TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28

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A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S

Statement

Let $V$ be a finite-dimensional real or complex inner product space and let $T : V \to V$ be linear. The following are equivalent:

$T$ is non-negative.
$T$ is self-adjoint and every eigenvalue of $T$ is a nonnegative real number.
Some orthonormal basis makes the matrix of $T$ diagonal with nonnegative real diagonal entries.
There exists a linear map $S : V \to V$ such that $T = S^{*} S$ .

Facts & Assumptions

Given: A finite-dimensional real or complex inner product space $V$ and a linear endomorphism $T : V \to V$ .

[L1]

A self-adjoint operator on a finite-dimensional real inner product space has an orthonormal eigenbasis (Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis).

[L2]

A normal operator on a finite-dimensional complex inner product space has an orthonormal eigenbasis (Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely).

[L3]

Every nonnegative real number has a unique nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Proof

technique · direct

1.1L1L2algebra

Assume $T$ is non-negative. Then $T$ is self-adjoint by Non-negative and positive operators. If $V$ is real, [L1] gives an orthonormal eigenbasis; if $V$ is complex, the identity $T = T^{*}$ yields $T T^{*} = T^{*} T$ , so [L2] gives an orthonormal eigenbasis. For any eigenvector $v \neq 0$ with $T v = λ v$ , the equality $⟨ T v, v ⟩ = λ ∥ v ∥^{2}$ and non-negativity give $λ \geq 0$ , and self-adjointness makes $λ$ real. Thus claim 1 implies claim 2.

1.2L3algebra

Assume claim 3. If the diagonal entries are $λ_{1}, \dots, λ_{n} \geq 0$ , then [L3] gives square roots $\sqrt{λ_{j}} \geq 0$ ; the diagonal operator $S$ with diagonal entries $\sqrt{λ_{j}}$ satisfies $S^{*} S = T$ . Thus claim 3 implies claim 4.

2.1step 1.1

In the orthonormal eigenbasis from step 1.1, the matrix of $T$ is diagonal with those nonnegative real eigenvalues on the diagonal. Thus claim 2 implies claim 3.

3.1algebra∎

Assume claim 4, so $T = S^{*} S$ . Then for every $v \in V$ one has $⟨ T v, v ⟩ = ⟨ S^{*} S v, v ⟩ = ⟨ S v, S v ⟩ \geq 0$ , and for all $u, v$ one has $⟨ T u, v ⟩ = ⟨ S u, S v ⟩ = ⟨ u, T v ⟩$ . Hence $T$ is self-adjoint and non-negative. Thus claim 4 implies claim 1.

Depends on

Used by

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)