Try
A coupled energy can be positive even when some cross terms are negative. Consider the real symmetric matrix A(t) with rows (2,1,0), (1,2,1), (0,1,t). For a column vector z=(x,y,w), its energy is the real number zᵀA(t)z. Expand it and complete squares:
2x²+2xy+2y²+2yw+tw² = 2(x+y/2)² + (3/2)(y+2w/3)² + (t−2/3)w².
Check each coefficient on the right. The first square contributes y²/2, the second contributes 3y²/2 and 2w²/3, and the final term corrects the coefficient of w². This identity exposes directions that the original entries hide. When t is greater than 2/3, every nonzero input has positive energy. At t=2/3 there is a nonzero invisible direction. Below that threshold, a negative-energy direction exists.
Find that direction before reading further: make the first two squares vanish and choose w=3. You obtain z=(1,−2,3), whose energy is 9t−6. At the boundary A(2/3)z=0, which you can verify row by row. A determinant sign alone would not supply this useful witness.
Predict
Does “all leading determinants are nonnegative” certify nonnegative energy? Test diag(0,−1). Its leading determinants are 0 and 0, but its energy at (0,1) is −1. The first zero has hidden the negative entry from the leading determinant chain.
Now predict whether positive leading determinants work without symmetry. The matrix with rows (1,4),(0,1) has leading determinants 1 and 1. Yet at (1,−1) its quadratic expression is −2. The criterion we will prove requires symmetry over the real numbers, or its conjugate analogue over the complex numbers. These are mathematical hypotheses, not cosmetic matrix formatting.
Experiment
Use the completed squares as a manual energy experiment. First set t=1, then 2/3, then 1/3. For each setting predict the sign at (1,−2,3), calculate it, and compare with the determinant. Next choose another vector to check that a single positive evaluation cannot establish positivity in every direction.
| t | Diagonal square weights | Energy at (1,−2,3) | Conclusion |
|---|
| 1 | 2, 3/2, 1/3 | 3 | Positive definite |
|---|
| 2/3 | 2, 3/2, 0 | 0 | Positive semidefinite, rank 2 |
|---|
| 1/3 | 2, 3/2, −1/3 | −3 | Not positive semidefinite |
|---|
Here A=LDLᵀ with L having rows (1,0,0), (1/2,1,0), (0,2/3,1) and D=diag(2,3/2,t−2/3). Multiplication verifies the factorization even at the zero final pivot. The three leading principal minors are 2, 3 and 3t−2. Their consecutive ratios, starting with the empty minor Δ₀=1, are exactly the displayed weights. This particular singular factorization does not authorize dividing by a zero pivot in another matrix.
Understand
Let A be an n×n matrix, with n≥1. Over the complex numbers, A∗ denotes conjugate transpose: transpose the matrix and conjugate each entry. A is Hermitian when A=A∗. Then x∗Ax is real, since its complex conjugate equals itself. Over the real numbers this condition is symmetry and ∗ becomes transpose. A is positive definite when x∗Ax>0 for every nonzero vector x, and positive semidefinite when x∗Ax≥0 for every vector. The squared Euclidean length is ||x||²=x∗x. All assertions here are exact; a rounded near-zero value is not a sign certificate.
A principal submatrix selects the same index set for rows and columns. A leading principal submatrix selects indices 1 through k; write its determinant as Δₖ. Principal minors and leading principal minors are different collections. The empty principal determinant is 1.
Completing a block square
Write a Hermitian matrix in blocks as A=(H,B;B∗,C), with H positive definite. Such H is invertible: a nonzero vector in its kernel would have zero energy. Its inverse is Hermitian because taking the adjoint of HH⁻¹=I proves (H⁻¹)∗=(H∗)⁻¹=H⁻¹. Define the Schur complement S=C−B∗H⁻¹B. For vectors x and y of the corresponding block sizes, direct expansion gives
(x,y)∗A(x,y) = (x+H⁻¹By)∗H(x+H⁻¹By) + y∗Sy.
The cross terms on the right are x∗By and y∗B∗x; the added y∗B∗H⁻¹By cancels the subtraction in S. The coordinate change (x,y)↦(x+H⁻¹By,y) is invertible. Consequently A is positive definite exactly when S is positive definite. For necessity set x=−H⁻¹By; for sufficiency separate y≠0 from y=0,x≠0. With non-strict inequalities the same argument says A is positive semidefinite exactly when S is, still under the hypothesis H positive definite. We have not inverted a singular semidefinite block.
Deriving the positive leading-minor test
The scalar version uses A=(a,r∗;r,C), where a is real. If A is positive definite, a>0 by testing the first coordinate vector, and its Schur complement S=C−rr∗/a is positive definite by the square identity. Induct on dimension. The one-dimensional case is a positive scalar. If S=LₛDₛLₛ∗ with Lₛ unit lower triangular and Dₛ positive diagonal, then
L=(1,0;r/a,Lₛ), D=diag(a,Dₛ), and A=LDL∗.
