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Let $M_n(\mathbb{C})$ denote the n times n matrices over the complex number field. N be a subspace of $M_n(\mathbb{C})$.

  1. If all the matrices in N are non-invertible , what is the maximum the dimension of N can be?

  2. If all the matrices in N commute with each other, what is the maximum the dimension of N can be?

  3. If all the matrices in N are nilpotent, what is the maximum the dimension of N can be?

  4. If all the non-zero matrices in N are invertible, what is the maximum the dimension of N can be?

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    $\begingroup$ Could you provide some motivation or context to allay the coming worries that this is a homework problem? $\endgroup$
    – Pete L. Clark
    Commented Mar 29, 2010 at 19:05
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    $\begingroup$ Also, why don't you tell us what you have tried already? For most of these, there are some fairly obvious lower bounds. The question is whether one can do better. $\endgroup$
    – Pete L. Clark
    Commented Mar 29, 2010 at 19:18
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    $\begingroup$ You might have better luck posting this kind of question on artofproblemsolving.com. I say this not because the question is inappropriate for MO but because I know there are a lot of strong problem-solvers there who like to think about this kind of question, although a few of them are here... $\endgroup$
    – Qiaochu Yuan
    Commented Mar 29, 2010 at 20:06
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    $\begingroup$ for problem 4 , if the base field is C , the answer is 1. But if the base field is R, the answer may be greater than 1 ,right? $\endgroup$
    – zhaoliang
    Commented Mar 29, 2010 at 20:52
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    $\begingroup$ Sure. One can reduce to subalgebras without loss of generality since the product and inverse of invertible matrices is invertible, and then you're just looking at a division algebra over R with a finite-dimensional representation. For that see en.wikipedia.org/wiki/… . $\endgroup$
    – Qiaochu Yuan
    Commented Mar 29, 2010 at 21:29

4 Answers 4

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1. "Non-invertible" means rank $\leq n-1$, and thus the upper bound $n\left(n-1\right)$ follows from the Theorem in paragraph 8.3 in Victor Prasolov's "Problems and theorems in linear algebra". (Scroll to page 58.) The reference given there is Flanders H., On spaces of linear transformations with bound rank, J. London Math. Soc. 37 (1962), pp. 10-16.

2. We can WLOG assume that our subspace $N$ is actually a subalgebra of $\mathrm{M}_n\left(\mathbb C\right)$ (because otherwise, we can replace it by the subalgebra it generates, and it will still have the property that any two of its elements commute), so the question is how large a commutative subalgebra of $\mathrm{M}_n\left(\mathbb C\right)$ can get. The answer is that the maximum possible dimension of such a subalgebra is $\left\lfloor \dfrac{n^2}{4} \right\rfloor + 1$, and this is a result of I. Schur (see the 2 links in that topic). A (relatively) short proof can be found in M. Mirzakhani, A Simple Proof of a Theorem of Schur, The American Mathematical Monthly, Vol. 105, No. 3 (Mar., 1998), pp. 260--262.

4. Here the maximal dimension is $1$, and Petya has told why.

As for 3., I can prove the upper bound $\frac{n^2}{2}$ (strangely enough, for $\mathbb C$ only), but unfortunately there is room between it and the lower bound $\frac{n\left(n-1\right)}{2}$.

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  • $\begingroup$ great thanks! but what if the field is R , the real number field ,not C ? $\endgroup$
    – zhaoliang
    Commented Mar 29, 2010 at 20:35
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    $\begingroup$ Interestingly enough, I can solve 3. over R, but not over C. $\endgroup$
    – darij grinberg
    Commented Mar 29, 2010 at 20:47
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    $\begingroup$ For 3, it's a theorem of Gerstenhaber that $n(n−1)/2$ is best possible and this holds only for conjugates of the strictly upper triangular matrices - jstor.org/pss/2372773 $\endgroup$
    – dke
    Commented Mar 29, 2010 at 21:00
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    $\begingroup$ Thanks for the link,dke (I had never seen the original article): I was writing my answer when you posted it! Did you notice that Gerstenhaber's second sentence (which states "the main purpose of this paper") is completely false, because he forgot to say that his k matrices are linearly independant ! $\endgroup$
    – Georges Elencwajg
    Commented Mar 29, 2010 at 21:45
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For problem 4 over the real field, the answer is the Radon-Hurwitz function at $n$. See for instance Petrovic, "On nonsingular matrices and Bott periodicity." The Radon-Hurwitz function is defined to be $\rho(n)=8a+2^b$, where the largest power of 2 dividing $n$ is $2^{4a+b}$, $a\geq 0$, $0\leq b\leq 3$.

