I'm currently getting into studying optimization problems over symmetric cones (NSCP) and I'm having some trouble to understand something.

Let me first give some context, sorry if it is repetitive to you: An Euclidean Jordan algebra $\mathbb{J}$ is a vector space equipped with a Jordan product $\circ$ and an inner product $\langle\cdot ,\cdot\rangle$. Moreover, if we consider the linear operator $L_x(y) = x\circ y$ for each $x,y\in\mathbb{J}$, we have that $$\langle L_x(y),z\rangle = \langle y, L_x(z)\rangle \text{ for each } x,y,z\in\mathbb{J}.$$ That is, the operator $L_x(\cdot)$ is self-adjoint witih respect to $\langle\cdot,\cdot\rangle$.

My question is how do we relate the spectral decomposition of an element $x\in\mathbb{J}$ with the spectral decomposition of the operator $L_x(\cdot)$. I've been investigating the main examples ($\mathbb{R}^n$, $\mathbb{S}^n$, and $\mathbb{R}^n\times\mathbb{R}$) and couldn't find any general rule regarding this topic.

More generally, what is the relation between Jordan frames and basis from the underlying vector space, if there is any?

Has anyone wondered about these questions before and could figure it out? Any help, guidelines and references are welcome.

Thanks for your attention!


When $x$ is idempotent, the operator $L_x$ has three eigenvalues, $0$, $1$ and $\frac12$. This leads to a decomposition of the space into three subspaces, known as the Peirce decomposition. Now for any $x$ in an EJA we can write it as a linear combination of orthogonal idempotents, so this should give you most of what you would want to know about the spectral decomposition of $L_x$ (as $L_x$ is linear in $x$).

If your EJA is not $\mathbb{R}^n$, then a Jordan frame is not a basis of the underlying space. Rather a Jordan frame corresponds to an orthonormal basis of the "underlying" vector space. I.e. if our EJA is $M_n(\mathbb{R})_{\text{sa}}$, the symmetric $n\times n$ real matrices, then a Jordan frame corresponds to an orthonormal basis of $\mathbb{R}^n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.