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An approximate eigenvalue of a bounded operator $T$ is a scalar $\lambda$ admitting unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$. Every approximate eigenvalue belongs to the spectrum; for a normal operator, all spectral points are approximate eigenvalues.

The continuous functional calculus of a bounded normal operator $T$ is a representation of $C(K)$, with $K=\sigma (T)$. The Riesz-Markov-Kakutani representation theorem represents $f\mapsto ⟨f(T)v,v⟩$ by a positive regular measure ${\mu }_v$. The assignment $f\mapsto f(T)v$ is isometric for the $L^2({\mu }_v)$ norm, and extends to a unitary map onto the closed cyclic subspace it generates. Multiplication by $z$ and $\overline{z}$ represent $T$ and $T^{\ast }$. Orthogonal decomposition into cyclic reducing subspaces, allowing an uncountable family, constructs the Borel functional calculus for a normal operator without requiring separability.

For a normal operator with disconnected spectrum $K=K_1\bigsqcup K_2$, the indicator function of a nonempty proper clopen component union is continuous on $K$. The continuous functional calculus makes it a nonzero proper self-adjoint linear projection commuting with $T,T^{\ast }$. Its range is therefore a nontrivial closed reducing subspace. No eigenvector is needed.

A finite family of pairwise commuting normal operators on a finite-dimensional complex inner product space, commuting also with each other's adjoints, generates a commutative unital C-star algebra. Its finite maximal ideal space identifies the algebra with functions on finitely many points. The characteristic functions of those points pull back under the Gelfand transform to mutually orthogonal self-adjoint linear projections summing to the identity. Their ranges are the simultaneous eigenspaces, giving simultaneous diagonalization without choosing separate incompatible eigenbases.

A bounded operator $T:H\to H$ on a real or complex Hilbert space is Hermitian, or self-adjoint, when $T=T^{\ast }$. Equivalently,Its directional values $q_T(x)=⟨Tx,x⟩$ are real, since this identity makes $q_T(x)$ equal its complex conjugate. Reality allows either sign; a positive operator additionally requires nonnegative values in every direction.

The diagram compares Hermitian quadratic forms along paths of unit vectors. The Hermitian matrix $D=\mathrm{diag}(1,-1)$ takes the values $1$ and $-1$ on the standard basis vectors; $P=\mathrm{diag}(1,0)$ takes the values $1$ and $0$; andsatisfies $H{e}_1=e_1$ and $H{e}_3=3e_3$. These eigenvectors form an orthonormal basis, and every $x=a{e}_1+b{e}_3$ has $q_H(x)=|a{|}^2+3|b{|}^2$. Thus $H$ is a positive operator, with eigenvalues $1,3$ and operator norm $3$.

Three conditions describe different geometry. A Hermitian operator satisfies $T=T^{\ast }$; a normal operator satisfies $T^{\ast }T=T{T}^{\ast }$; a unitary operator satisfies $T^{\ast }T=T{T}^{\ast }=I$, where $I$ is the identity operator. Every Hermitian operator and every unitary operator is a normal operator. Concrete matrices separate the conditions:

The spectrum of a bounded Hermitian operator is real. In finite dimension, the spectral theorem supplies an orthonormal basis of eigenvectors, turning its action into independent real scalings. This makes the geometry easy to interpret: each eigenvalue records a signed response along an orthogonal direction. An arbitrary bounded Hermitian operator in infinite dimension need not have any eigenvectors; the general spectral theorem uses a projection-valued measure. MIT's lecture 21 develops the real spectrum and the orthogonality of eigenspaces.^[ref]

A historical connection is the study of forms with conjugate variables. In his 1901 account of Charles Hermite's work, Camille Jordan describes these forms and emphasizes their role in reducing forms with real or complex coefficients. In today's notation, $x^{\ast }Hx$ is the diagonal value of a Hermitian form; the conjugate variables make a real-valued expression possible even with complex coefficients.^[ref]

For a densely defined operator outside this bounded setting, the displayed pairing identity alone defines a symmetric operator. A self-adjoint operator must also have the same operator domain as its adjoint operator.