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    $\begingroup$ oh, I visited the webpage and downloaded the pdf version of the article. that's nice of you. now I have found the answers to all the four problems. thank everyone, thank MO! $\endgroup$
    – zhaoliang
    Commented Mar 30, 2010 at 5:22
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    $\begingroup$ By the way, are you professor Stanley, the author of the <<enumeration combinatorics>>? $\endgroup$
    – zhaoliang
    Commented Mar 30, 2010 at 5:24
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    $\begingroup$ zhaoliang, to read a file called filename.ps.gz on Linux or Unix, type "gunzip filename.ps.gz". This should give you a file called "filename.ps", which you can view in the usual way (e.g. by typing "gv filename.ps") or convert to pdf if you prefer (e.g. by typing "ps2pdf filename.ps"). Don't know what you do on other operating systems. $\endgroup$
    – Tom Leinster
    Commented Mar 30, 2010 at 7:46
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    $\begingroup$ Ah, but I see Douglas has edited the link, so you don't need my instructions after all. $\endgroup$
    – Tom Leinster
    Commented Mar 30, 2010 at 7:48
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    $\begingroup$ zhaoliang, try clicking on the username "Richard Stanley" and following the link to his homepage to answer your question ;) $\endgroup$
    – JBL
    Commented Mar 30, 2010 at 16:45
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Dear zhaoliang, here is the answer (from Gerstenhaber's thesis) to question 3.

a) The maximal dimension of a space of $n$ times $n$ nilpotent matrices is $\frac {n(n-1)}{2}$.

b) The subspaces of that dimension are exactly: the space of strictly upper triangular matrices and its conjugates.

Here is a fairly modern related article in the bibliography of which you will find the original references : http://www.win.tue.nl/~jdraisma/publications/NilpotentSubspacesv14.pdf

If you understand mathematical Portuguese (which is easy), here

http://ptmat.fc.ul.pt/~pedro/tese.pdf

is an interesting thesis attacking this kind of problem both with algebraic geometry and combinatorics: a combination that should warm the heart of many a MathOverflower...

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  • $\begingroup$ thank you , Elencwajg. I am a chinese student in Peking university. so Portuguese ......^_^ the first pdf file is nice for me. $\endgroup$
    – zhaoliang
    Commented Mar 29, 2010 at 21:41
  • $\begingroup$ also in problem 4, if we change the base field to R, the real number field , wether the dimension of N can be larger $\endgroup$
    – zhaoliang
    Commented Mar 29, 2010 at 21:55
  • $\begingroup$ Dear zhaoliang, I am very sorry for the poor formulation of my remark about Portuguese. What I meant was that Portuguese is an Indo-European language whose scientific vocabulary has much in common with that of English. Moreover it is a Romance language and so even closer to French, Spanish, Italian, Romanian and my remark was addressed to the non-negligible part of us who have some knowledge of one of those. This may not apply to you but I must tell you my sincere admiration for your command of English, a language from a linguistic family completely different to your mother tongue (mandarin?) $\endgroup$
    – Georges Elencwajg
    Commented Mar 29, 2010 at 22:04
  • $\begingroup$ thank you,Elencwajg.In fact, there are lots of fallacies in my spelling and grammer~^_^~ I re-read the second file, it seems I can guess the meaning of the text by its math formula. But this one still used some algebraic geometry, huh? I saw Zariski topology. $\endgroup$
    – zhaoliang
    Commented Mar 29, 2010 at 22:44
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no,it's not a homework. Do you think a homework can be so hard as these ones? :P.

I proposed them by myself. of course I don't konw whether others have solved them. They are very 'natrul' and easy to formulate, but I find it quite hard to deal with them.

for I, I guess it is n(n-1), the A satisfy Ax=0 for a given x not 0 is ok

for II, I guess it is n(n-1)/2, uptriangle with zero engenvalues will suffice

for III and IV I have no idea.

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  • $\begingroup$ When you say "for II" do you mean "for 3"? $\endgroup$
    – Jonas Meyer
    Commented Mar 29, 2010 at 19:39
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    $\begingroup$ Thanks, that's better. (I think you have mixed up II and III in your answer.) I have to run now and am not a matrix algebra expert anyway, but I'll tell you my guesses: I. the same as yours. II. n (keyword: Cartan subalgebra). III. the same as yours. IV: 1. (Hint: det = 0 is a hypersurface in affine $n^2$-space which intersects your subspace at the origin.) $\endgroup$
    – Pete L. Clark
    Commented Mar 29, 2010 at 19:42
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    $\begingroup$ Answer in 4 is one. Hint: look at the expression det(A+tB) and think when it equals to 0. $\endgroup$
    – Petya
    Commented Mar 29, 2010 at 19:48
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    $\begingroup$ You can do better than n for II (if n is at least 4). It came up at a recent question that you can have a subalgebra of dimension 1 + the floor of (n/2)^2 by considering scalars + 2-by-2 block strictly upper triangular matrices. When you only require subspace rather than subalgebra, perhaps you can do better. $\endgroup$
    – Jonas Meyer
    Commented Mar 29, 2010 at 19:49
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    $\begingroup$ oh, dke ,I didnt catch you. skew-symmetric matrices do not commute in general. $\endgroup$
    – zhaoliang
    Commented Mar 29, 2010 at 20:10

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