# Perturbation of matrices

Let $$A(t)$$ be a symmetric $$n\times n$$ matrix that continuously depend on $$t\in [0,1]$$. Let $$\lambda_1(t)$$ stand for the smallest eigenvalue for $$A(t)$$.

Question. Does there exist a Lebesgue measurable vector $$v_1(t)$$ on $$(0,1)$$ that is not zero for almost every t and satisfies $$A(t)v_1(t)=\lambda_1(t)v_1(t)$$ almost everywhere on $$(0,1)$$ such that $$v_1 \in L^{\infty}((0,1);\mathbb R^n)?$$

• I think the Courant minimax principle will probably help? Commented Oct 10, 2023 at 4:07
• Yes, $\lambda_1$ is continuous (see for example Kato's book), and then it's clear, though probably quite tedious to prove formally, that $v_1$ can be chosen as a measurable function. Being in $L^{\infty}$ is no extra requirement since we can normalize. Commented Oct 10, 2023 at 14:11

It should not be hard to prove - e.g. by some minmax-characterization - that $$\lambda_1$$ is measurable. (As already remarked in the comments by Christian Remling, $$\lambda_1$$ is actually continuous, but measurability of $$\lambda_1$$ is sufficient for the following argument.)
Then the existence of a measurable selection $$v_1$$ of the zeroes (e.g. on the unit sphere) of the superposition operator generated by the Caratheodory function $$f(t,v)=A(t)v-\lambda_1(t)v$$ is a trivial special case of a Filippov type implicit function theorem. See, for instance, Theorem 3' in https://core.ac.uk/download/pdf/81993694.pdf.