For a normal operator on a real or complex inner product space, equality of $\Vert Tv\Vert$ and $\Vert T^{\ast }v\Vert$ gives $\ker T=\ker T^{\ast }$. If $T^{2}v=0$, put $w=Tv$. Then $Tw=0$ forces $T^{\ast }w=0$, and the adjoint operator identity givesThus $Tv=0$ and $\ker T^2=\ker T$. Applying this implication successively to $T^{k-2}v,T^{k-3}v,\dots ,v$ provesIn particular a nilpotent normal operator is zero. The argument uses no spectral theorem.

On a real or complex inner product space of finite dimension, a normal operator has a square-free polynomial as its minimal polynomial. To prove this without the spectral theorem, suppose an irreducible polynomial $f$ occurs at least twice in $m_T$. Put $h=m_T/f^2$ and $N=f(T)$. Since polynomials of a normal operator are normal, $N$ is a normal operator. For every $v$,By kernel stabilization for a normal operator, $Nh(T)v=0$. Hence the smaller-degree monic polynomial $m_T/f$ annihilates $T$, contradicting minimality.

Over the complex numbers, all irreducible polynomials are linear, so the primary decomposition theorem for linear maps now gives a basis of eigenvectors. Over the real numbers, irreducible quadratics can remain: a nontrivial planar rotation need not have any real eigenvectors, although its minimal polynomial is square-free.

On the new subspace of cusp forms, a common eigenspace for the Hecke operators away from the level has dimension one. Distinct newforms, even at different primitive levels, have different systems of these eigenvalues. Consequently the commuting bad-prime operators also act by scalars on these lines. This is the multiplicity-one part of Atkin-Lehner-Li theory, not a consequence of normal operator diagonalization alone. See Stein's statement of the theorem.

A bounded normal operator admits a normal square root even when no continuous scalar square root exists on its Banach algebra spectrum. Choose a bounded Borel branch $s(z)=|z{|}^{1/2}\exp (i\mathrm{Arg}(z)/2)$, with $s(0)=0$. The Borel functional calculus for a normal operator gives $S=s(T)$; multiplicativity gives $S^2=T$ and preservation of the adjoint gives $S^{\ast }S=S{S}^{\ast }=|s{|}^2(T)$. Unlike a positive square root, this square root need not be unique.

For a normal operator, (a) gives $\Vert Tx\Vert =\Vert T^{\ast }x\Vert$, so one side is zero exactly when the other is. Hence $\boxed{\ker T=\ker T^{\ast }}$.

For real $\lambda$, the adjoint of $T-\lambda I$ is $T^{\ast }-\lambda I$. Expanding both products shows that their difference is $T^{\ast }T-T{T}^{\ast }=0$. Thus $T-\lambda I$ is normal, and applying the kernel of a linear map conclusion to it givesThis proves equality of each real eigenspace, hence the same eigenvalues and eigenvectors on the stated real vector space. If neither has real eigenvalues, the conclusion is vacuous. Upon complexification one instead pairs $\lambda$ with $\overline{\lambda }$ for the adjoint; no claim of the same complex eigenvalues is needed here.

Restricting $T^{\ast }T=T{T}^{\ast }$ to $V_i$, using (i), gives $T_i^{\ast }{T}_i=T_i{T}_i^{\ast }$, so the global normal operator property implies the normal operator property of every restriction. Conversely, if the restrictions are normal and the adjoint preserves each summand, then for $x={\sum }_i{x}_i$,Thus $\boxed{T\text{ is normal}⟺\text{every }{T}_i\text{ is normal}}$, with condition (i) automatic under the stated decomposition.

