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the heading is the question and here are my two approach, I want to know are they correct or not, if not I need to know the answer

1) They are path connected as $\gamma(t)=At+(1-t)B, t\in [0,1]$ where $A$ and $B$ are trace one matrices.But I am not sure all matrices in this path are trace $1$?

2) as trace $1$ so considering diagonal entries I get a hyperplane $x_{11}+\dots+x_{nn}=1$ and remaining other $n^2-n$ entries I can send to $\mathbb{R}^{n^2-n}$ and Thus they are Homeomorphic to $H\times \mathbb{R}^{n^2-n}$ as this ia product of two connected topological space is connected. So trace 1 matrices are connected.well here I have considered a matrix is just a point in $\mathbb{R}^{n^2}$

Thank you.

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For 1) Isn't the trace linear? $\text{tr} \gamma(t)=t\cdot\text{tr}(A)+(1-t)\cdot\text{tr}B=t+(1-t)=1, \forall t$ –  draks ... Oct 12 '12 at 10:27
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If you are not requiring that all matrices are invertible, then as your (2) points out, the trace 1 matrices is a hyperplane in $k^{n^2}$, hence homeomorphic to $k^{n^2-1}$, and hence connected. Here $k=\mathbb C$ or $\mathbb R$. –  Andrew Oct 12 '12 at 10:31
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1 Answer

up vote 2 down vote accepted

Partial answer for 1) $\text{tr} (\gamma(t))=t\cdot\text{tr}(A)+(1-t)\cdot\text{tr}(B)=t+(1-t)=1, \forall t$

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