Timeline for A question of invertibility of matrices

6 events

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Jun 25, 2020 at 15:50	comment	added	Christian Remling		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.
Jun 25, 2020 at 10:03	comment	added	Federico Poloni		@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.
Jun 25, 2020 at 9:49	history	edited	Denis Serre	CC BY-SA 4.0	added 1 character in body
Jun 25, 2020 at 8:17	vote	accept	A beginner mathmatician
Jun 25, 2020 at 8:17
Jun 25, 2020 at 7:43	comment	added	A beginner mathmatician		I have edited the question. I guess under my hypothesis we must have $Tr(A)=Tr(B)=0$. Your example falls into this category.
Jun 25, 2020 at 6:48	history	answered	Denis Serre	CC BY-SA 4.0