A question of invertibility of matrices Let $A$ and $B$ be self-adjoint $n \times n$ matrices. Let $A$ be diagonal. Suppose $A+tB$ and $tA+B$ are invertible for all $t \in \mathbb R$. What can we say about $A$ and $B$?
My guess is that $\mbox{Tr}(A) = \mbox{Tr}(B) = 0$. Definitely when $n$ is odd there exist no $A, B$ satisfying the hypothesis. Because in this case $\det(A+tB)$ is an odd-degree polynomial. I do not know what happens if $n$ is even.
If $B$ is also diagonal, then it cannot happen that $A+tB$ and $tA+B$ are invertible for all $t \in \mathbb R$. This is easy to see.
 A: What about
$$A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\qquad B=\begin{pmatrix} 0 & 1 \\ 1 & 0  \end{pmatrix}\quad ?$$
A: There is a Jordan-like canonical form for symmetric matrix pairs $(A,B) = (A^*,B^*) \in \mathbb{C}^{n\times n} \times \mathbb{C}^{n\times n}$ under the transformation $(A,B) \to (M^*AM,M^*BM)$, with $M$ square invertible. You can find it stated, for instance, in Lemma~3 of Thompson's paper https://doi.org/10.1016/0024-3795(76)90021-5 . This canonical form is known today as even Kronecker canonical form (in a slightly different variant when $B$ is anti-Hermitian, but you can just multiply $B \gets iB$ to fix this).
Note that if you are interested in a canonical form under that transformation the requirement that $A$ is diagonal becomes superfluous, because you can always reduce to that case with another transformation of the same kind.
Each block in this canonical form determines a polynomial factor of $\det(A+tB)$, apart from type IV which is present only in pairs for which $\det(A+tB)\equiv 0$. So you just need to check which blocks correspond to factors that have no real zeros to find a complete characterization of these matrix pairs in terms of their blocks. If I am not mistaken, the allowed blocks are those of type II and III.
