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Counterexample

Is this true?

det A > 0 implies A is positive definite

No — it is false.

The counterexample

A = diag(−1, −1)

Why it fails

det = 1 > 0 but both eigenvalues are negative. All leading principal minors must be positive.

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Invariants don't determine

Two objects sharing an invariant were treated as the same object.

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Go deeper· Explained in full in Quadratic forms, signature and definiteness· Asked in CSIR NET June 2025· From Inner Product Spaces and Forms › Quadratic forms, positive definiteness, Sylvester's law

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