False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A tensor is not determined by its values on diagonal tuples without symmetry

Statement

False claim: a tensor is determined by its values on diagonal tuples $(v, \dots, v)$ without any symmetry hypothesis.

Facts & Assumptions

Given: On $V = R^{2}$ , the bilinear forms $B_{1} (x, y) = x_{1} y_{2}$ and $B_{2} (x, y) = x_{2} y_{1}$ .

[F1]

Bilinear forms are type $(0, 2)$ tensors (A type $(r, s)$ tensor on a finite-dimensional vector space).

Refutation

technique · direct

1.1

By [F1], both $B_{1}$ and $B_{2}$ are tensors of the same type. They are distinct because $B_{1} (e_{1}, e_{2}) = 1$ while $B_{2} (e_{1}, e_{2}) = 0$ .

F1given

1.2

For every $v = (a, b) \in R^{2}$ , one has $B_{1} (v, v) = a b = B_{2} (v, v)$ . Thus the two tensors agree on every diagonal pair $(v, v)$ .

givenalgebra

2.1

Therefore diagonal values alone do not determine a general tensor.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)