Supposing you have two SPD matrices $A,B\in\mathbb{R}^{n\times n}$ are there any known results on the existence or non-existence of a unitary matrix $Q$ such that $Q^\top A Q=T_A$ and $Q^\top B Q=T_B$ are both tridiagonal. If such a transformation exists in general, it is not required for my purposes that it be computable in finitely many steps.

I am aware of non-orthogonal congruence transformations which tridiagonalize two matrices.

Thanks!

Supposing you have two SPD matrices $A,B\in\mathbb{R}^{n\times n}$ are there any known results on the existence or non-existence of a unitary matrix $Q$ such that $Q^\top A Q=T_A$ and $Q^\top B Q=T_B$ are both tridiagonal. If such a transformation exists in general, it is not required for my purposes that it be computable in finitely many steps.

I am aware of non-orthogonal congruence transformations which tridiagonalize two matrices.

Thanks!

Edit:

Thanks for the response. I am familiar with the papers of Tisseur and Garvey et. al, but they are using non-orthogonal transformations. In one paper they use alternating 1D Householder reflectors and matrices of the form $L=I+xy^\top$ to force portions of the leading columns to be in the same space.

I tried finding a counter-example from the 3x3 case, but it looks like I have plenty of degrees of freedom to play with and higher dimensions become treacherously difficult to manage individual elements.

Maybe this question is equivalent to finding a $Q$ such that for an arbitrary matrix $V$ that $Q^\top V$ is bidiagonal, which certainly looks hopeless to me.