Why do we need to determine the definiteness of the Hessian to decide what a critical point is? In univariate calculus, if we know that $f'(c)=0$, we can determine if the function $f$ has a minimum at $c$ by checking that $f''(c) > 0$. The multivariate analogue of the second derivative is the Hessian matrix. I now learned that to decide between extreme and saddle points in this case, it has to be checked whether the Hessian is positive definite, negative definite or indefinite. This can be achieved by checking its eigenvalues.
I have several questions regarding this:


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*Why is it not sufficient to check the sign of the values in the Hessian, but we need to check for definiteness?

*Does the definiteness just make sure some convexity or concavity properties check out, or is there a more meaningful interpretation of that?

*How do the eigenvalues of a matrix tell us its definiteness?

*Addendum: What do the off-diagonal entries in the Hessian even mean? How the slope in a certain dimension changes by making changes in a different dimension?

When you Taylor expand a multivariable function, it looks like:
$$f(x+h)=x+(Df)\cdot h+h^T(D^2f)h+o(h^Th)...,$$
so locally around a critical point, it looks like $x+h^T(D^2f)h.$ 
It's clear now that if $(D^2f)$ is positive definite, then locally, $f(x)$ increases in any direction away from $x$. The opposite occurs when it's negative definite. Otherwise, when it's indefinite or semidefinite, you get either a saddle or need to look at higher order derivatives to conclude.
