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I can see that invertible matrices are a differentiable manifold however I don't know how to show that something is not a differentiable manifold so easily.

Is it ever the case that singular matrices form a differentiable manifold?

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    The matrixes with rank at least k form an open manifold, but I don't know how it helps in there.2012-05-13
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    I interpret your question as asking whether the set of *all* singular matrices is a submanifold and I have given an answer below. Of course if you take the set of multiples of a singular nonzero matrix you will obtain a line, which is a submanifold, consisting only of singular matrices2012-05-13
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    I want to congratulate the user who posted an answer and deleted it after some commenters pointed at a flaw in it: everybody makes mistakes but not everybody reacts so gracefully .2012-05-13
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    @GeorgesElencwajg I deleted because I considered post a comment and discuss the problematic points on the argument.2012-05-13
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    I was thinking on this: if we take matirces of form $$\left( \begin{array}{ccc} 0 & \times & \times \\ 0 & \times & \times \\ 0 & \times & \times \end{array} \right)$$ and of form $$\left( \begin{array}{ccc} \times & 0 & 0 \\ \times & 0 & 0 \\ \times & 0 & 0 \end{array} \right)$$ the union of these two tipes would be like linear subspaces of different dimensions intersecting transversally, but I thought it could not be a manifold, because of the unconstancy of dimension2012-05-13

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