# How to find the spectral decomposition of $A\otimes A + B\otimes B$ without forming the Kronecker product?

Given symmetric, positive definite $A$ and $B$ that don't necessarily share the same eigenspace, is there a good way to find spectral decomposition of $A\otimes A + B\otimes B$ without forming the Kronecker product directly?

Certainly, if just needed the spectral decomposition of $A$ or $B$, we could play a trick such as $VDV^T=A$, which yields $(V\otimes V)(D\otimes D)(V\otimes V)^T=A\otimes A$. I'd like to know if a similar trick can be played with $A\otimes A + B\otimes B$.

The closest that I can come up with is solving the generalized eigenvalue problem, $AV = BVD$, where $V^TAV=D$ and $V^TBV=I$. Then, $$A\otimes A + B\otimes B = (V^{-T}\otimes V^{-T})(D\otimes D + I\otimes I)(V^{-1}\otimes V^{-1})$$ The problem here is that $V$ is no longer orthonormal and I'd like that property.

$U=A\otimes A+B\otimes B$ is the sum of $2$ symmetric $>0$ matrices and, therefore, is symmetric $>0$. If you know the spectra of $A,B$, then one can deduce inequalities concerning $spectrum(U)$: cf. the Horn conjecture