Connect positive energy with spectral square roots

Try

Consider the energy q(x,y)=5x²+2xy+2y². Completing the square gives q=5(x+y/5)²+(9/5)y², so it is positive for every nonzero real input. Its matrix [[5,1],[1,2]] is symmetric. Now compare x²+6xy+y²: both diagonal coefficients are positive, but at (1,−1) its value is −4. Cross terms matter, and inspecting only diagonal entries cannot certify positive energy.

Predict

Predict how a positive square root should act on an eigenvector. If T scales a direction by λ≥0, its positive square root should scale it by √λ. Negative choices also square to λ, but they are excluded by positivity. For diag(4,0,9), the positive root is diag(2,0,3). The zero direction remains a genuine kernel; taking a square root does not recover information lost by T.

Experiment

Use the following exact table as a spectral energy diagram. For each basis direction, compare the energy coefficient and its square-root factor. Then change the second eigenvalue from zero to −1 and explain why a positive self-adjoint square root is impossible.

Direction	Eigenvalue of T	Eigenvalue of positive root	Energy at the unit vector
First	4	2	4
Second	0	0	0
Third	9	3	9

Understand

A self-adjoint T is positive semidefinite when ⟨Tx,x⟩≥0 for every x, and positive definite when the inequality is strict for nonzero x. In an orthonormal eigenbasis this expression is Σλⱼ|cⱼ|², proving that positivity is equivalent to nonnegative eigenvalues, with strict positivity equivalent to all eigenvalues positive. Define T¹ᐟ² by replacing each λⱼ with √λⱼ. This definition does not depend on a basis chosen inside a repeated eigenspace, since the root acts there as the same scalar. For uniqueness, any positive square root R commutes with T=R² and preserves each T-eigenspace. The self-adjoint restriction of R has nonnegative eigenvalues whose squares are λ, so it equals √λI on that space.

Apply

For a Hermitian matrix A, the form q(x)=x*Ax equals ⟨Ax,x⟩ under our linear-first convention. Substitution x=Sz gives the congruent matrix S*AS, whereas operator similarity is S⁻¹AS. Congruence preserves the numbers of positive, negative and zero squares, not their numerical eigenvalues. To see the positive count is invariant, diagonalize the form. Its positive coordinate span has dimension p. Any subspace of larger dimension intersects the nonpositive coordinate span and therefore cannot be positive definite. Thus p is an intrinsic maximal dimension, preserved by invertible changes; the negative count follows by replacing the form with its negative, and nullity is preserved by invertibility. This proves the inertia classification over R, with the analogous Hermitian statement over C.

Unique positive root does not mean unique square root

The identity has a unique positive semidefinite square root, namely itself. But reflection across a line also squares to the identity. For a real or complex self-adjoint S with S²=I, the spectral theorem says its eigenvalues are +1 and −1. Put E=(S+I)/2. Then E²=E and E is self-adjoint, so E is the orthogonal projection onto its image: its image and kernel are orthogonal, because ⟨Ex,y⟩=⟨x,Ey⟩ vanishes when Ey=0. Conversely any orthogonal projection E gives a self-adjoint root S=2E−I. These two constructions are inverse. Thus all such roots correspond to subspaces, not just choices of diagonal signs in one fixed basis.

In R², apart from I and −I, choose a unit vector u=(cos θ,sin θ). The root 2uuᵀ−I fixes the line spanned by u and negates its perpendicular line. Replacing θ by θ+π gives the same matrix; different lines give different roots. This continuum of self-adjoint roots does not contradict positive-root uniqueness: any nonzero negative eigenspace violates positive semidefiniteness.

Count roots over a finite field, checking the characteristic

Now discard the inner product and count all A with A²=I on a two-dimensional space over a field F with q elements. Positivity has not been defined here. When the characteristic is not two, define E₊=(I+A)/2 and E₋=(I−A)/2. Their sum is I, product zero, and each is a projection onto the corresponding eigenspace. Every vector therefore splits uniquely as a +1 vector and a −1 vector. Besides the two scalar roots I and −I, a root is specified by an ordered pair of distinct lines: the first is its positive eigenspace, the second its negative eigenspace. There are q+1 lines, from (q²−1)/(q−1), and q choices of a different second line. The total is q(q+1)+2.

