Spectral Theorems

positive square root

Every nonnegative number has exactly one nonnegative square root. The spectral theorem gives operators the same privilege. If T is a positive operator, there is one and only one positive operator S with S^2 = T, written S = sqrt(T) or T^(1/2). The construction is the functional calculus applied to the function f(x) = sqrt(x): take the spectral resolution T = sum lambda_i P_i and set sqrt(T) = sum sqrt(lambda_i) P_i.

This makes sense exactly because T is positive: every lambda_i >= 0, so sqrt(lambda_i) is a genuine nonnegative real number and the projections are untouched. The result is self-adjoint (real eigenvalues), positive (nonnegative eigenvalues), commutes with T, and squares back to T. Uniqueness among positive operators is the subtle and important part: there are many operators whose square is T (any sign choice on eigenvalues, or non-self-adjoint roots), but only one of them is itself positive.

The positive square root is a quiet workhorse behind two headline factorizations. In the polar decomposition every operator A factors as A = U P with U a partial isometry and P = sqrt(A^* A) the positive square root of A^* A, the operator analogue of writing a complex number as (magnitude)(phase). And the singular values of A are exactly the eigenvalues of P, which is why the SVD and the spectral theorem for A^* A are two views of one structure. Whenever you need to take a clean half-step of a positive transformation, this is the tool.

T = sum lambda_i P_i, lambda_i >= 0 -> sqrt(T) = sum sqrt(lambda_i) P_i, P = sqrt(A^* A) in A = U P

The positive square root via functional calculus, and its role as the positive factor in the polar decomposition.

Uniqueness is only within the positive operators. Drop that requirement and a positive T has many square roots, just as 4 has square roots 2 and -2; only the positive one is meant by sqrt(T).

Also called
square root of a positive operatorsqrt(T)