# Is the Hessian Matrix anything more than a mnemonic?

1. Sep 6, 2014

### wotanub

Several questions I have been thinking about... let me know if you have thoughts on any of them I added numbers to for coherence and readability.

So, the Hessian matrix can be used to determine the stability of critical points of functions that act on $\mathbb{R}^{n}$, by examining its eigenvalues, but is there any interpretation of what it means to diagonalize the Hessian?

For example, in diagonalizing the Jacobian of a function $f: \mathbb{R}^{n} → \mathbb{R}^{m}$, there are implications that can be drawn about the transformation, and the matrix itself defines how the points transform under $f$

(1)Since the Hessian is second derivatives, maybe it says something about the Jacobian? But I'm not sure about this since there are more second derivatives than first derivatives.

(2) I'm wondering that since the Hessian is a matrix, can be thought of as an operator? What space does it act on? It takes functions of $\mathbb{R}^{n}$ into a matrix of functions? How do I denote that? Something like... $H:f(\mathbb{R}^{n}) → f(\mathbb{R}^{n})^{n \times n}$? What does it mean to be an eigenfunction of the Hessian?

(3a) what does the Hessian being diagonal imply? I think it means any term in the function can only involve one of the $x_{i}$ (after thinking for about second, let me know if you can think of a counterexample).

i.e. $H(f) = \mathrm{diag}(\frac{∂^{2}f}{∂x_{1}^{2}},\cdots,\frac{∂^{2}f}{∂x_{n}^{2}}) ⇔ f(x_{1},\cdots,x_{n}) = \sum g_{i}(x_{i})$, where $g_{i}$ is some arbitrary function.

(3b) If the Hessian can be diagonalized, does this imply there is some change of coordinates that can be applied to $\mathbb{R}^{n}$ points so that that function or some class of functions can be decomposed into some $g_{i}$'s that I was describing in the last part?

Last edited: Sep 6, 2014
2. Sep 7, 2014

### mathwonk

isn't the hessian symmetric? if so, (e.g. if your functions are C^2), then I seem to recall it can be diagonalized. (see "spectral theorem".)

3. Sep 7, 2014

### wotanub

It can be diagonalized, but usually when we diagonalize a $n \times n$ matrix, there is some eigenbasis that we can express the $n$ dimensional vectors it acts on. But this is strange here because the Hessian acts on scalar functions?