Block multiplication verifies the equality. Every diagonal entry of L is 1, so it is invertible. Its leading blocks give Δₖ=d₁⋯dₖ>0, where dⱼ are the real diagonal entries of D. In particular positive definiteness implies positive leading minors without needing eigenvalues.
Conversely suppose every leading minor is positive. Then a=Δ₁>0. On each leading (j+1)-block, subtract r/a times the first row from the lower rows. These determinant-preserving operations leave a block triangular matrix with first diagonal entry a and the leading j-block of S. Thus that minor of S is Δⱼ₊₁/a>0. Induction makes S positive definite, and the square identity makes A positive definite. This proves Sylvester’s criterion, including its construction rather than merely naming it.
Finally, any factorization A=LDL∗ with invertible L and positive diagonal D gives positive energy because L∗x is nonzero whenever x is nonzero. In the unit lower normalization, consecutive leading products give dₖ=Δₖ/Δₖ₋₁. We have proved the equivalence of positive definiteness, positive leading principal minors and positive normalized LDL∗ pivots. The empty-dimensional convention has determinant 1 and an empty factorization; the stated induction starts at dimension 1.
Why semidefiniteness needs all principal minors
Necessity uses the existing Hermitian spectral theorem and positive forms, rather than reproving diagonalization here. If A is positive semidefinite, any principal block is too: extend its vectors by zeros in the omitted coordinates. Its Hermitian eigenvalues are nonnegative, so their product, its determinant, is nonnegative.
For sufficiency assume every principal minor of A is nonnegative. Let ε be any positive real number. For an m×m principal block M, column multilinearity gives
det(M+εI) = Σ over J⊆❴1,…,m❵ of ε^(m−|J|) det(M[J]).
Here |J| counts selected indices. To prove the coefficient identity, choose columns from M at indices in J and coordinate columns εeⱼ elsewhere. In the permutation formula every coordinate column forces its own row. The remaining row and column indices are both J, and the permutation sign is that of the restricted permutation: adjoining fixed points does not change parity. The surviving sum is exactly det(M[J]). The empty choice contributes εᵐ, with coefficient 1.
All summands are nonnegative and that empty term is positive. Hence every leading minor of A+εI is positive. Sylvester’s criterion makes A+εI positive definite. For fixed nonzero x this says x∗Ax+ε||x||²>0 for every ε>0. If x∗Ax were negative, taking ε smaller than −x∗Ax/||x||² would contradict it. Therefore x∗Ax≥0. The zero vector already has zero energy, completing the proof.
This is why checking only leading minors is insufficient in the semidefinite case. For diag(0,−1) the missing singleton principal minor at index 2 detects the failure. Another consequence is useful before elimination: if a diagonal entry of a positive semidefinite Hermitian matrix is zero, every off-diagonal entry in that row must vanish. The principal 2×2 determinant is then −|aᵢⱼ|²≥0, forcing aᵢⱼ=0. A nonzero coupling beside a zero diagonal entry is a direct obstruction.
Apply
Change the worked matrix to B(t) with rows (3,1,0), (1,1,1), (0,1,t). Construct the squares yourself before revealing the result. Your certificate must include factors, a parameter boundary, and a vector that distinguishes the boundary from smaller parameters.
Review the changed certificate
The energy is 3(x+y/3)²+(2/3)(y+3w/2)²+(t−3/2)w². Thus the boundary is t=3/2. The leading minors are 3,2,2t−3. The vector (1,−3,2) has energy 4t−6 and is a kernel vector at the boundary. Above the boundary all three square weights are positive; at the boundary the energy is nonnegative with exactly one free zero direction. Below it the displayed vector proves failure.
Now test your complex convention. For H(s) with rows (2,i),(−i,s), the square is 2|x+iy/2|²+(s−1/2)|y|². The vector (−i,2) has energy 4s−2. Using i² in place of |i|² would produce the wrong threshold. Explain aloud where conjugation entered the expansion.
Continue
A positive-definite factorization supplies a verified energy decomposition and prepares a square-root triangular factor. A semidefinite boundary requires separate care: a zero weight can represent a genuine kernel, while a zero leading pivot with a nonzero coupling can expose failure of positivity. Compare the exact square identity, all-principal-minor criterion and a concrete witness before choosing a solver.
Return to exact LU existence and normalization for the algebra behind pivot ratios. Use positive operators and spectral roots to connect these elimination certificates with the already proved spectral viewpoint. This lesson supplies an alternative proof of the strict criterion; it does not claim floating-point elimination alone certifies the sign of an uncertain matrix.
