Let $\mathbf{k},\mathbf{h}\in \mathbb{R}^n$ and $\mathbf{M}$ is a $n \times n$ positive semi-definite matrix.
- Is this hold?
$$\mathbf{k}^T \mathbf{M} \mathbf{h} = \mathbf{h}^T \mathbf{M} \mathbf{k}$$
- Can I replace $\mathbf{h}$ with $\mathbf{k}$ in the following?
$$\mathbf{h}^T \mathbf{M} \mathbf{k} \mathbf{k}^T \mathbf{M} \mathbf{h} = \mathbf{h}^T \mathbf{M} \mathbf{h} \mathbf{k}^T \mathbf{M} \mathbf{k}$$
I appreciate if you could answer this with mathematical proof. And by the way, is there a special name for $\mathbf{k}^T \mathbf{M} \mathbf{h}$ combination?
Thank you!
Point 1: First of all, I have seen you agree that $\mathbf{h}$ and $\mathbf{k}$ are column vectors, i.e., $n \times 1$ vectors:
A sufficient condition for $\mathbf{k}^T \mathbf{M} \mathbf{h} = \mathbf{h}^T \mathbf{M} \mathbf{k}$ to hold is that $M$ is symmetrical. Indeed, as the LHS and the RHS are real numbers, and by the fact that a number is equal to its transpose :
$$\mathbf{k}^T \mathbf{M} \mathbf{h}=(\mathbf{k}^T \mathbf{M} \mathbf{h})^T = \mathbf{h}^T \mathbf{M}^T \mathbf{k}=\mathbf{h}^T \mathbf{M} \mathbf{k}\tag{1}$$
Point 2: If you are used to notations of bilinear/quadratic forms, your second relationship reads :
$$B(\mathbf{h},\mathbf{k})^2=B(\mathbf{h},\mathbf{h}).B(\mathbf{k},\mathbf{k})\tag{2}$$
If relationship (2) holds (and (1) holds as well), the answer is "yes, you can exchange $h$ and $k$". But if your question is "does (2) holds whatever $k$ and $h$", the answer is "no, in general" (unless for example when $h=k$).