most recent 30 from http://mathoverflow.net 2013-05-26T05:05:50Z http://mathoverflow.net/feeds/question/24849 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24849/path-connected-set-of-matrices Portland 2010-05-16T01:49:56Z 2010-05-20T13:31:06Z

<p>Let $N \in \mathfrak{M}_n(\mathbb{C})$ nilpotent, such that there exists $X \in \mathfrak M_n(\mathbb{C})$ with $X^2=N$ (take for instance $n>2$ and $N(1,n)=1$; $N(i,j)=0$ otherwise).</p> <p>Denote by $\mathcal{S}_N$ the set of $X \in \mathfrak M_n(\mathbb{C})$ such that $X^2=N$. Is $\mathcal{S}_N$ connected or path-connected ? What happens when we change $2$ by $3,4,\ldots $?</p>

http://mathoverflow.net/questions/24849/path-connected-set-of-matrices/24876#24876 Xandi Tuni 2010-05-16T10:55:43Z 2010-05-16T12:32:21Z

<p>Edit: This is just half an answer: I can only show that the sets matrices with $X^2=N$ and fixed Jordan type are path connected.</p> <p>Every nilpotent matrix is conjugate to a nilpotent matrix in Jordan form, which is unique up to permutation of Jordam blocks. So we have a bijection</p> <p>$$\mathrm{Nilp}_n(\mathbb C)/\mathrm{conjugation} \quad \cong \quad \mbox{integer partitions of }n$$</p> <p>associating with a conjugacy class of a nilpotent matrix $X$ the sizes of its Jordan blocks $(a_1, \ldots, a_r)$ which sum up to $n$. The max of the $a_i$'s is the nilpotency-degree of $X$. To the class of $X^2$ is associated the partition $$(\lfloor (a_1 +1)/2 \rfloor, \lfloor a_1/2\rfloor ,\lfloor (a_2 +1)/2 \rfloor ,\lfloor a_2/2\rfloor, \ldots, \lfloor a_r/2 \rfloor)$$</p> <p>From here we can derive a necessary and sufficient condition on a nilpotent matrix to be a square.</p> <p>Now fix your preferred nilpotent matrix $N$. Let $X$ be a matrix with $X^2=N$ and Jordan type $(a_1, \ldots, a_r)$. Conjugating the whole setup, we may assume $X$ is in Jordan form.</p> <p>Let $Y$ be a matrix with $Y^2=X^2 = N$ having the same Jordan type as $X$, and let us construct a path from $X$ to $Y$. Since $X$ and $Y$ have the same Jordan type, there exists an invertible matrix $S$ with $Y=SXS^{-1}$. Because $Y^2=X^2=N$ the matrix $S$ commutes with $N$. It is enough to construct a path from the identity matrix to $S$ in the set $\mathcal C_N$ of invertible matrices that commute with $N$.</p> <p>I claim $\mathcal C_N$ is path connected (for just any $N$). Indeed, the set $[N]$ of all commutators of $N$ is linear subspace of the vector space $\mathrm M_n(\mathbb C)$. The determinant, as a function on $[N]$ is a polynomial function which is not identically zero since the identity matrix belongs to $[N]$. Thus, the zero set of the determinant is Zariski closed, so $\mathcal C_N$ is Zariski open in $[N]$. Any Zariski-open in a complex vector space is path connected.</p> <p>What remains is to connect different Jordan types. We certainly can connect $(5,2)$ with $(5,1,1)$ by the 1 in the $2\times 2$--block with $t$ and vary $t$ from $1$ to zero. The problem that remains is to connect, say, type $(4,2)$ with type $(3,3)$ as pointed out in the comments.</p>

http://mathoverflow.net/questions/24849/path-connected-set-of-matrices/24956#24956 algori 2010-05-17T00:38:13Z 2010-05-20T13:31:06Z

<p>This is a memorial to an incorrect solution that used to be here. Unfortunately, I can't delete it since it was accepted.</p>