Is the subset of symmetric $ 3\times 3 $ matrices with given eigenvalues a manifold?

Question

For $ \mu_1>\mu_2>\mu_3 $ are three real numbers. Consider th set $$ S(\mu_1,\mu_2,\mu_3)=\{A\in M(3,\mathbb{R}):,A^T=A,\,\,\lambda_1(A)=\mu_1,\lambda_2(A)=\mu_2,\lambda_3(A)=\mu_3\}, $$ where $ \lambda_1(A)>\lambda_2(A)>\lambda_3(A) $ are eigenvalues of $ A $ and $ M(3,\mathbb{R}) $ denotes the sets of $ 3\times 3 $ matrices. Then the questions is that if $ S(\mu_1,\mu_2,\mu_3) $ is a manifold.

Here is my try, define $$ f:\{A\in M(3,\mathbb{R}):A^T=A\}\to\mathbb{R}^3\quad(f(A)=(\lambda_1(A),\lambda_2(A),\lambda_3(A))). $$ I want to show that for any $ \mu_1>\mu_2>\mu_3 $ the point $ (\mu_1,\mu_2,\mu_3) $ is a regular point of $ f $ and then $ S(\mu_1,\mu_2,\mu_3) $ is a manifold. However I cannot deal with the derivatives of $ f $. Can you give me some hints or references?

Answer 1

This answer expands on your idea, although is not the simplest way to prove the statement. We consider the general dimension $n$. Let us first show the following general statement:

Let $M \in M_n(\Bbb R)$ be a real matrix with $n$ distinct eigenvalues. Then there exist a neighbourhood $U \subset M_n(\Bbb R)$ of $M$ and a smooth map $\Lambda \colon U \to \Bbb R^n$ such that for every $M' \in U$, $M'$ has $n$ distinct eigenvalues given by $\Lambda(M')$.

Proof. Let $V = \{(t_1,\ldots,t_n) \in \Bbb R^n \mid t_1 < \ldots < t_n \}$, which is open. Consider the map $$ \begin{array}{r|ccc} F \colon & M_n(\Bbb R) \times V & \longrightarrow & \Bbb R^n \\ & (M,t_1,\ldots,t_n) & \longmapsto & \left(P_M(t_1),\ldots, P_M(t_n) \right) \label{eq:star} \tag{$\star$} \end{array} $$ where $P_M(X) = \det(M-XI_n)$ is the characteristic polynomial of $M$. Notice that $F$ polynomial in all its entries, so that $F$ is smooth. Moreover, the differential of $F$ in the second factor at $(M_0,\lambda_1,\ldots,\lambda_n)$ is $$ \begin{array}{r|ccc} D_2F_{(M_0,\lambda_1,\ldots,\lambda_n)} \colon & \Bbb R^n & \longrightarrow & \Bbb R^n \\ & (t_1,\ldots,t_n) & \longmapsto & (P'_{M_0}(\lambda_1)t_1,\ldots,P'_{M_0}(\lambda_n)t_n). \end{array} $$ Assume that $(\lambda_1,\ldots,\lambda_n) \in V$ is the spectrum of $M_0$, that is to say that $F(M_0,\lambda_1,\ldots,\lambda_n) = 0$. Then the eigenvalues of $M_0$ are simple, and $P'_{M_0}(\lambda_j) \neq 0$ for all $j$. It follows that $D_2F_{(M_0,\lambda_1,\ldots,\lambda_n)}\colon \Bbb R^n \to \Bbb R^n$ is an isomorphism. By the implicit function Theorem, there exists an open neighbourhood $U_0 \subset M_n(\Bbb R)$ of $M_0$, an open neighbourhood $V_0 \subset V$ of $(\lambda_1,\ldots,\lambda_n)$, and a smooth function $\Lambda \colon U_0 \to V_0$ such that $$ \forall (M,t_1,\ldots,t_n) \in U_0\times V_0,\quad F(M,t_1,\ldots,t_n) = 0 \iff (t_1,\ldots,t_n) = \Lambda(M). $$ This proves the statement. $\blacksquare$

