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


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

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

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

*If all the non-zero matrices in N are invertible, what is the maximum the dimension of N can be?
 A: 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 $2^{4a+b}$, $a\geq 0$, $0\leq b\leq 3$. 
A: 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...
A: 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 $\frac{n\left(n-1\right)}{2}$.
A: 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.
