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.
Yes, in general, there is no "god-given" (canonical, natural,...) Riemannian metric on a smooth manifold. However, sometimes, there is one! Much of the research in the modern differential geometry is about finding such canonical metrics. For instance, this is how Poincare conjecture was solved in dimension 3 by Perelman. See my answer [here](https://math.stackexchange.com/questions/475018/define-a-metric-based-on-a-topology) for details and references.