In fact the adjoint invariance follows from the orthogonal decomposition and the invariance under $T$, without using the normal operator property. For $x\in V_i$ and $y\in V_j$ with $j\ne i$,since $Ty\in V_j$. Hence $T^{\ast }x$ is orthogonal to every other summand and therefore lies in $V_i$. Moreover, for $x,y\in V_i$, the adjoint identity restricted to that subspace shows

Let $e_1,\dots ,e_d$ be the given orthonormal basis, with $T{e}_j={\lambda }_j{e}_j$. From the defining identity of the adjoint operator,These coordinates prove $T^{\ast }{e}_j={\overline{\lambda }}_j{e}_j$ without assuming a matrix formula. Thereforefor every basis vector. Thus $T$ is a normal operator, andThe condition of possessing an orthonormal basis of eigenvectors is already part of the hypothesis; real eigenvalues alone would not imply self-adjointness for an arbitrary linear transformation.

For a polynomial $f(x)={\sum }_{j=0}^d{a}_j{x}^j$, the defining identity and uniqueness of the adjoint operator giveIndeed, the adjoint operator of a sum is the sum of the adjoint operators, the adjoint of $aT$ is $\bar{a}{T}^{\ast }$, and the adjoint of a product $AB$ is $B^{\ast }{A}^{\ast }$. Since the normal operator $T$ commutes with $T^{\ast }$, each power of $T$ commutes with each power of $T^{\ast }$. Multiplying the two finite sums therefore givesHence polynomials of a normal operator are normal, including constant polynomials.

For the normal operator $T$, the adjoint operator identity and $T^{\ast }T=T{T}^{\ast }$ giveFor the last equality, $⟨T{T}^{\ast }v,v⟩=⟨T^{\ast }v,T^{\ast }v⟩$ and complex conjugation allow reversal of the two arguments. Taking the nonnegative square roots provesIn particular the two kernels coincide.

Suppose some exponent $q_j\ge 2$. DefinePart (iii) makes $N$ a normal operator. For every $v\in V$, put $w=h(T)v$. The minimal polynomial identity givesPart (ii), applied to $N$, forces $Nw=0$. As this holds for every $v$,This contradicts minimality: $m_T/f_j$ is a monic polynomial of smaller degree annihilating $T$. Therefore the minimal polynomial of a normal operator is a square-free polynomial over either of the given fields:This argument does not require the irreducible polynomials to be linear, so it also covers the real case.

Take $N=T-\lambda I$. It is a polynomial in the normal operator $T$, so part (b)(iii) makes $N$ a normal operator. Also $N^{\ast }=T^{\ast }-\bar{\lambda }I$. If $Tv=\lambda v$, part (b)(i), applied to $N$, givesThe norm vanishes only at the zero vector, soThis is the adjoint eigenvector identity for a normal matrix expressed independently of coordinates.

Let $K=\sigma (T)$ for a bounded normal operator on a nonzero complex Hilbert space. The zero Hilbert space has the trivial calculus. Take the inner product linear in its first argument. The continuous functional calculus is an isometric unital star-representation $\pi :C(K)\to \mathcal{B}(H)$ with $\pi (z)=T$. The Borel calculus will act on $B_b(K)$, the algebra of actual bounded Borel measurable functions. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.

Here is a construction that also covers a nonseparable Hilbert space. For $v\in H$, the positive functional $f\mapsto ⟨\pi (f)v,v⟩$ has a finite regular Borel measure ${\mu }_v$ by the Riesz-Markov-Kakutani representation theorem. On continuous functions,Continuous functions are dense in $L^2(K,{\mu }_v)$. Hence $f\mapsto \pi (f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi (C(K))v}$. This subspace reduces $T$ and $T^{\ast }$, which correspond to multiplication by $z$ and $\overline{z}$. Decompose $H$ as an orthogonal direct sum of such cyclic reducing subspaces, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the cyclic multiplication model for a normal operator.

