-Radon numbers $\rho(n)$, which is defined as follows: if $n=2^{4a+b}c$ where $0\le b\le3$ and $c$ is odd, then $\rho(n)=8a+2^b$. See

J. F. Adams (1962), Vector fields on spheres, Annals of Mathematics, 75(3): 603-632,

J. F. Adams, P. Lax and R. Phillips (1965), On matrices whose real linear combinations are non-singular, Proc. Amer. Math. Soc., 16:318-322,

J. F. Adams, P. Lax and R. Phillips (1966), Corrections to "On matrices whose real linear combinations are non-singular", Proc. Amer. Math. Soc., 17: 945-947.

The result of Adams (1962) essentially says that the maximal number of linearly independent vector fields on the $(n-1)$-dimensional sphere $S^{n-1}\subset\mathbb R^n$ is $\rho(n)-1$. The following presentation briefly explains the connection of these vector fields with invertible matrix subspace:

Rachel Quinlan, Special spaces of matrices, IMS Meeting 2013, NUI Maynooth.

Here are some obvious properties of the Hurwitz-Radon numbers: (a) $\rho(n)=1$ when $n$ is odd, (b) $\rho(n)\le n$ for every $n$, (c) $\rho(n)=n$ iff $n=1,2,4,8$. From (b) and (c), it follows that there is an $n$-dimensional subspace of real $n\times n$ invertible matrices if and only if $n=1,2,4,8$.

A proof of the weaker statement that the maximal dimension is bounded above by $n$ is given in

Zoran Z. Petrović (1999), On nonsingular matrices and Bott periodicity, Publications de l'Institut Mathématique 65(79).85: 97-102.

Clearly, the results of Adams et al. apply to real invertible matrices only. When the matrix space $M_n(\mathbb C)$, the maximal dimension of a subspace of invertible matrices is obviously $1$, as $A-\lambda B$ is singular when $\lambda$ is an eigenvalue of $AB^{-1}$.

Community
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user1551
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As to why it's important to work over $\mathbb{R}$, user1551 gives a first indication, but it goes in the "wrong" direction (it shows that there are even less subspaces with invertible matrices over $\mathbb{C}$).

If you take $\mathbb{Q}$, you can find $n$-dimensional subspaces of $M_n(\mathbb{Q})$ consisting of invertible matrices (except for $0$ of course) for arbitrarily large $n$. This is because just as over $\mathbb{R}$ you have the well-know Hamilton quaternions, over $\mathbb{Q}$ you have division algebras of all degrees (the degree is the squareroot of the dimension in this case, so quaternions are of degree $2$ and dimension $4$). To see that, you may for instance use the Brauer-Hasse-Noether theorem, but it is probably overkill (I just don't immediately see an elementary argument).

If $D$ is such a division algebra, it embeds in $End_\mathbb{Q}(D)$ (where $D$ is seen as a vector space) by multiplication on the left, and the resulting matrices are invertible except for $0$.

Captain Lama
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