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I have problem showing that this sets are path connected. It's a little abstract for me to work in $Gl_n$.

I have to prove that the following sets are path connected in $C^{n^2}$. a) diagonalizable complex $nxn$ and invertibles matrices.

b) hermitian matrices

c) unitary matrices ( Hint : it's similar to a diagonal matrix)

d) Using the fact that every complex matrix $nxn$ it's the propduct of an hermitian matrix positive definite and an unitary matrix, prove that $ Gl_n (C)$ it's path connected in $C^{n^2}$

My solution I realized that in some cases (maybe all) this sets are open, so it's enough to prove connectedness. And for this it's enough to show a continuous surjection from a connected set (like a subset of R^n) for example. In a) the surjection is given by $(a_1,...a_n) $ to the matrix with that diagonal , with $a_i \ne 0$ it's continuous but not connected, but I think that I can fix the problem defining a lot of maps (restricting the domain in several ways such that all the new domains are connected), such that the image of all of them always intersect, and then the union will be the set of matrices .

But the others? I have no idea what can I do!!

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Let A and B be two matrices in $GL_n(\mathbb{C})$ Define $$g(t)=At+(1-t)I$$ let $t_1,\dots,t_n$ be the roots of the polynomial $$det[g(t)=At+(1-t)I]=0$$ Now choose $t\in \mathbb{C}\setminus\{t_1,\dots,t_n\}$, As $\mathbb{C}\setminus\{t_1,\dots,t_n\}$ is path connected(why?) I can choose a path $$f:[0,1]\rightarrow \mathbb{C}\setminus\{t_1,\dots,t_n\}$$ such that $$f(0)=0$$ and $$f(1)=1$$ Now choose $$G(t)=f(t)A+(1-f(t))I$$ So clearly $G(t)$ is a path from A to I so in the same way there will be path from I to B and hence A to B, This shows $GL_n(\mathbb{C})$ is path connected. Use the fact that for every matrix $U\in U(n)$ there is an invertible matrix $g$ such that $U = gDg^{−1}$ where $D = diag(e^{ix_1},\dots,e^{ix_n})$ (here $x_1,\dots,x_n$ are real numbers) to show that the topological group $U(n)$ is that every point of $U(n)$ can be joined by a path with the identity matrix.

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@Masuaki Thanks! Your answers it's very beatiful!! Sorry for ask again And now how can I use the given hypothesis? like being similar to a diagonal matrix ? I have no idea :S! – Matias May 2 '12 at 16:13
I proved that the hermitian matrix are also path connected ( in fact they are convex) – Matias May 2 '12 at 16:25
could write how did you proof that – Un Chien Andalou May 2 '12 at 16:39
Given two hermitian matrix A,B the new matrix it's clearly hermitian ( if t is real) $ tA+(1-t)B$ so in particular for $t \in [0,1] $ thus is convex ) And similarly holds for simetric matrix, and also hermitian positive, and simetric positive matrices ! :D! – Matias May 2 '12 at 16:42
exactly! done! :) – Un Chien Andalou May 2 '12 at 16:44

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