On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a unital star-homomorphismThus $(fg)(T)=f(T)g(T)$, $\overline{f}(T)=f(T{)}^{\ast }$, and positivity of $f$ gives positivity of $f(T)$. The continuous functional calculus is recovered on $C(K)$.

Set $E(S)=\Phi (1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an orthogonal projection, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the strong operator topology. Indeed on every multiplication summand the indicators add pointwise, and the tail squared norm is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a projection-valued measure, andFor $v,w\in H$, polarization gives scalar measures ${\mu }_{v,w}(S)=⟨E(S)v,w⟩$ satisfying $⟨f(T)v,w⟩=\int fd{\mu }_{v,w}$. Regularity of these finite scalar measures and uniqueness in the Riesz-Markov-Kakutani representation theorem make the projection-valued measure unique: its values on continuous functions determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.

The correct convergence assertion is bounded pointwise convergence in the strong operator topology. If $f_n\to f$ pointwise and ${\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$, thenby the dominated convergence theorem. This need not be convergence in operator norm. For multiplication by $t$ on $L^2[0,1]$, the functions $1_{[0,1/n]}$ converge pointwise to $1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.

A Borel measurable function can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\epsilon \})=0$ for every $\epsilon >0$, andThe upper norm bound follows by integrating $|f{|}^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\Vert f(T)\Vert \ge C$. The induced map on the quotient by these spectral-null Borel measurable functions is isometric. Every nonempty relatively open subset of $K$ has nonzero spectral projection, since otherwise a nonzero continuous function supported there would contradict faithfulness of the continuous functional calculus. This full support gives the usual continuous spectral mapping theorem, but for a Borel measurable function the exact statement is insteadIf an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\Vert (f(T)-\lambda )v\Vert$ arbitrarily small, contradicting invertibility. This is spectral essential range in Borel functional calculus. For example $1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.

Spectral projections provide reducing subspaces, and $E(\{\lambda \})H=\ker (T-\lambda I)$: the squared norm of $(T-\lambda )v$ is $\int |z-\lambda {|}^{2}d{\mu }_v$. In finite dimension the calculus is simply $f(T)={\sum }_{j}f({\lambda }_j)P_j$, the sum over eigenspaces. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ with $s(0)=0$, giving a normal operator $s(T)$ with $s(T{)}^2=T$ even when a continuous scalar branch is unavailable.

Any bounded operator commuting with both $T$ and $T^{\ast }$ commutes with their continuous functional calculus and with all $E(S)$. To justify the latter, regularity approximates $1_S$ by uniformly bounded continuous functions in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the Borel calculus lies in the Von Neumann algebra generated by $T$. On a separable Hilbert space, a countable dense list of vectors gives a controlling measure $\mu ={\sum }_j{2}^{-j}{\mu }_{v_j}/(1+\Vert v_j{\Vert }^2)$; spectral-null sets are exactly its null sets, and the range is the represented $L^{\infty }(K,\mu )$, equal to that Von Neumann algebra. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in [0,1]$, the generated Von Neumann algebra has arbitrary bounded diagonal functions, while the original Borel calculus has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the Borel functional calculus for a normal operator from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.

For any complex $\lambda$, the shifted linear operator $T=\alpha -\lambda I$ is normal: expansion of $T^{\ast }T-T{T}^{\ast }$ leaves just ${\alpha }^{\ast }\alpha -\alpha {\alpha }^{\ast }=0$. Apply part (i) to this shifted normal operator:Hence a nonzero vector is an eigenvector of $\alpha$ with eigenvalue $\lambda$ if and only if it is an eigenvector of ${\alpha }^{\ast }$ with eigenvalue $\bar{\lambda }$. The eigenspaces coincide, with conjugate eigenvalues.

A normal operator satisfies ${\alpha }^{\ast }\alpha =\alpha {\alpha }^{\ast }$. Moving an operator across the inner product givesOne of these norms is zero exactly when the other is zero. Therefore