Let $n$ and $r$ be positive integers with $3 \leq r \leq n$. Let $X$ be the set of real $n \times n$ symmetric matrices $A$ such that the submatrix of each $A$ formed by columns $1,2$ and rows $3,...,r$ has rank $1$. How does $X$ form a smooth manifold?
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This is easier than your previous question. In this case your $X$ is the same as all the space of all $r-2 \times 2$ matrices with rank one. (The choice of row and column just say that there are no contraints on the entries) – Mar 02 '15 at 05:31
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Thanks. In what way are they the same? Are their topologies homeomorphic? – SorTheene Mar 02 '15 at 16:32
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I would say that $X$ as a set is bijective to the space of all such matrices. While the space of all matrices can be given a topology (induced from $\mathbb R^{2(r-2)}$ in this case), $X$ also has a topology. – Mar 03 '15 at 05:56
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@John $X$ can be given the subspace topology, but what's the smooth structure? If $X$ is open in the space $S(n)$ of real $n \times n$ symmetric matrices then it would be a smooth manifold, but I'm not sure how to show that $X$ is open (or not open). – SorTheene Mar 03 '15 at 15:19