How do I see that $G_n(\mathbb{R}^m)$ is diffeomorphic to the smooth manifold consisting of all $m \times m$ symmetric, idempotent matrices of trace $n$?
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For starters, do you have a geometric interpretation of an idempotent matrix? A symmetric idempotent matrix...? – Andrew D. Hwang Dec 19 '15 at 22:42
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This seems curious as it would make $G_n(\mathbb{R}^m)$ an affine variety. – Matt Samuel Dec 20 '15 at 01:41
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Related – C.F.G Nov 27 '20 at 12:17
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An idenpotent symmetric matrix of trace $n$ is nothing else but an orthogonal projector onto a subspace of dimension $n$. This gives a bijection between these matrices, and $n$ dimensional subspaces of $\bf R^m$, i.e. the Grassmanian.
Thomas
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Do you by any chance have a reference where I can read about this statement? – Math Student 020 Nov 03 '16 at 17:31
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1Idenpotent matrix of trace $n$ = projection onto an $n$dimensional space is certainly explained in every textbook on linear algebra. The fact that the projection is othorgonal iff the matrix is symmetric follows form the formula $<PX,Y>= <X, P^tY>$ (exercise) – Thomas Nov 04 '16 at 13:33