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For commuting real symmetric matrices $A,B$, each eigenspace of $A$ is invariant under $B$: $A(Bv)=B(Av)=\lambda Bv$. The restriction of $B$ to that eigenspace is again symmetric. The spectral theorem therefore supplies an orthonormal eigenbasis for $B$ inside each eigenspace of $A$. Combining these bases gives a common orthonormal eigenbasis of the whole space. Conversely, two symmetric matrices diagonal in the same orthonormal basis commute.

For a qubit control and finite-dimensional target with unitary operator V, the controlled unitary gate $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$ is entangling if and only if V is not scalar. If $V=e^{i\theta }I$, the controlled operation is a phase gate on the control tensored with the target identity and preserves every product state.

Otherwise the spectral theorem supplies two orthonormal eigenvectors with distinct eigenvalues ${\lambda }_0,{\lambda }_1$. Set $|b⟩=(|v_0⟩+|v_1⟩)/\sqrt{2}$ and use product input $|+⟩|b⟩$. The two conditional target vectors $|b⟩$ and $V|b⟩$ are linearly independent because their eigenbasis coefficients would otherwise require ${\lambda }_0={\lambda }_1$. The conditional-vector product-state criterion shows that their coherent controlled superposition is entangled. Thus even a target gate arbitrarily close to scalar can be entangling, although the resulting entanglement need not be large.

If a matrix with orthonormal columns is divided into two row blocks, each having at least as many rows as there are columns, it admits simultaneous thin factorizations $A_1=U_{1}C{V}^T$, $A_2=U_{2}S{V}^T$, with orthonormal $U_1,U_2$, an orthogonal matrix $V$, and nonnegative diagonal matrices $C,S$ satisfying $C^2+S^2=I$. To prove this, apply the spectral theorem to $A_1^T{A}_1$; the same orthonormal eigenbasis diagonalizes $A_2^T{A}_2=I-A_1^T{A}_1$. Normalize the nonzero columns of $A_{i}V$ and complete the zero columns to orthonormal sets. The entries of $C,S$ can be written as cosines and sines of angles in $[0,\pi /2]$.

For a real random vector $X=(X_1,\dots ,X_p{)}^T$ with finite second moments, its covariance matrix collects all pairwise covariances:The diagonal entries are variances. The covariance matrix is a symmetric matrix and a positive semidefinite matrix, because for every real vector $a$,Thus one matrix describes the variance of every linear combination of the coordinates. It is a positive-definite matrix exactly when no nonzero linear combination is constant almost surely. A deterministic linear relation produces a zero eigenvalue.

For example,has unit coordinate variances, covariance $0.8$, and eigenvalues $1.8$ and $0.2$. Its unit eigenvectors $(1,1{)}^T/\sqrt{2}$ and $(1,-1{)}^T/\sqrt{2}$ describe directions of greatest and least projected variance. For a unit vector $a=(\cos \theta ,\sin \theta {)}^T$, direct multiplication gives $a^T\Sigma a=1+0.8\sin (2\theta )$: the variation in the sum direction is nine times that in the difference direction.

The spectral theorem supplies an orthonormal basis of these directions. This connects covariance matrices to principal component analysis: its leading eigenvector identifies the unit-direction projection with the largest variance. For a linear transformation $A$, the transformed random vector has covariance matrix $A\Sigma A^T$, so uncertainties propagate through a linear model by matrix multiplication.

For symmetric negative-definite matrices, put $a={\lambda }_{\max }(A)$, $b={\lambda }_{\max }(B)$ and $c={\lambda }_{\max }(A+B)\le a+b$. The spectral theorem gives exact exponential norm bounds, and $\Vert [e^{sA},B]{\Vert }_2\le 2\Vert B{\Vert }_2{e}^{sa}$. Integration in the exponential product defect identity proves the displayed estimate if $c<a+b$. At equality its continuous limit is $2t\Vert B{\Vert }_2{e}^{tc}$; the quotient must not be interpreted as division by zero.

