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For closed subspaces of a Hilbert space, set $T=P_{V^{\perp }}{|}_W$. The two positive cosines bound $T$ and its adjoint operator $P_W{|}_{V^{\perp }}$ below. The first bound gives a closed range and injectivity; the second makes the range dense by image-kernel orthogonality for an adjoint. Thus $T$ is a bounded bijection, and $w=T^{-1}{P}_{V^{\perp }}f$ yields the unique direct sum decomposition $f=w+v$. In equal finite dimensions, a lower bound on $T$ alone suffices by the rank-nullity theorem.

For a bounded self-adjoint operator $L$ on a complex Hilbert space, defineThis equals the spectrum of a bounded operator after removing its isolated eigenvalues of finite multiplicity. To see isolation when the kernel is finite and the shifted range is closed, restrict the shifted operator to its kernel complement: it is invertible there by image-kernel orthogonality for an adjoint. A Neumann-series perturbation leaves nearby nonzero shifts invertible on that complement and on the finite kernel. The closed range must concern the shifted operator, not the original one.

Again put $A=\lambda I-T$. Surjectivity gives, by image-kernel orthogonality for an adjoint,If $\lambda$ is real, $A=A^{\ast }$, so $\ker A=\{0\}$. If $\lambda$ is nonreal and $Ax=0$, then $Tx=\lambda x$; reality of $⟨Tx,x⟩=\lambda \Vert x{\Vert }^2$ again implies $x=0$. Thus $A$ is injective as well as surjective, without needing to assume separately that $\lambda$ is not an eigenvalue.

Since $A$ is a bounded linear operator on the Banach space $X$, the bounded inverse theorem makes $A^{-1}$ bounded. ThereforeThis is exactly membership in the resolvent set of an operator.

Put $A=\lambda I-T$. Since $T$ is a self-adjoint operator, $A^{\ast }=\overline{\lambda }I-T$. Applying the preceding image-kernel orthogonality for an adjoint identities to $A^{\ast }$ givesIf $\lambda$ is real, $A^{\ast }=A$ and $\ker A=\{0\}$ because $\lambda$ is not an eigenvalue. If $\lambda$ is nonreal and $A^{\ast }x=0$, then $Tx=\overline{\lambda }x$. The scalar $⟨Tx,x⟩$ is real by self-adjointness, while it equals $\overline{\lambda }\Vert x{\Vert }^2$. Its imaginary part forces $x=0$. Thus $\ker A^{\ast }=\{0\}$ in both cases, andThe conjugation on $\lambda$ in the adjoint operator is essential when $\lambda$ is nonreal.

Use the convention that the inner product is linear in its first argument. By the defining identity for the adjoint operator,For the last implication choose $y=Tx$. This proves the first image-kernel orthogonality for an adjoint identity. Taking orthogonal complements and using the double orthogonal complement theorem for the vector subspace $\mathrm{Im}{T}^{\ast }$ gives

For a bounded self-adjoint operator, $g$ is orthogonal to its range exactly when $⟨Lf,g⟩=⟨f,Lg⟩=0$ for every $f$, or equivalently $Lg=0$. ConsequentlyThis is image-kernel orthogonality for an adjoint. Taking the double orthogonal complement gives the closure; it does not assert that the range itself is closed.

If unit vectors $f_n$ satisfy $(L-\lambda I)f_n\to 0$, a bounded inverse would give $1\le \Vert (L-\lambda I{)}^{-1}\Vert \Vert (L-\lambda I)f_n\Vert \to 0$. Hence $\lambda$ belongs to the spectrum of a bounded operator.

Conversely a spectral point is real by the preceding argument. Put $A=L-\lambda I$, again a self-adjoint operator. If ${\inf }_{\Vert f\Vert =1}\Vert Af\Vert >0$, then $A$ is injective with closed range. Its range is dense by image-kernel orthogonality for an adjoint, so it is bijective with a bounded inverse, a contradiction. Thus this infimum is zero; choose unit $f_n$ with $\Vert A{f}_n\Vert <1/n$. We have provedSuch a sequence is a spectral Weyl sequence. No weak convergence is required here; nonreal $\lambda$ are ruled out by the lower bound in the preceding solution.

For $u\in U$, define $\varphi (u)\in U^{\ast }$ by $\varphi (u)(u^{\prime })=⟨u,u^{\prime }⟩$. If $\varphi (u)=0$, then in particularso $u=0$. Thus $\varphi$ is injective. Since a finite-dimensional vector space and its dual space have the same dimension, $\varphi$ is also surjective and hence an isomorphism. This is the finite-dimensional real case of the Riesz representation theorem.

The adjoint operator of $\alpha :V\to W$ is the unique linear map ${\alpha }^{\ast }:W\to V$ satisfyingfor every $v\in V$ and $w\in W$. In the stated orthonormal bases, if $v$ and $w$ denote coordinate columns, thenTherefore the matrix of ${\alpha }^{\ast }$ is the matrix transpose $A^T$.

For $w\in W$,Hence $\ker {\alpha }^{\ast }=(\mathrm{im}\alpha {)}^{\perp }$. Taking orthogonal complements in the finite-dimensional space $W$ proves the image-kernel orthogonality for an adjoint

Put $r=\alpha (v_0)-w_0$. For any $h\in V$,This is minimized at $h=0$ exactly when $r\perp \mathrm{im}\alpha$, equivalently when ${\alpha }^{\ast }r=0$. Thus the normal equation for a linear inverse problem is

For the given linear least-squares problem,The normal equations $2x+y=3$ and $x+2y=5$ have the unique solution

If $T$ is invertible, the bounded inverse theorem makes $T^{-1}$ bounded, andMoreover $T^{\ast }$ is invertible with inverse $(T^{-1}{)}^{\ast }$, so the same argument shows that $T^{\ast }$ is bounded below.

Conversely, supposefor some $c,d>0$. The first bound makes $T$ injective. It also makes $\mathrm{ran}T$ closed: if $T{x}_n$ is Cauchy, then $x_n$ is Cauchy, and continuity gives its limit in the range. The second bound gives $\ker T^{\ast }=\{0\}$. By image-kernel orthogonality for an adjoint,so the range is dense. Being both closed and dense, it equals $H$. Thus $T$ is bijective, and the first lower bound gives $\Vert T^{-1}\Vert \le c^{-1}$. Therefore

Expand an arbitrary vector in the orthonormal eigenbasis:Its image has no component along ${\mathbf{e}}_0$, soAlternatively, symmetry gives $(A\mathbf{r})\cdot {\mathbf{e}}_0=\mathbf{r}\cdot A{\mathbf{e}}_0=0$. This is image-kernel orthogonality for an adjoint, with $A^{\mathsf{T}}=A$.

The support of a positive operator $A$ is the closed subspaceThe range identity follows from image-kernel orthogonality for an adjoint and $A=A^{\ast }$. Equivalently, it is the range of the spectral projection $E_A((0,\infty ))$. In finite dimension, it is the span of eigenvectors with positive eigenvalues, and the range is already closed.

Closure matters in infinite dimension. On the l2 sequence space, $A{e}_n=e_n/n$ has zero kernel and hence full support, but is not onto: $(1/n)$ is square-summable and its formal inverse image $(1,1,\dots )$ is not. In quantum relative entropy, containment of the first state's support in the reference state's support is necessary for finiteness; finite dimensions also make it sufficient, while infinite dimensions can have divergence despite that containment.