Now, for your concern, consider $M$ symmetric with distinct eigenvalues. From the preceding study, the map $f$ you are considering is locally given by $$ \begin{array}{r|ccc} f \colon & U\cap S_n(\Bbb R) & \longrightarrow & \Bbb R^n \\ & M & \longmapsto & \Lambda(M), \end{array} $$ where $U$ and $\Lambda \colon U \to \Bbb R^n$ are given by the above result. The differential of $f$ at $M$ is then $$ df_M \colon H \in S_n(\Bbb R) \longmapsto df_M(H) = d\Lambda_M(H) \in \Bbb R^n. $$ We thus need to compute the differential of the map $\Lambda \colon U\to \Bbb R^n$. Notice that the map $$ M \in U \mapsto F(M,\Lambda(M)) \in \Bbb R^n $$ identically vanishes, where $F$ is defined by \eqref{eq:star}. Differentiating, this yields $$ \forall M \in U, \quad D_1F_{(M,\Lambda(M))} + D_2F_{(M,\Lambda(M))} \circ d\Lambda_M = 0, $$ and since $D_2F_{(M,\Lambda(M))}$ is invertible, we obtain the following expression for $d\Lambda_M$: $$ \forall M \in U,\forall H \in M_n(\Bbb R),\quad d\Lambda_M(H) = - (D_2F_{(M,\Lambda(M))})^{-1}\circ D_1F_{(M,\Lambda(M))}(H). $$ Since $D_2F_{(M,\Lambda(M))}$ is an isomorphism, to show that $f \colon S_n(\Bbb R)\cap U \to \Bbb R^n$ is a submersion at $M$, one only needs to show that $D_1F_{(M,\Lambda(M))} \colon S_n(\Bbb R) \to \Bbb R^n$ is surjective. By Jacobi's formula, we have $$ \newcommand{\adj}{\mathrm{adj}} \newcommand{\trace}{\mathrm{trace}} D_1F_{(M,\lambda_1,\ldots,\lambda_n)}(H) = (\trace(\adj(M-\lambda_1I_n)H),\ldots,\trace(\adj(M-\lambda_n I_n)H). $$ This last expression is a bit annoying, but here is a trick to overcome this difficulty. Note that there exists $O\in O_n(\Bbb R)$ such that $$ \newcommand{\diag}{\mathrm{diag}} O^{-1}MO = \diag(\lambda_1,\ldots,\lambda_n), $$ and that since $f(M) = f(O^{-1}MO)$, one has $$ df_M(H) = df_{O^{-1}MO}(O^{-1}HO). $$ Since $H\in S_n(\Bbb R) \mapsto O^{-1}HO \in S_n(\Bbb R)$ is an isomorphism, it suffices to show the surjectivity of $df_M$ for $M = \diag(\lambda_1,\ldots,\lambda_n)$ with $\lambda_1 < \cdots <\lambda_n$. But then, one has $$ \adj(\diag(\lambda_1,\ldots,\lambda_n) - \lambda_jI_n) = \adj(\diag(\lambda_1-\lambda_j,\ldots, \lambda_n-\lambda_j)) = \diag(0,\ldots, 0,\alpha_j, 0,\ldots,0) $$ with $\alpha_j = \prod_{i\neq j}(\lambda_i-\lambda_j) \neq 0$. Choose $H' = \diag(\frac{t_1}{\alpha_1},\ldots,\frac{t_n}{\alpha_n})$, and notice that $$ df_{\diag(\lambda_1,\ldots,\lambda_n)}(H') = (t_1,\ldots,t_n). $$ The result now follows.

Answer 2

This second answer gives a shorter proof, more algebraic, which does not expand on your ideas. $\newcommand{\O}{\mathcal{O}}$

We will use the following fact twice, whose proof is given by elementary computations:

If $D$ is a diagonal matrix with distinct entries, then its commutator $\mathcal{C}(D)$ is precisely the set of diagonal matrices.

Let $D = \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 &\lambda_3 \end{pmatrix}$ with $\lambda_1 < \lambda_2 < \lambda_3$. Consider the action $\O_3(\Bbb R) \curvearrowright S_3(\Bbb R)$ given by $O\cdot M = OMO^T$. The spectral Theorem yields that $S(\lambda_1,\lambda_2,\lambda_3)$ is exactly the orbit $S(\lambda_1,\lambda_2,\lambda_3) = \mathrm{Orb}(D)$.

Let $g\colon \O_3(\Bbb R) \to S_3(\Bbb R)$ be given by $g(O) = O\cdot D$, which is smooth. Notice that $g$ is equivariant: $$ \forall O_1,O_2 \in \O_3(\Bbb R), \quad g(O_1O_2) = O_1\cdot g(O_2). $$ This implies that if $O\in \O_3(\Bbb R)$ and $H \in T_O \O_3(\Bbb R)$ then $$ dg_O(H) = O(dg_{I_3}(O^TH)) O^T, $$ and $g$ has constant rank, equal to its rank at $I_3$. Notice that $T_{I_3}\O_3(\Bbb R) = A_3(\Bbb R)$ is the space of antisymmetric matrices, so that we have $$ dg_{I_3} \colon H \in A_3(\Bbb R) \longmapsto DH - HD \in S_3(\Bbb R). $$ Hence, $\ker dg_{I_3} = A_3(\Bbb R) \cap \mathcal{C}(D)$, where $\mathcal{C}(D)$ is the set of matrices that commute with $D$, which are the diagonal matrices. Since the only antisymmetric diagonal matrix is the zero matrix, one obtains that $\ker dg_{I_3} = \{0\}$. It follows that $g$ is and immersion.

As a consequence, if $\mathrm{Stab}(D) = g^{-1}(D)$ is the stabilizer subgroup of $D$, then $g$ induces a smooth injective immersion $$ \bar{g} \colon \O_3(\Bbb R) / \mathrm{Stab}(D) \to S_3(\Bbb R). $$ We compute $\mathrm{Stab}(D)$: \begin{align} g(O) = D \iff OD = DO \iff O \in \mathcal{C}(D)\cap \O_3(\Bbb R) \iff O = \begin{pmatrix} \pm 1 & 0 & 0 \\ 0 & \pm 1 & 0 \\ 0 & 0 & \pm 1 \end{pmatrix}. \end{align} Hence, $\mathrm{Stab}(D)$ is discrete, and $\O_3(\Bbb R) / \mathrm{Stab}(D)$ has dimension $\dim \O_3(\Bbb R) = 3$. Since $\O_3(\Bbb R)$ is compact, $\O_3(\Bbb R)/\mathrm{Stab}(D)$ is also a compact manifold. Hence, $\bar{g} \colon \O_3(\Bbb R)/\mathrm{Stab}(D) \to S_3(\Bbb R)$ is a smooth injective immersion from a compact manifold, and is thus an embedding. It follows that its image $S(\lambda_1,\lambda_2,\lambda_3)$ is a submanifold of $S_3(\Bbb R)$ of dimension $\dim\O_3(\Bbb R) = 3$.