Problem:
Let $\lambda_1,\dots,\lambda_n$ be distinct real numbers; set $\Lambda$ to be the matrix with $\Lambda_{ii}=\lambda_i$ and $$H = \{Q\in\mathrm{Mat}(n,\mathbb{R}):Q\text{ has eigenvalues }\lambda_1,\dots,\lambda_n\}\\=\{Q\in\mathrm{Mat}(n,\mathbb{R}):Q\text{ is similar to }\Lambda\}$$ I want to show that $H$ is a smooth submanifold of $\mathrm{Mat}(n,\mathbb{R})$.
Previous Work:
Since the $\lambda_i$ are distinct, then any two matrices are similar to $\Lambda$ if and only if they have the same distinct eigenvalues $\lambda_1,\dots,\lambda_n$.
Let $G=\mathrm{GL}(n,\mathbb{R})$ and $M=\mathrm{Mat}(n,\mathbb{R})$. Consider the smooth left action $\alpha:G\times M\to M$ given by conjugation, i.e., $\alpha(g,m)=gmg^{-1}$. Considering the orbit of $\Lambda$, we note that $H=G\cdot\Lambda$. I am trying to show that $H$ is an embedded submanifold via a more direct approach:
First, I want to show that $\alpha_{\Lambda}:G\to H\subset M$ is an immersion. Here is my attempt:
Fix $g\in G$ and $X\in \mathrm{T}_gG=M$. Let $\gamma_g:\mathbb{R}\to G$ be the curve given by $\gamma_g(t)=ge^{g^{-1}X}$ so that $\gamma_g(0)=g$ and $\gamma_g'(0)=X$. We may use this curve to compute the differential of $\alpha_{\Lambda}$ at $X$. Working with $\alpha_{\Lambda}\circ\gamma_g$ first, we have \begin{align*} (\alpha_{\Lambda}\circ\gamma_g)(t) &=\gamma_g(t)\Lambda\gamma_g(t)^{-1}\\ &=\gamma_g(t)\Lambda\gamma_g(-t)\\ &=ge^{g^{-1}Xt}\Lambda ge^{-g^{-1}Xt}\\ &=e^{Xg^{-1}t}\alpha_{\Lambda}(g)e^{-Xg^{-1}t} \end{align*} so that \begin{align*} \mathrm{D}_g\alpha_{\Lambda}(X)=\frac{\mathrm{d}}{\mathrm{d}t}e^{Xg^{-1}t}\alpha_{\Lambda}(g)e^{-Xg^{-1}t}\bigg|_{t=0}=Xg^{-1}\alpha_{\Lambda}(g)-\alpha_{\Lambda}(g)Xg^{-1} \end{align*} Now, for this differential to be injective everywhere, i.e., $\mathrm{D}_g\alpha_{\Lambda}(X)=0\implies X=0$, we have the condition $$Xg^{-1}\alpha_{\Lambda}(g)=\alpha_{\Lambda}(g)Xg^{-1}$$ $$\text{ or equivalently}$$ $$g^{-1}X\Lambda = \Lambda g^{-1}X$$ that is, $g^{-1}X$ commutes with $\Lambda$. Since $\Lambda$ is a diagonal matrix with distinct entries, then $g^{-1}X$ has to be a diagonal matrix.
This is where my progress stops.
Alternative:
I have found this math stack exchange post outlining a proof of a more general statement, that if Lie group acts smoothly on a manifold, then the orbits are immersed submanifolds. Moreover, if the action is proper, then the orbits are embedded submanifolds. The question I would have in this case is: how do I show that conjugation is a proper action?
Question:
I would be willing to give up on my previous work above, but not without asking someone else first. I am asking for a second look by someone with some more insight. Am I making an error anywhere in my previous work? Is there some connection I am not making to show that $X$ is zero?
Any help is appreciated. Thank you.