A finite-dimensional density operator $\rho$ represents a quantum state on a Hilbert space. It is Hermitian, positive semidefinite, and has trace one:The density matrix is its matrix in a chosen orthonormal basis. In dimension $N$ it has $N^2-1$ real parameters: a Hermitian matrix has $N^2$, and fixing its trace removes one; positivity restricts the allowed values.

A quantum state ensemble of normalized pure states $|{\psi }_i⟩$ with positive probabilities $p_i$ givesZero-weight terms can be omitted. Conversely, the spectral theorem gives such an ensemble for every density operator, using its orthonormal eigenvectors and nonnegative eigenvalues, with zero eigenvalues omitted. A pure state has $\rho =|\psi ⟩⟨\psi |$; a mixed state needs at least two nonzero eigenvalues. ^[ref]

For a projective measurement with orthogonal projections ${\Pi }_a$, the Born rule isThis is the average of the component pure state probabilities. Positivity makes each outcome probability nonnegative, and ${\sum }_a{\Pi }_a=I$ makes them sum to one. The same density operator therefore predicts every measurement on the system without requiring a unique preparation history. ^[ref]

For a closed system, Schrödinger picture evolution under a unitary operator iswhen ${\rho }_H={\rho }_S(0)$ denotes the time-independent Heisenberg picture state and $U(0)=I$. In the Schrödinger picture the density operator evolves; in the Heisenberg picture the observables evolve instead, with identical measurement predictions. This unitary operator preserves the eigenvalues and hence the purity of a density operator. ^[ref]

A historical source for the probabilistic formulation of quantum mechanics is von Neumann's 1927 paper Wahrscheinlichkeitstheoretischer Aufbau der Quantenmechanik. ^[ref]

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

Every self-adjoint operator on a finite-dimensional real or complex inner product space has an orthonormal eigenbasis, and all its eigenvalues are real. In orthonormal coordinates, a real symmetric matrix has the form $A=Q\Lambda Q^T$, with $Q$ an orthogonal matrix and $\Lambda$ a real diagonal matrix. A complex Hermitian matrix instead has $A=U\Lambda U^{\ast }$, with $U$ a unitary matrix and $U^{\ast }$ its conjugate transpose. These are the self-adjoint cases of the spectral theorem; Axler gives the real and complex finite-dimensional statements in Section 7B.^[ref]

An orthonormal eigenbasis turns a seemingly coupled linear map into independent scalar multiplications while preserving lengths and angles. If $x={\sum }_j{c}_j{e}_j$ and $A{e}_j={\lambda }_j{e}_j$, thenThe last identity removes every cross term from the quadratic form. The eigenvalues tell us whether it is positive, negative, or mixed in sign, and the eigenvectors identify the corresponding directions. This connects positive-definite matrices, quadratic forms, and the Hermitian Rayleigh quotient range.

For a fully computed example, takeDirect multiplication gives $A{e}_{+}=3e_{+}$ and $A{e}_{-}=e_{-}$. Their dot product is zero and each has norm one. ThusWriting $x=c{e}_{+}+d{e}_{-}$ changes $x^{T}Ax=2x_1^2+2x_1{x}_2+2x_2^2$ into $3c^2+d^2$. Its level set $x^{T}Ax=1$ is therefore an ellipse with semiaxis lengths $1/\sqrt{3}$ and $1$, along $e_{+}$ and $e_{-}$. The coordinates change, but the ellipse does not: orthogonal diagonalization reveals its existing geometry.

For distinct eigenvalues, orthogonality follows at once from self-adjointness: $\lambda ⟨u,v⟩=⟨Au,v⟩=⟨u,Av⟩=\mu ⟨u,v⟩$. Within a repeated eigenspace, choose an orthonormal basis; arbitrarily chosen eigenvectors of the same eigenvalue need not be orthogonal. One outline of the full argument finds an eigenvector, observes that its orthogonal complement is invariant, and repeats in that smaller space. Merely having real eigenvalues is weaker: $\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has two real eigenvalues, but its eigenvectors $(1,0{)}^T$ and $(-1,1{)}^T$ are not orthogonal.

