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I am sure the answer to this is (kind of) well known. I've searched the web and the site for a proof and found nothing, and if this is a duplicate, I'm sorry.

The following question was given in a contest I took part. I had an approach but it didn't solve the problem.

Consider $V$ a linear subspace of the real vector space $\mathcal{M}_n(\Bbb{R})$ ($n\times n$ real entries matrices) such that $V$ contains only singular matrices (i.e matrices with determinant equal to $0$). What is the maximal dimension of $V$?

A quick guess would be $n^2-n$ since if we consider $W$ the set of $n\times n$ real matrices with last line equal to $0$ then this space has dimension $n^2-n$ and it is a linear space of singular matrices.

Now the only thing there is to prove is that if $V$ is a subspace of $\mathcal{M}_n(\Bbb{R})$ of dimension $k > n^2-n$ then $V$ contains a non-singular matrix. The official proof was unsatisfactory for me, because it was a combinatorial one, and seemed to have few things in common with linear algebra. I was hoping for a pure linear algebra proof.

My approach was to search for a permutation matrix in $V$, but I used some 'false theorem' in between, which I am ashamed to post here.

Beni Bogosel
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    I'm a little curious about the combinatorial proof myself. – anon Sep 23 '11 at 05:42
  • I'll try and write it once I remember it. :) – Beni Bogosel Sep 23 '11 at 05:52
  • I am looking forward to it too. What's the basis for singular matrices then? –  Sep 23 '11 at 08:30
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    Some relevant papers: http://arxiv.org/pdf/1004.0298 ; http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.104.1615&rep=rep1&type=pdf ; http://www.math.technion.ac.il/~meshulam/eprints/maxrank.pdf. The first published proofs seem to be in J. Dieudonné, Sur une généralisation du groupe orthogonal à quatre variables, Arch. Math., 1 (1949) 282-287 and H. Flanders, On spaces of linear transformations with bounded rank, J. Lond. Math. Soc., 37 (1962) 10-16; I couldn't find either of those two with free access. – joriki Sep 23 '11 at 11:43

2 Answers2

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We can show more generally that if $\mathcal M$ is a linear subspace of $\mathcal M_n(\mathbb R)$ such that all its element have a rank less than or equal to $p$, where $1\leq p<n$, then the dimension of $\mathcal M$ is less than or equal to $np$. To see that, consider the subspace $\mathcal E:=\left\{\begin{pmatrix}0&B\\^tB&A\end{pmatrix}, A\in\mathcal M_{n-p}(\mathbb R),B\in\mathcal M_{p,n-p}(\mathbb R) \right\}$. Its dimension is $p(n-p)+(n-p)^2=(n-p)(p+n-p)=n(n-p)$. Let $\mathcal M$ a linear subspace of $\mathcal M_n(\mathbb R)$ such that $\displaystyle\max_{M\in\mathcal M}\operatorname{rank}(M)=p$. We can assume that this space contains the matrix $J:=\begin{pmatrix}I_p&0\\0&0\end{pmatrix}\in\mathcal M_n(\mathbb R)$. Indeed, if $M_0\in\mathcal M$ is such that $\operatorname{rank}M_0=p$, we can find $P,Q\in\mathcal M_n(\mathbb R)$ invertible matrices such that $J=PM_0Q$, and the map $\varphi\colon \mathcal M\to\varphi(\mathcal M)$ defined by $\varphi(M)=PMQ$ is a rank-preserving bijective linear map.

