Question: I have a block matrix of size PN x PN of the form:

![enter image description here](https://i.stack.imgur.com/SVMdg.jpg)

Where A and C are P x P matrices. I would like to find the eigenvalues of the matrix B, that is

![enter image description here](https://i.stack.imgur.com/LPYZW.jpg)

where 

![enter image description here](https://i.stack.imgur.com/1pfNk.jpg)

Answer: 
"Why is there no natural metric on manifolds?"

Think of three physical variables, like $p$, $V$, $T$, dependent on each other via some natural law
$$W(p, V, T)=0\ ,\tag{1}$$
where $W$ is an explicit expression in three variables (say, van der Waals' law) chosen by a professor of theoretical physics. The equation $(1)$ defines a two-dimensional manifold $\Omega\subset{\mathbb R}^3$, maybe with singularities. The theory of differentiable manifolds then can describe how the physical variables $p$, $T$, $V$, and their mutual partial derivatives are dependent on each other, etc., and it can even describe the nature of $\Omega$'s singularities in terms of algebraic properties of the expression $W$.

But any physicist would laugh at the idea of a "natural metric" on this $(p,V,T)$-manifold $\Omega$. Note that the actual values of the variables $p,\ V,\ T$ are heavily dependent on the chosen units, and it is by no means God-given whether $1^\circ$ difference in temperature is of the same order of magnitude as a volume increase of $1\>{\rm cm}^3$. But any metric on $\Omega$ at once creates such an identification of scales, which is utterly arbitrary.
Metric is not a topological structure. You can only say a metric defines a metric topology, or a topological manifold is metrizable (the [metrization theorems](http://en.wikipedia.org/wiki/Metrization_theorem#Metrization_theorems)). 

1) 2) 3) do not ensure a differentiable manifold structure, you also need the [second countability axiom](http://en.wikipedia.org/wiki/Second-countable_space).

You will benefit from local Euclidean property the following: 

* [**locall connectedness**](http://en.wikipedia.org/wiki/Locally_connected_space): which ensure that a [connected](http://en.wikipedia.org/wiki/Connectedness#Connectedness_in_topology) manifold is also [path connected](http://en.wikipedia.org/wiki/Path_connected#Path_connectedness);
* [**local compactness**](http://en.wikipedia.org/wiki/Locally_compact): which, together with hausdorff axiom and 2nd countability ensures [paracompactness](http://en.wikipedia.org/wiki/Paracompact_space);
* [**regular space**](http://en.wikipedia.org/wiki/Regular_space): as a result of being locally compact and Hausdorff;
* [**normal space**](http://en.wikipedia.org/wiki/Normal_space): as a result of being regular and [Lindelof](http://en.wikipedia.org/wiki/Lindel%C3%B6f_space);
* [**metrizable**](http://en.wikipedia.org/wiki/Metrization_theorem): as a result of being 2nd countable and regular ([Urysohn's metrization theorem](http://en.wikipedia.org/wiki/Metrization_theorem))

It says nothing about the metric being unique. In fact, a common way to construct a global Riemannian metric from metrics on local coordinate charts is to use the [partition of unity](http://en.wikipedia.org/wiki/Paracompact_space#Partitions_of_unity), whose existence is ensured by the manifold being Hausdorff and paracompact. It seems to me the non-uniqueness of Riemannian metric arises as a result of non-uniqueness of POU.

Yes, coordinate mappings are differentiable by definition. However, the diffeomorphism is obviously not global, otherwise you have every $n$-dimension manifold is homeomorphic (topologically equivalent) to $\mathbb R^n$ (topologists will starve).