4.5.39 · D5Linear Algebra (Full)
Question bank — Quadratic forms — positive definite, negative definite, indefinite
Prerequisites worth having open: Eigenvalues and Eigenvectors, Symmetric Matrices, Spectral Theorem, Determinants, Hessian Matrix, Second Derivative Test.
True or false — justify
A quadratic form can be built from a non-symmetric matrix.
True — any matrix gives , but only its symmetric part affects the output, so we always pick the unique symmetric .
If every entry of is positive, then is positive definite.
False — entries and eigenvalues are different things. has all-positive entries but eigenvalues , so it is indefinite.
If is positive definite then every diagonal entry is positive.
True — by feeding in the standard basis vector, which is nonzero.
The converse holds: positive diagonal entries force positive definiteness.
False — diagonals positive is necessary but not sufficient; has positive diagonal yet eigenvalues of mixed sign.
A matrix with is positive definite.
False — for , only says the eigenvalues share a sign; two negative eigenvalues also give positive determinant (that would be negative definite).
Positive definite .
True — and all makes the product strictly positive (in particular is invertible).
Every indefinite matrix is invertible.
False — "indefinite" only requires one positive and one negative eigenvalue; a third eigenvalue could be zero (e.g. ), making .
If is positive definite, so is .
True — the eigenvalues of are , and reciprocals of positive numbers stay positive.
If is positive definite, so is .
True — eigenvalues become , and squares of nonzero reals are positive; in fact is PD even when is merely invertible and symmetric.
Swapping to turns positive definite into negative definite.
True — every eigenvalue flips sign, so all-positive becomes all-negative; this is exactly why the ND minor test is the PD test applied to .
Spot the error
" is negative definite because all its leading minors are negative."
Error: ND needs the minors to alternate , not all be negative. The correct condition is , since (which must be PD) has leading minors equal to .
" with , so is indefinite."
Error: hitting zero at a nonzero vector without ever going negative means positive semidefinite, not indefinite. Indefinite requires strictly positive and strictly negative outputs.
" is symmetric with , so it is positive definite."
Error: the zero eigenvalue kills strictness — equals along the second eigenvector. This is positive semidefinite.
"Sylvester's criterion says proves PSD."
Error: the leading-minor test cleanly certifies only strict PD (). Semidefiniteness needs all principal minors (every same-index diagonal sub-block), not just the leading (top-left) ones — has yet is negative semidefinite (NSD).
" has , so it starts out looking positive definite."
Error: instantly rules out PD, and a negative (with no zeros) means indefinite; you must read all the minors before concluding anything.
"The cross term means ."
Error: symmetry splits the cross-coefficient in half, so . Only the sum equals the coefficient of .
"Since the Hessian has a positive trace, the critical point is a minimum."
Error: trace is the sum of eigenvalues; it can be positive while one eigenvalue is negative (e.g. ), giving a saddle. You need all eigenvalues positive.
Why questions
Why must be symmetric for the eigenvalue test to work at all?
The Spectral Theorem guarantees real eigenvalues and an orthogonal (rotation) diagonalization only for symmetric ; without symmetry eigenvalues can be complex and the sign classification breaks.
Why does the change of variable let us "read off" definiteness?
It rotates axes so becomes a pure sum of squares; each , so the whole sign is dictated purely by the signs of the .
Why doesn't the rotation change whether is zero?
is orthogonal, hence invertible, so ; the "for all " condition survives the coordinate change intact.
Why do mixed-sign eigenvalues guarantee both output signs actually occur?
Plug in the eigenvector for a positive to get a positive value and the eigenvector for a negative to get a negative value — the sum-of-squares form makes each direction reachable.
Why does the Second Derivative Test reduce to definiteness of the Hessian Matrix?
Near a critical point ; the quadratic term dominates, so a PD curves up everywhere (min), ND curves down (max), indefinite gives a saddle.
Why can Sylvester's criterion fail to detect negative definiteness if you only check "all minors negative"?
Because ND is PD of , and each minor scales by ; even-sized minors flip back to positive, producing the alternating pattern rather than all-negative.
Why is a sum of squares like automatically positive definite?
Every squared term is , and the only way all of them vanish simultaneously is ; nonzero input forces at least one positive term, so .
Why is Cholesky Decomposition evidence of positive definiteness?
with real invertible gives for ; such a factorization exists exactly for PD matrices, so success of Cholesky is itself a PD certificate.
Edge cases
Is the zero matrix positive semidefinite, negative semidefinite, or both?
Both — for every satisfies and simultaneously; it is the degenerate case sitting on every boundary at once.
Can a matrix be indefinite?
No — a single eigenvalue cannot be both positive and negative, so a scalar is only PD (), ND (), or the degenerate zero case; indefiniteness needs at least two eigenvalues.
What is ?
Indefinite — eigenvalues are and ; note is negative at and positive at even though it has zero diagonal.
If , what definiteness types are still possible?
PSD, NSD, or indefinite — a zero determinant means a zero eigenvalue, which rules out strict PD and ND but leaves every "semi" or mixed case open.
Does adding a tiny to a PSD matrix make it positive definite?
Yes — it shifts every eigenvalue up by , so any zero eigenvalues become strictly positive; this "regularization" trick is exactly how Least Squares stabilizes a singular normal-equations matrix.
If and are both positive definite, is positive definite?
Yes — is a sum of two strictly positive numbers for ; definiteness is preserved under addition (though not under general products).
If is PD and is a scalar, what is ?
Negative definite — scaling multiplies every eigenvalue by the negative , flipping all-positive into all-negative.
Can a positive definite matrix have a negative off-diagonal entry?
Yes — off-diagonals may be any sign; e.g. has eigenvalues and is PD despite the negative entries.
Recall One-line survival guide
PD = all = all leading minors = Cholesky exists. ND = flip signs, minors alternate. Semidefinite = a zero sneaks in. Indefinite = both signs appear — and beware: positive entries, positive trace, or positive determinant alone prove none of these.