Note: this question is still unanswered! (I will provide an answer if I come to one on my own).
Background
Say we have a continuous twice differentiable function $f:\mathbb{R}^n \rightarrow \mathbb{R}$. We know that this function has a symmetric, positive definite Hessian matrix, $\nabla^2 f$. This Hessian can be decomposed as,
$$\nabla^2 f = R \Lambda R^T, $$
where $\Lambda: \mathbb{R}^n \rightarrow \mathbb{R}^{n \times n}$ is a diagonal matrix of eigen valuaes and $R:\mathbb{R}^n \rightarrow \mathbb{R}^{n \times n}$ is a matrix of eigen vectors. Note that here $\Lambda$ and $R$ are themselves functions of the arguments to $f$.
Question
Given the function $f$ has a Hessian with decomposition $R \Lambda R^T$, is there a way to compute all functions $g:\mathbb{R}^n \rightarrow \mathbb{R}$ such that, $$\nabla^2 g = R D R^T, $$ where $D$ is a diagonal matrix?
Updates:
- John provided a method for computing $g$ so that its Hessian has the correct decomposition at a chosen point. Unfortunately this does not hold over all of $\mathbb{R}^n$.
Sure: $$ g(v) = f(v) + v^t R(D - \Lambda)R^t v + h(v), $$ where $h$ is any function with $\nabla^2 h = 0$.
I'm assuming that you either mean $\nabla^2f(0)$ is $R\Lambda R^t$, and simply omitted the argument "0", or you mean for $R$ and $\Lambda$ to be parameterized by the argument to $f$, in which case mine will need that as well.