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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.

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    $\begingroup$ @People who are trying to close the question. Hi! The question looks very innocent and may be foolish. But look already it is leading to something nontrivial. I do not know why I am still getting negative votes! $\endgroup$
    – A beginner mathmatician
    Commented Jun 26, 2020 at 13:32

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much. Actually that is what was looking for. I think in your paper A and B are real matrices . Right? Also could not see E_4. I can see E_1,E_2 and E_3 conditions. $\endgroup$
    – A beginner mathmatician
    Commented Jun 26, 2020 at 3:17
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    $\begingroup$ @Abeginnermathmatician Oops, I wrote this answer looking at the Arxiv preprint, where we had a variant with a fourth block type E4 (EKCF vs. EWCF). I have fixed it, as well as the theorem and table number according to the published version. $\endgroup$
    – Federico Poloni
    Commented Jun 26, 2020 at 6:02
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    $\begingroup$ @Abeginnermathmatician You are correct that in my paper the theorem is phrased for real symmetric matrices only, but it holds without changes for complex Hermitian ones. I have now edited my answer and references to point to the original paper by Thompson that we cite; that reference treats directly the Hermitian-Hermitian case. $\endgroup$
    – Federico Poloni
    Commented Jun 26, 2020 at 6:12
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    $\begingroup$ Thank you again. For the similarity can we take $X$ in the paper to be unitary? The matrix $X$ is referred to as "nonsingular constant matrix". $\endgroup$
    – A beginner mathmatician
    Commented Jun 26, 2020 at 13:31
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    $\begingroup$ @Abeginnermathmatician no, $X$ is not unitary in general, unfortunately. Thompson's setting uses matrix pencils, i.e., linear expressions $\lambda A + \mu B$ in the variables $\lambda, \mu$, so he has to specify that $X$ does not depend on $\mu$ nor $\lambda$, i.e., it is "constant". However, this setting is equivalent to considering matrix pairs $(A,B)$, and in that case $X$ is just a general invertible complex matrix. $\endgroup$
    – Federico Poloni
    Commented Jun 26, 2020 at 14:06
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What about $$A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\qquad B=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\quad ?$$

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  • $\begingroup$ I have edited the question. I guess under my hypothesis we must have $Tr(A)=Tr(B)=0$. Your example falls into this category. $\endgroup$
    – A beginner mathmatician
    Commented Jun 25, 2020 at 7:43
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    $\begingroup$ @Abeginnermathmatician Replace $A \leftarrow A-\frac12I$ to get a counterexample without zero trace. More generally, $\det(A+xB) = -x^2-1$, so small perturbations to $A$ and $B$ will give a small perturbation of this polynomial, which still has no real zero. This is an open condition. $\endgroup$
    – Federico Poloni
    Commented Jun 25, 2020 at 10:03
  • $\begingroup$ In complete generality, since $tA+B = t(A+(1/t)B)$, the second assumption is only adding the single requirement that $B$ is invertible. We can rephrase and say that we're looking at $A+tB$, $t\in\mathbb R_{\infty}$ (to interpret this rigorously, take advantage of the fact that we can always multiply by a non-zero number), which is a compact space, and the invertible matrices form an open set, so a small perturbation of an example always is an example too. $\endgroup$
    – Christian Remling
    Commented Jun 25, 2020 at 15:50

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