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Is this true?

det A > 0 implies A is positive definite

No — it is false.

The counterexample

A = diag(−1, −1)

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

The kind of mistake this is

Execution slip

The idea was right. The computation was not.

Drill statements like this
linear algebraquadratic forms

Others that fail the same way

From Inner Product Spaces and Forms › Quadratic forms, positive definiteness, Sylvester's law

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