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.

A bounded Hermitian operator determines a Hermitian form $b_T(x,y)=⟨Tx,y⟩$ and its real-valued diagonal expression $q_T(x)=b_T(x,x)$. On a complex Hilbert space,Thus it is a quadratic form on the underlying real vector space; it is generally not a quadratic form over the complex field, whose scaling rule would involve ${\alpha }^2$. This distinction explains why replacing $x^{\ast }$ by $x^{\mathrm{T}}$ loses the geometry of a complex inner product.

For a finite-dimensional Hermitian operator with an orthonormal basis of eigenvectors $e_j$ and corresponding eigenvalues ${\lambda }_j$, the spectral theorem givesFor a unit vector, this is an average of the eigenvalues with nonnegative weights summing to one. It is also the Rayleigh quotient, so its minimum and maximum are the smallest and largest eigenvalues.

In the Hermitian operator diagram, take $x(\theta )=(\cos \theta ,\sin \theta {)}^{\mathrm{T}}$ for $D,P$, and $x_H(\theta )=\cos \theta e_1+\sin \theta e_3$ for $H$. The computed curves areThese paths illustrate the extremal values; the spectral theorem supplies the bounds for every complex unit vector, including those outside the plotted paths.

A real symmetric matrix $A$ is indefinite when $x^{T}Ax$ is an indefinite quadratic form. Equivalently, $A$ has both a positive and a negative eigenvalue, by the spectral theorem. A zero eigenvalue is permitted, so an indefinite matrix need not be an invertible matrix.

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.

If every eigenvalue of a finite square matrix $A$ has negative real part, its matrix exponential decays and its integral converges. The Jordan normal form shows that polynomial factors in $t$ are dominated by exponential decay. Integrating $d(e^{tA})/dt=A{e}^{tA}$ gives $A{\int }_0^{\infty }{e}^{tA}dt=-I$, proving the displayed matrix inverse formula. In particular, for a Hermitian matrix with negative eigenvalues, the spectral theorem reduces the integral to scalar exponentials with integrals $-1/{\lambda }_j$. The minus sign is essential: the integral is positive definite in this Hermitian case, while $A^{-1}$ has negative eigenvalues.

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.

For a bounded linear operator, the series $e^A={\sum }_{n=0}^{\infty }{A}^n/n!$ converges in operator norm. For an unbounded self-adjoint operator $H$, the spectral theorem instead defines the bounded unitary operator $e^{-iHt/\hslash }$ by applying the scalar function $e^{-iEt/\hslash }$ to its projection-valued measure. This yields unitary time evolution without assuming that every vector lies in the domain of every power of $H$. In finite dimensions it is the matrix exponential.

For a self-adjoint Hamiltonian operator bounded below with ground-state energy $E_0$, every admissible nonzero trial wavefunction has a Rayleigh quotient at least $E_0$. To see this explicitly in a complete orthonormal eigenbasis, write $|\psi ⟩={\sum }_n{c}_n|n⟩$ with $H|n⟩=E_n|n⟩$. ThensoFor continuous spectral components, the spectral theorem replaces the sum by an integral with the same nonnegative integrand. Equality requires the trial quantum state to lie entirely in the ground-state subspace. The variational method chooses a tractable family, evaluates the quotient and minimizes over its parameters. This gives an upper bound; enlarging the family can only improve that bound. The trial must belong to the operator domain for the operator expression above, or have finite energy in the corresponding quadratic-form version of the quantum variational principle.

For the oscillator, use the standard kinetic operator $T=-{\hslash }^2{d}^2/(2md{x}^2)$: the original PDF omits the superscript $2$ on the derivative in its numerator. Take $a>0$, and normalize the scaled exact ground state asThe factor $\sqrt{\lambda }$ is optional in a quotient but makes the calculation transparent. Changing variable to $u=\lambda x$ and differentiating the scaled wavefunction givesAt $\lambda =1$ this trial is the exact minimizing state, so its energy $\mathcal{E}(\lambda )={\lambda }^2{T}_0+{\lambda }^{-2}{V}_0$ has zero derivative there. Hence $2T_0-2V_0=0$. Since $E_0=T_0+V_0$,This is the quantum virial identity from scaling for a quadratic potential, and hence the oscillator form of the virial theorem.

