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For a bounded linear operator $T:H\to K$ between real or complex Hilbert spaces, its adjoint is the unique bounded linear operator $T^{\ast }:K\to H$ satisfyingThe inner product here is linear in its first argument. The adjoint operator transfers an output measurement back to the input: measuring $Tx$ against the vector $y$ gives exactly the same scalar as measuring $x$ against $T^{\ast }y$. It depends on both inner products.

In this real example,Both inner products equal $3$. The dashed lines collect all vectors giving that same measurement: the output line $u_1=3$ becomes the input line $2x_1+x_2=3$. The adjoint operator therefore moves the direction that describes the measurement. It generally differs from the inverse matrix: here $A^{-1}y=(1/2,0{)}^{\mathrm{T}}$.

For an explicitly complex example, use the standard inner product $⟨u,v⟩={\sum }_j{u}_j\overline{v_j}$ on ${\mathbb{C}}^2$. In the standard orthonormal basis, the adjoint operator is represented by the conjugate transpose:For $x=(1+i,2{)}^{\mathrm{T}}$ and $y=(1,i{)}^{\mathrm{T}}$, direct matrix multiplication givesand henceThe complex conjugate is essential to this agreement.

Existence follows from the Riesz representation theorem: for fixed $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ has a unique representing vector $T^{\ast }y$. The resulting adjoint operator has the same operator norm as $T$, andMIT's lecture 18 develops the representation argument and the conjugate transpose example.^[ref]

This transfer explains the normal equation for a linear inverse problem: a least-squares solution has residual $Tx-b$ orthogonal to every attainable direction $Th$, so $⟨h,T^{\ast }(Tx-b)⟩=0$ for every $h$, equivalently $T^{\ast }Tx=T^{\ast }b$. An output residual becomes an input condition through the same adjoint operator. All identities in this section concern bounded operators defined on the whole source Hilbert space; for a densely defined operator that is not bounded, use the adjoint of a densely defined operator and check its operator domain.

A linear operator $Z$ on a Banach space is called a dissipative operator whenOn a Hilbert space, using the inner product linear in its first argument,If $\mathrm{Re}⟨Zf,f⟩\le 0$, this proves the defining inequality. Conversely that inequality implies $2\lambda \mathrm{Re}⟨Zf,f⟩\le \Vert Zf{\Vert }^2$ for every $\lambda >0$; letting $\lambda \to \infty$ gives

For the next spectral assertion, dissipativity must be retained as a hypothesis. Let $Z$ be closed, densely defined and dissipative. For $\alpha >0$, the dissipative bound makes $\alpha I-Z$ injective. Its range is closed: if $(\alpha I-Z)f_n\to h$, the bound makes $f_n$ Cauchy, and then $Z{f}_n=\alpha f_n-(\alpha I-Z)f_n$ converges. Closedness produces a preimage of $h$. If the range is all of $H$, the same bound gives a bounded inverse, so $\alpha$ is in the resolvent set. If the closed range is proper, there is a nonzero vector in its orthogonal complement, and the definition of the adjoint of a densely defined operator givesThis proves that the positive spectrum of a dissipative operator is adjoint point spectrum:Without dissipativity the PDF's assertion is false. For example, multiplication by $x$ on $L^2(0,2)$ is bounded, closed and densely defined. The value $1$ is spectral: normalized functions supported closer and closer to $x=1$ make $\Vert (Z-I)f\Vert$ tend to zero. But $Z^{\ast }=Z$ has no eigenvector for eigenvalue $1$, since a square-integrable function supported on the singleton $\{1\}$ is zero.

For the given maximal-domain operator $Q$, finite-support sequences belong to $D(Q)$, so the domain is dense. Coordinatewise limits also show it is closed: if $f^{(j)}\to f$ and $Q{f}^{(j)}\to h$ in ${\ell }^2$, each defining coordinate equation passes to the limit, giving $Qf=h\in {\ell }^2$ and hence $f\in D(Q)$.

Put $g_n=2^n{f}_n$ and $h=Qf$. For $n\ge 1$,Since ${\sum }_{n\ge 1}{2}^{-n}|h_n|<\infty$ by Cauchy-Schwarz inequality, $g_n$ has a finite limit ${\ell }_f$. Its increments are square summable as well. Direct summation givesThe original inner-product series converges because $Qf,f\in {\ell }^2$. Taking $N\to \infty$ therefore givesThus $Q$ is dissipative, including its full maximal domain; the limiting boundary term has not been discarded.

To decide generation, solve $(\alpha I-Q)f=h$ for arbitrary $h\in {\ell }^2$ and fixed $\alpha >0$. The first coordinate gives $g_0=f_0=h_0/(\alpha +1)$, and the rest giveBecause all denominators are at least one,Hence $g_n$ is bounded, so $f_n=2^{-n}{g}_n$ belongs to ${\ell }^2$. The coordinate equations give $Qf=\alpha f-h\in {\ell }^2$, proving $f\in D(Q)$ and surjectivity of $\alpha I-Q$. Together with density, closedness and dissipativity, the Lumer-Phillips theorem proves the contraction generator from an exponentially weighted difference operator conclusion:

The domain is crucial. The formal transposed matrix suggests a positive-eigenvalue solutionwhich is square summable because the products converge. It is not a true adjoint eigenvector: the admissible test vector $f_n=2^{-n}$ has $Qf=(-1,0,0,\dots )$, so $⟨Qf,u⟩=-\overline{u_0}$, while $\alpha ⟨f,u⟩=\alpha \overline{u_0}{\sum }_n{4}^{-n}{\prod }_{k<n}(1+\alpha 4^{-k})$. These cannot agree for $\alpha >0$ unless $u_0=0$. This exemplifies how boundary terms exclude formal adjoint eigenvectors, and explains why a calculation using only finite-support test vectors would give the wrong generation answer.