# Hessian of f(x)^T*Q*y

Gold Member
Hey all. Let me just get right to it! Assume you have a function $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ and we know nothing else except the following equation:
$\triangledown_x\triangledown_x^Tf(x)^TQy=0$
where $\triangledown_x$ is the gradient with respect to vector $x$ (outer product of two gradient operators is the hessian operator). Also let the dimensions of $Q$ and $y$ conform.

Using the information provided above what can you conclude about $f(x)$ (if anything)? Can you infer that $f(x)$ is linear?

Thank you : )

andrewkirk
Homework Helper
Gold Member
The role of ##Q## and ##y## in this equation is unclear. The most natural interpretation, which I will adopt, pending clarification, is that the equation is implicitly prefixed by ##\forall y## and ##\forall Q##. If so then the equation is equivalent to the simpler equation (writing ##u## for ##Qy##):

$$\forall u:\ H \langle f(x),u\rangle=0$$
where ##H## denotes the Hessian operator.

This in turn can be written:
$$\forall u\forall i:\ \sum_j\sum_k u_k\frac{\partial}{\partial x_i\partial x_j}f_k(x)=0$$

By letting ##u## be each of the basis vectors in turn, we can get:
$$\forall i\forall k:\ \sum_j\frac{\partial}{\partial x_i\partial x_j}f_k(x)=0$$

Note that there are ##m## separate Hessian matrices involved here, indexed by ##k## in this formula. The formula tells us that, in each such matrix, all row sums are zero. I think that will make each Hessian singular, but they need not be zero. For instance we could have a Hessian ##\pmatrix{1&-1\\-1&1}##.

So there can still be curvature (ie ##f## is not necessarily linear) but there would be some sort of constraining relationship within that curvature.

• perplexabot
Gold Member
The role of ##Q## and ##y## in this equation is unclear. The most natural interpretation, which I will adopt, pending clarification, is that the equation is implicitly prefixed by ##\forall y## and ##\forall Q##. If so then the equation is equivalent to the simpler equation (writing ##u## for ##Qy##):

$$\forall u:\ H \langle f(x),u\rangle=0$$
where ##H## denotes the Hessian operator.

This in turn can be written:
$$\forall u\forall i:\ \sum_j\sum_k u_k\frac{\partial}{\partial x_i\partial x_j}f_k(x)=0$$

By letting ##u## be each of the basis vectors in turn, we can get:
$$\forall i\forall k:\ \sum_j\frac{\partial}{\partial x_i\partial x_j}f_k(x)=0$$

Note that there are ##m## separate Hessian matrices involved here, indexed by ##k## in this formula. The formula tells us that, in each such matrix, all row sums are zero. I think that will make each Hessian singular, but they need not be zero. For instance we could have a Hessian ##\pmatrix{1&-1\\-1&1}##.

So there can still be curvature (ie ##f## is not necessarily linear) but there would be some sort of constraining relationship within that curvature.
Hmmm. I find your post interesting. I do not understand how you achieved your equations. Maybe my question was badly worded, or maybe I have truncated too much information from the question. Your final answer, f not necessarily being linear, is what I also think. Would it be wise to link or post the paper of which my question stems from? It has something to do with taking the hessian of the log of a multivariate normal distribution.

Thank you for your help : )