subset of hermitian matrices given by eigenvalues form a submanifold Let $\mathcal{O}_\lambda$ be the set of hermitian $n+1 \times n+1$ matrices with Eigenvalues $\lambda = (\lambda_1, \dots, \lambda_{n+1})$.
and $\mathcal{O}^\mu$ the set of hermitian $n \times n$ matrices with Eigenvalues $\mu = (\mu_1, \dots, \mu_n)$.
Let $\pi \colon \mathcal{O}_\lambda \to \mathbb{C}^{n\times n}, \ 
\begin{pmatrix}
a_0 & \alpha^T \\
\overline{\alpha} & A
\end{pmatrix} \mapsto A$ be the projection on the lower right "corner".
How could I prove that $\pi^{-1}(\mathcal{O}^\mu)$ is a submanifold of $\mathcal{O}_\lambda$ and that $(d_x\pi)^{-1}T_{\pi(x)}\mathcal{O}^\mu=T_x (\pi^{-1}(\mathcal{O}^\mu))$?
My idea was to show that the map, assigning each hermitian matrix the corresponding eigenvalues in non-decreasing order, is a differentiable map and is a submersion. But I'm not quite sure if thats true or how to prove this.
Edit: Thanks to Robert Bryant we now know, that the condition for the tangentspaces doesn't have to be fulfilled, if $\lambda_j < \lambda_{j+1}$ and $\mu_j = \lambda_j$ or $\mu_j = \lambda_{j+1}$ for some $j$. But is it fufilled, if we look only at the generic cases, where for $\lambda_j < \lambda_{j+1}$ we have $\lambda_j < \mu_j < \lambda_{j+1}$?
 A: The set $\pi^{-1}(\mathcal{O}^\mu)$ is always a smooth submanifold of $\mathcal{O}_\lambda$ (though it may well be empty).  When it is not empty, it is a single orbit of $\mathrm{U}(n)\subset \mathrm{U}(n{+}1)$, where $\mathrm{U}(n)$ is the subgroup of $\mathrm{U}(n{+}1)$ consisting of those unitary $(n{+}1)$-by-$(n{+}1)$ matrices whose top row is $(1,0,\ldots,0)$.
To see this, note that any element $B\in\pi^{-1}(\mathcal{O}^\mu)$ can be written in the form
$$
B = \begin{pmatrix}a_0& v^T\bar P^T\\ Pv & PM\bar P^T\end{pmatrix}
= \begin{pmatrix}1& 0\\ 0 & P\end{pmatrix} \begin{pmatrix}a_0& v^T\\ v & M\end{pmatrix} \begin{pmatrix}1& 0\\ 0 & \bar P^T\end{pmatrix}
$$
where $P\ \bar P^T = I_n$ and where $M = \mathrm{diag}(\mu_1,\ldots,\mu_n)$ and $v$ lies in $\mathbb{R}^n$ and has nonnegative real entries.  Writing
$$
v = \begin{pmatrix} \sqrt{a_1}\\ \vdots \\ \sqrt{a_n}\end{pmatrix},
$$
where $a_i\ge 0$, we can now compute the characteristic polynomial of $B$ and set it equal to $(\lambda_0-t)(\lambda_1-t)\cdots(\lambda_n-t)$ to arrive at the equation
$$
(a_0{-}t) - \frac{a_1}{\mu_1{-}t} - \cdots - \frac{a_n}{\mu_n{-}t}
= \frac{(\lambda_0-t)(\lambda_1-t)\cdots(\lambda_n-t)}{(\mu_1-t)\cdots(\mu_n-t)}.
$$
Note that it follows from this that, if $\mu_i$ occurs with multiplicity $r_i$ in the list $(\mu_1,\ldots,\mu_n)$ then it must occur with multiplicity at least $r_i-1$ in the list $(\lambda_0,\ldots,\lambda_n)$.  For example, suppose that $\mu_1 = \mu_2 = \cdots \mu_{r_1}$ with $\mu_i\not=\mu_1$ for $i>r_1$ and number the $\lambda_i$ so that $\lambda_k = \mu_1$ for $0\le k\le r_1-2$.  Then we find that the above equation implies
$$
a_1 + a_2 + \cdots + a_{r_1} = 
-\frac{(\lambda_{r_1-1}-\mu_1)(\lambda_{r_1}-\mu_1)\cdots(\lambda_n-\mu_1)}{(\mu_{r_1+1}-\mu_1)\cdots(\mu_n-\mu_1)}.
\tag 1
$$
Of course, the right hand side of this equation must be nonnegative, or there is no solution.  In particular, $v$ is a sum of eigenvectors of $M$ and the norm of the eigenvector component of $v$ with eigenvalue $\mu_i$ is uniquely determined by the above equation. Of course, we also have 
$$
a_0 = \lambda_0 + \cdots + \lambda_n - \mu_1 - \cdots - \mu_n\ .
$$
Thus, $B$ lies in a $\mathrm{U}(n)$-orbit of a single element of $\pi^{-1}(\mathcal{O}^\mu)$, as claimed.  Since $\pi^{-1}(\mathcal{O}^\mu)$ is a single $\mathrm{U}(n)$-orbit, it is a smooth submanifold of $\mathcal{O}_\lambda$ (when it is non-empty).  
Moreover, the map $\pi:\pi^{-1}(\mathcal{O}^\mu)\to \mathcal{O}^\mu$ is a smooth submersion when $\pi^{-1}(\mathcal{O}^\mu)$ is not empty (i.e., when the multiplicity conditions and inequalities mentioned above are satisfied), and the fibers are products of spheres $S^{2r_i-1}$ for each eigenvalue $\mu_i$ of multiplicity $r_i\ge 1$ for which the corresponding expression on the right hand side of the above equation $(1)$ is positive.
As for the characterization of the tangent space that the OP wanted, that does not always hold.  It suffices to consider $n=1$ and the case that $(\lambda_0,\lambda_1) = (1,2)$, when the manifold $\pi^{-1}(\mathcal{O}^\mu)$ is empty unless $1\le \mu_1\le 2$, and is a point when $\mu_1 = 1$ or $\mu_1 = 2$, but is a circle when $1 < \mu_1 < 2$.  The manifold $\mathcal{O}_\lambda$ is a $2$-sphere in this case and the map $\pi:\mathcal{O}_\lambda\to \mathbb{R}$ is the projection onto the interval $[1,2]$, so the desired tangent characterization fails when $\mu_1 = 1$ or $\mu_1 = 2$.
