In functional analysis, the Grothendieck trace theorem is an extension of Lidskii's theorem about the trace and the determinant of a certain class of nuclear operators on Banach spaces, the so-called ${\displaystyle {\tfrac {2}{3}}}$-nuclear operators. The theorem was proven in 1955 by Alexander Grothendieck. Lidskii's theorem does not hold in general for Banach spaces.

The theorem should not be confused with the Grothendieck trace formula from algebraic geometry.

Given a Banach space ${\displaystyle (B,\|\cdot \|)}$ with the approximation property and denote its dual as ${\displaystyle B'}$.

Let ${\displaystyle A}$ be a nuclear operator on ${\displaystyle B}$, then ${\displaystyle A}$ is a ${\displaystyle {\tfrac {2}{3}}}$-nuclear operator if it has a decomposition of the form $${\displaystyle A=\sum \limits _{k=1}^{\infty }\varphi _{k}\otimes f_{k}}$$ where ${\displaystyle \varphi _{k}\in B}$ and ${\displaystyle f_{k}\in B'}$ and $${\displaystyle \sum \limits _{k=1}^{\infty }\|\varphi _{k}\|^{2/3}\|f_{k}\|^{2/3}<\infty .}$$

Let ${\displaystyle \lambda _{j}(A)}$ denote the eigenvalues of a ${\displaystyle {\tfrac {2}{3}}}$-nuclear operator ${\displaystyle A}$ counted with their algebraic multiplicities. If $${\displaystyle \sum \limits _{j}|\lambda _{j}(A)|<\infty }$$ then the following equalities hold: $${\displaystyle \operatorname {tr} A=\sum \limits _{j}|\lambda _{j}(A)|}$$ and for the Fredholm determinant $${\displaystyle \operatorname {det} (I+A)=\prod \limits _{j}(1+\lambda _{j}(A)).}$$