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Part BCSIR NET December 2023leading-principal-minors

Leading principal minors

For a∈Ra \in \mathbb{R}, let A_a = [[2, −1, 0], [−1, 2, −1], [0, −1, a]]. Which one of the following statements is true?

  1. A.A_a is positive definite for all a < 3.
  2. B.A_a is positive definite for all a > 3.✓ Correct
  3. C.A_a is positive definite for all a ≥ −2.
  4. D.A_a is positive definite only for finitely many values of a.

The idea, in plain words

For a symmetric matrix to be positive definite, every top-left square block must have positive determinant. The first two do not involve the unknown at all, so the whole question comes down to where the last determinant turns positive.

Solution

Leading principal minors: 2, 3 and det A_a = 3a − 2.

Positive definite iff a > 2/3

which contains all a > 3.

Why each option is right or wrong

Checked against 4 symbolic computations
  • A.WrongBoundary and endpoint

    False.

    Sylvester's criterion needs all leading principal minors positive: they are 2, 3, and det A_a

    which SymPy evaluates as 3a − 2.

    So A_a is positive definite exactly when a > 2/3.

    The range a < 3 includes a = 0, where the determinant is −2 (verified) — not positive definite.

  • B.Correct

    True.

    The condition is a > 2/3 (SymPy: 3a − 2 = 0 at a = 2/3), and every a > 3 satisfies it.

    Note the option does not claim to describe all positive-definite cases — only that this range works

    which it does.

    The trigger: for a tridiagonal symmetric matrix with one unknown, compute the leading principal minors in order; only the last one carries the parameter.

  • C.WrongBoundary and endpoint

    False. a ≥ −2 includes a = −2, where the determinant is 3(−2) − 2 = −8 (verified)

    so the matrix is indefinite there.

    The threshold is 2/3, well above −2.

  • D.Wrong

    False.

    The positive-definite set is the open ray (2/3,∞)(2/3, \infty)

    which is infinite.

    Positive definiteness is an open condition given by strict inequalities

    so it is never satisfied by only finitely many values of a continuous parameter.

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Every real symmetric matrix is positive definite if det > 0

More on this topic

The chapter behind this: Quadratic forms, signature and definiteness — free to read

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

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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