# 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 $\lambda_j < \mu_j < \lambda_{j+1}$?

• I have a remark. You have to be careful that the map sending an hermitian matrix to the corresponding eigenvalues in non-decreasing order is only piecewise smooth. Indeed, consider the $1$-parameter family of hermitian $2$ by $2$ matrices $diag(1-t, t)$, for $t \in (0,1)$, say. You can see that your eigenvalue map restricted to this smooth curve (a line segment) is not smooth at the "crossing" time $t = 1/2$. – Malkoun Jan 7 '16 at 21:53

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 $\mu_1 = 2$.

• Unless I am wrong, one needs the condition $\lambda_1\le\mu_1\le\lambda_2\le\dots$ for the orbit to be nonempty if the eigenvalues are ordered. This should follow from the variational characterisation of eigenvalues. I haven't checked if the condition is also sufficient - one would have to construct an example for each tuple of eigenvalues satisfying those inequalities. – Sebastian Goette Jan 8 '16 at 16:32
• @SebastianGoette: That's probably correct, and it is probably equivalent to the above characterization in terms of the non-negativity of the expressions that I wrote above, but I haven't had time to check that. – Robert Bryant Jan 8 '16 at 16:55
• Thank you very much for your answer. I think I understood most of the things you wrote. So if we have $B$, there exists a "unique" element $X$ in $\pi^{-1}(diag(\mu))$, with $v$ and $M$ as you claimed (if we assume that the eigenvalues $\mu_j$ are nondecreasing and unique up to equation (1)). It's possible to show that $U(r_i)$ acts transitively on the set of $(a_1, \dots, a_{r_i})$, which are solving $(1)$.We see by your proof that $v$ and $M$ dependent solely on $\lambda$ and $\mu$ and not on $B$. – Olorin Jan 8 '16 at 17:46
• So we find for every $B$ a $U \in U(n)$, s.th. $UB\overline{U}^T=X$. And therefore $\pi^{-1}(\mathcal{O}^\mu)$ is a single $U(n)$-orbit. Is that the correct thought? – Olorin Jan 8 '16 at 17:47
• @Olorin: Yes, if $$\lambda_1 < \mu_1 < \lambda_2 < \cdots <\lambda_j < \mu_j<\lambda_{j+1} < \cdots <\lambda_n<\mu_n<\lambda_{n+1}\ ,$$ then everything you would want to be true is true. The pre-image is then a $\mathbb{T}^n$-bundle over $\mathcal{O}^\mu$ and the map is a submersion. This follows from the characterization of both of them as $\mathrm{U}(n)$-orbits. – Robert Bryant Jan 11 '16 at 1:07