If we take $M\in\mathcal M\cap \mathcal E$, then we can show, considering $M+\lambda J\in\mathcal M$, that $M=0$. Therefore, since $$\dim (\mathcal M+\mathcal E)=\dim(\mathcal M)+\dim(\mathcal E)\leq \dim(\mathcal M_n(\mathbb R))=n^2, $$
we have $$\dim (\mathcal M)\leq n^2-n(n-p)=np.$$

Jose Brox
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Davide Giraudo
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  • I have one question: Why can we assume that $J \in \mathcal{M}$? – Beni Bogosel Sep 23 '11 at 18:30
  • @beni: we can find a matrix $M_0\in\mathcal M$ whose rank is $p$. Now, we take $P\in\mathcal M_n(\mathbb R)$ invertible such that $J=PM_0P^{-1}$, and we consider the map $\varphi\colon \mathcal M\to\varphi(\mathcal M)$ defined by $\varphi(M)= PMP^{-1}$. It's an isomorphism, hence $\varphi(\mathcal M)$ and $\mathcal M$ have the same dimension. – Davide Giraudo Sep 23 '11 at 18:34
  • @DavideGiraudo: Ok. Now is clear. – Beni Bogosel Sep 23 '11 at 18:54
  • @DavideGiraudo How can we prove easily that $rank(M+λJ)\leq p$ for all $\lambda$ implies $M=0$? I see a proof computing the $p+1$ minors via the Schur complement, is there some cleaner argument? – Jose Brox Aug 16 '18 at 08:41
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    @JoseBrox I considered the same thing, here's what I came up with. We have $$ M = \pmatrix{0 & B\B^T & A} \neq 0; \qquad \lambda J + M = \pmatrix{\lambda I & B\B^T & A}. $$ If $\lambda \neq 0$, we can apply block row/column operations to show that $M + \lambda J$ has the same rank as $$ \pmatrix{\lambda I & 0\0 & A - \frac{B^TB}{\lambda}}. $$ The rank of this matrix is $p + \operatorname{rk}(A - \frac{B^TB}{\lambda})$, so it suffices to show that there exists a non-zero $\lambda$ such that $A - \frac{B^TB}{\lambda} \neq 0$. – Ben Grossmann Jul 19 '19 at 17:44
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    An important detail in the above: if $B$ is non-zero, then so is $B^TB$. In fact, $B$ and $B^TB$ necessarily have the same rank. – Ben Grossmann Jul 19 '19 at 17:54
  • The name seems to me very Italian :-) Congratulations on the profile. – Sebastiano Oct 31 '20 at 16:50
  • @DavideGiraudo,In your reply to Beni you certainly mean $\exists P,Q\in GL_n(\mathbb{K})$ such that $J=PM_0Q$. That's what you explained in your answer. $M_0$ and $J$ have the same rank but they are not necessarily similar. Thank you very much for your eminent response. – Mohamed May 27 '23 at 23:54
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If you consider a subspace $\mathcal{M}$ of symmetric singular matrices of rank $p$, Davide's argument can be reused to prove $\text{dim}(\mathcal{M})\le \frac{p(p+1)}2$.

Proof

Consider

$$ \mathcal E:=\left\{\begin{pmatrix}0&B\\B^T&A\end{pmatrix}, A\in \text{Sym}_{n-p}(\mathbb R),B\in\mathcal M_{p,n-p}(\mathbb R) \right\}. $$ which is a subspace of the symmetric $n\times n$ matrices $\text{Sym}_n(\mathbb{R})$ with dimension $$ \text{dim}(\mathcal E) = p(n-p) + \frac{(n-p)(n-p+1)}2 = \frac{n-p}2\bigl(2p + (n-p+1)\bigr) = \frac{(n-p)(n+p+1)}2 $$ With similar arguments, $\mathcal{M}\cap \mathcal{E} = \emptyset$, which results in $$ \text{dim}(\mathcal E) + \text{dim}(\mathcal{M}) \le \text{Sym}_n(\mathbb R) = \frac{n(n+1)}2 $$ And thus $$ \text{dim}(\mathcal{M}) \le \frac{n(n+1)}2 - \frac{(n-p)(n+p+1)}2 = \frac{n^2 + n}2 - \frac{(n^2 - p^2) + (n-p)}2 = \frac{p^2+p}2 $$

Felix Benning
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