I've been trying to prove this result:
Let $f:V\rightarrow\mathbb R^2$ be a $C^2$ function, and $b\in V$ a critical point of $f$. If $\phi:U\rightarrow V$ is a $C^2$-diffeomorphism with $\phi(a)=b$, then the Hessian of $f$ in $b$ has the same eigenvalues of the Hessian of $f\circ\phi$ in $a$.
I've calculate the derivative $(f\circ\phi)''=f''(\phi)\phi'^2+f'(\phi)\phi''$, then using that $\phi(a)=b$ and $b$ is a critical point, $(f\circ\phi)''(a)=f''(b)\phi'(a)^2$. And $f''(b)=H_bf$, so I think that the solution is close, but I don't see how.
So what is the trick to see that they have the same eigenvalues from that expression?
Thanks in advance.
The Hessian is not the matrix of a linear map, but of a quadratic function. This matrix depends on the coordinates used to express the function $f$. If $b$ is a critical point of $f$, as in the case at hand, then Hessian matrices of $f$ corresponding to different coordinate systems are congruent (and not similar!). You have shown this in an abbreviated way. This means that the numbers of positive, zero, and negative eigenvalues are well defined, but not the values of these eigenvalues. These facts play an essential rôle in the setup of Morse theory.
I suggest you look at this question on the mathoverflow site:
https://mathoverflow.net/questions/34799/index-of-a-morse-function-via-the-hessian-tensor