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This is a memorial to an incorrect solution that used to be here. Unfortunately, it was accepted so I can't delete it.

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

Complementing Xandi's posting: the answer to the question as it is stated is no. It is easy to check that the closure of the set of nilpotent matrices with a given partition $P$ is the set of all nilpotent matrices whose partitions refine $P$. So we can't connect a matrix of type $(4,2)$ with a matrix of type $(3,3)$ without changing the square. In each path joining two such matrices, there is either a matrix of type $(6)$ (the only partition that $(3,3)$ is a refinement of) or a matrix whose partition refines $(3,3)$. In both cases we have a matrix with the square of type other than $(2,1,2,1)$.

This can be generalized as follows: two matrices that square to a nilpotent matrix $N$ are in the same path component of the square roots of $N$ if and only if the partition of the one can be obtained from the partition of the other by subdividing and/or merging in which only blocks of size $\leq 2$ participate (more precisely, we can merge a block $(1,1)$ or split a block $(2)$). A similar statement holds if we consider $n$-th powers instead of squares, in which case only manipulations involving blocks of size $\leq n$ are allowed.

This is a memorial to an incorrect solution that used to be here. Unfortunately, it was accepted so I can't delete it.

Complementing Xandi's posting: the answer to the question as it is stated is no. It is easy to check that the closure of the set of nilpotent matrices with a given partition $P$ is the set of all nilpotent matrices whose partitions refine $P$. So we can't connect a matrix of type $(4,2)$ with a matrix of type $(3,3)$ without changing the square. In each path joining two such matrices, there is either a matrix of type $(6,0)$ (the only partition that $(3,3)$ is a refinement of) or a matrix whose partition refines $(3,3)$. In both cases we have a matrix with the square of type other than $(2,1,2,1)$.

This can be generalized as follows: two matrices that square to a nilpotent matrix $N$ are in the same path component of the square roots of $N$ if and only if the partition of the one can be obtained from the partition of the other by subdividing and/or merging in which only blocks of size $\leq 2$ participate (more precisely, we can merge a block $(1,1)$ or split a block $(2)$). A similar statement holds if we consider $n$-th powers instead of squares, in which case only manipulations involving blocks of size $\leq n$ are allowed.

Complementing Xandi's posting: the answer to the question as it is stated is no. It is easy to check that the closure of the set of nilpotent matrices with a given partition $P$ is the set of all nilpotent matrices whose partitions refine $P$. So we can't connect a matrix of type $(4,2)$ with a matrix of type $(3,3)$ without changing the square. In each path joining two such matrices, there is either a matrix of type $(6)$ (the only partition that $(3,3)$ is a refinement of) or a matrix whose partition refines $(3,3)$. In both cases we have a matrix with the square of type other than $(2,1,2,1)$.

This can be generalized as follows: two matrices that square to a nilpotent matrix $N$ are in the same path component of the square roots of $N$ if and only if the partition of the one can be obtained from the partition of the other by subdividing and/or merging in which only blocks of size $\leq 2$ participate (more precisely, we can merge a block $(1,1)$ or split a block $(2)$). A similar statement holds if we consider $n$-th powers instead of squares, in which case only manipulations involving blocks of size $\leq n$ are allowed.