When the two eigenvalues merge

Predict the number of matrices A with A²=I before changing fields. This experiment counts all matrices, without imposing an inner product or self-adjointness. Entries are residues modulo a prime.

Field F2F3F5 Verified matrix 1: rows (0, 1); (1, 0)2: rows (0, 2); (2, 0)3: rows (1, 0); (0, 1)4: rows (2, 0); (0, 1)5: rows (2, 1); (0, 1)6: rows (2, 2); (0, 1)7: rows (2, 0); (1, 1)8: rows (2, 0); (2, 1)9: rows (1, 0); (0, 2)10: rows (2, 0); (0, 2)11: rows (1, 1); (0, 2)12: rows (1, 2); (0, 2)13: rows (1, 0); (1, 2)14: rows (1, 0); (2, 2)

Predicted 14; enumerated 14 roots among 81 matrices. Selected matrix rows (0, 1) and (1, 0).

Dimension of ker(A−I): 1; dimension of ker(A+I): 1. The two eigenspaces are complementary; their dimensions sum to two.

N=A−I has entries [2, 1, 1, 2]; N² has entries [2, 1, 1, 2]. N need not square to zero: use the distinct factors (t−1)(t+1) instead.

Input v	Av	Fixed?
(0, 0)	(0, 0)	Yes
(1, 0)	(0, 1)	No
(2, 0)	(0, 2)	No
(0, 1)	(1, 0)	No
(1, 1)	(1, 1)	Yes
(2, 1)	(1, 2)	No
(0, 2)	(2, 0)	No
(1, 2)	(2, 1)	No
(2, 2)	(2, 2)	Yes

Reset

At most 625 matrices and 25 vector images are displayed or inspected per choice. Exact finite enumeration checks these examples; the general field-counting proof is separate. State is temporary and awards no mastery.

In characteristic two the division by two is unavailable and +1=−1. Set N=A−I; then A²=I exactly when N²=0. If N is nonzero in dimension two, im(N) is contained in ker(N), and rank–nullity forces both to be the same line L. Such a map is specified uniquely by L and a nonzero linear map V/L→L: first take the quotient, then apply that map and include L in V. There are q+1 choices of L and q−1 nonzero maps between the one-dimensional spaces. Thus there are q²−1 nonzero N, plus N=0, for q² roots A. These formulas apply to prime-power fields with the stated characteristic; arithmetic modulo a composite integer is not a substitute for an extension field.

For example F₂ has four roots: I, the two elementary shears and the coordinate swap. Each nonidentity root has a one-dimensional fixed space and is not diagonalizable: a diagonal matrix whose only eigenvalue is 1 would be I. Over F₃ there are fourteen roots, all diagonalizable. Do not infer the field-wide theorem solely from the enumerated small examples.

Transfer the count by constructing the data, not memorizing a formula

Over a field of odd characteristic in dimension n, a root is an ordered complementary pair of subspaces. If its positive eigenspace has dimension k, choose that subspace, then one complement. A fixed k-dimensional subspace has q^(k(n−k)) complements: after fixing one complement W, every other complement is the graph of a unique linear map W→U. Prove this using the projection onto W; it restricts to an isomorphism on any complement. This explains both the exponent and why complements cannot be counted as arbitrary disjoint lists.

In dimension three over F₂, the square-zero map N has rank at most one. For rank one choose an image line L and a kernel plane K containing L; then choose a nonzero map V/K→L. There are seven lines, three planes containing each line (lines in V/L), and one nonzero map between the remaining one-dimensional F₂ spaces. The 21 maps, plus zero, produce 22 roots. Here the image and kernel have different dimensions, so the two-dimensional formula q² must not be reused.

Continue

Positive forms express weighted geometry, curvature and energy, while semidefinite forms permit invisible directions. They also lead to the factor T*T and singular values, which apply to every map rather than only self-adjoint ones. Do not confuse existence of some algebraic square root with existence of a positive self-adjoint root; the latter is the unique object constructed here. Numerical positivity near zero needs a stated error scale.

Review Linear Algebra I when a prerequisite needs practice. Continue to Preserve geometry and diagonalize compatible operators.

For a complementary elimination-based certificate, try positive matrices, factors and energy. It develops leading-principal-minor tests, Cholesky construction and singular energy cases; a triangular Cholesky factor is distinct from the self-adjoint positive square root above.