Finite support square roots and generalized inverses

Let $H$ be a finite-dimensional complex inner-product space, and let $A:H\to H$ be a linear endomorphism that is positive semidefinite: $A=A^{\ast }$ and $⟨Ax,x⟩\ge 0$ for every $x\in H$. Here $A^{\ast }$ denotes its adjoint. Write $A={\sum }_j{\lambda }_j{P}_j$ using an orthonormal eigenbasis, where $P_j$ is orthogonal projection onto the line spanned by its $j$th basis vector. Define the support $\mathrm{supp}A=(\ker A{)}^{\perp }$, the positive square root $A^{1\mathrm{/}2}={\sum }_j\sqrt{{\lambda }_j}{P}_j$, and the support inverse square root $A^{-1\mathrm{/}2}={\sum }_{{\lambda }_j>0}{\lambda }_j^{-1\mathrm{/}2}{P}_j$. Thus $A^{-1\mathrm{/}2}$ is zero on $\ker A$. The orthogonal projector onto $\mathrm{supp}A$ is denoted $P_{\mathrm{supp}A}$, and $A^{-1\mathrm{/}2}A{A}^{-1\mathrm{/}2}=P_{\mathrm{supp}A}$. All operators here are finite-dimensional; no unbounded inverse is asserted.

Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely

The adjoint T^*:W→ V is characterised by ⟨ Tv,w⟩_W=⟨ v,T^*w⟩_V

Every linear map between finite-dimensional inner product spaces has a unique adjoint

Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space

The orthogonal complement W^⊥={v:⟨ v,w⟩=0 for all w∈ W}

The orthogonal projection P_Wv is the W-component in V=W⊕ W^⊥

Real and complex inner product spaces, with the inner product linear in the first argument

Orthogonal vectors and subspaces, orthogonal and orthonormal sets, and orthonormal bases

Eigenvalues, eigenvectors, eigenspaces E_λ(T)=ker(T-λ I), and the spectrum σ_F(T) of an endomorphism

Kernel and image of a linear map

Finite-dimensional vector space, and its dimension dim_F V; infinite-dimensional means having no finite basis

Linear map between vector spaces over the same field

Linear subspace of a vector space

Linear combination of a finite list, and the span span(S) as the smallest linear subspace containing S

Proof technique: direct.

1.1 By [F2], $A^{\ast }$ exists uniquely. Since $A=A^{\ast }$, [F3] makes $A$ normal, so [F1] supplies an orthonormal eigenbasis. For an eigenvector $v\mathrm{\ne }0$ with eigenvalue $\lambda$, self-adjointness makes $\lambda$ real and positivity gives $\lambda \ge 0$. The kernel is the span of the eigenvectors with eigenvalue zero, and its orthogonal complement is the span of those with positive eigenvalues. [F1, F2, F3, F4, given, algebra]

2.1 For each distinct eigenvalue $\lambda$, the orthogonal projection onto its eigenspace equals ${\prod }_{\mu \mathrm{\ne }\lambda }(A-\mu I)\mathrm{/}(\lambda -\mu )$, where $\mu$ ranges over the other distinct eigenvalues; the empty product is $I$. This follows by applying the polynomial to each vector in an orthonormal eigenbasis. Hence each eigenspace projection, and therefore both sums in the definition, is independent of the choice of eigenbasis. The zero-dimensional case has only the zero endomorphism and empty sums. [F5, step 1.1, algebra]

3.1 On an eigenvector with eigenvalue $\lambda$, the square of $A^{1\mathrm{/}2}$ acts by $\lambda$, while $A^{-1\mathrm{/}2}A{A}^{-1\mathrm{/}2}$ acts by $1$ if $\lambda >0$ and by $0$ if $\lambda =0$. Consequently $(A^{1\mathrm{/}2}{)}^2=A$, $A^{1\mathrm{/}2}$ is positive semidefinite, and the latter product is the orthogonal projector onto $(\ker A{)}^{\perp }$. This establishes the stated construction and identity, including when $A=0$. [F2, F3, F4, F5, step 1.1, step 2.1, algebra] ∎