For the inverse-power attraction, take $a>0$ and suppose, for contradiction, that an exact ground state exists with separately finite kinetic energy $T_0>0$ and potential mean $U_0<0$, admitting the same dilation variations. Homogeneity now givesStationarity at $\lambda =1$ would require $2T_0+p{U}_0=0$. Its second derivative would then beFor $p>2$ this is negative, contradicting a ground-state minimum. More strongly, one does not need a presumed exact state: choose any normalized smooth test function supported away from the singular origin. Its $T$ and $U$ are finite, and its scaled quotient satisfiesThe unregularized attraction therefore exhibits fall to the centre for a supercritical inverse-power potential and has no stable ground state.

There is a necessary qualification to the PDF's broader wording about “no bound state”. Under the usual finite-energy virial theorem assumptions, a putative negative-energy bound state would satisfywhich contradicts being below the continuum threshold at zero. Thus no negative-energy bound state satisfying those regularity assumptions is possible. But this singular differential expression needs a specified domain at $x=0$; separately divergent kinetic and potential means or boundary conditions not preserved by dilation invalidate that ordinary calculation. The domain assumptions in singular-potential virial arguments must therefore be stated: the scaling proof establishes absence of a semibounded ground-state problem, rather than a domain-independent theorem about every formal square-integrable solution. A short-distance regularization would be a different potential.

Insert the complete orthonormal basis of energy eigenstates on either side of the Boltzmann factor. The spectral theorem givesAt low temperature, the gap $\hslash \omega$ suppresses excited energy eigenstates. The harmonic oscillator eigenfunction of the ground state isThereforeThe common zero-point energy factor cancels. Dividing the kernel by its trace would give the position probability density of the canonical ensemble; at zero temperature this tends to $|{\phi }_0(x){|}^2$, whose variance is $\hslash /(2m\omega )$. Thus zero temperature projects onto the ground state and retains its quantum spread, rather than concentrating at the minimum of the classical potential energy.

There is also a check on the first correction. With $q=\sqrt{m\omega /\hslash }x$, the first excited harmonic oscillator eigenfunction satisfies $|{\phi }_1(x){|}^2=2q^2|{\phi }_0(x){|}^2$ and ${\phi }_1(0)=0$. Hence the spectral theorem gives $R=e^{-q^2}[1+2q^2{e}^{-z}+O(e^{-2z})]$, in agreement with part (c).

Proceed by induction on the dimension. For positive dimension, the fundamental theorem of algebra gives a root of the characteristic polynomial, hence a nonzero eigenvector $v$. Normalize it to have norm one. Its span is an invariant subspace, and part (ii) makes its orthogonal complement invariant too:The restriction of $T$ to the complement remains skew-adjoint, since the same inner product identity holds there. The induction hypothesis supplies an orthonormal eigenbasis for that smaller space. Adjoining $v$ gives an orthonormal eigenbasis for $V$.

The zero-dimensional base case is immediate. Thus $T$ is diagonalizable, indeed unitarily diagonalizable. This proves the finite-dimensional case of the spectral theorem for a skew-adjoint operator without assuming that distinct eigenvalues have multiplicity one.

For a self-adjoint operator on a finite-dimensional complex inner product space, every eigenvalue is real and eigenspaces for distinct eigenvalues are orthogonal. The spectral theorem gives an orthonormal eigenbasis. Accordinglyand the $E_i$ are orthogonal projections satisfying $E_i^{\ast }=E_i$. Since $T$ is nonsingular, none of its eigenvalues is zero. Its representing matrix in the orthonormal eigenbasis is real diagonal.