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Choose a real unit eigenvector $e_1$ of the real symmetric matrix $Q$, with eigenvalue ${\lambda }_1$. Its orthogonal complement is an invariant subspace: for any $v$ with $e_1^{T}v=0$,The restriction of $Q$ to this $(n-1)$-dimensional space is again real and symmetric. Applying the same argument inductively produces an orthonormal basis $e_1,\dots ,e_n$ of real eigenvectors. The base case $n=1$ is immediate. This proves the spectral theorem for real symmetric matrices without assuming that distinct eigenvalues are available.

One can also check that eigenvectors with different eigenvalues are automatically orthogonal: ${\lambda }_i{e}_i^T{e}_j=(Q{e}_i{)}^T{e}_j=e_i^{T}Q{e}_j={\lambda }_j{e}_i^T{e}_j$. Within a repeated eigenspace, the Gram-Schmidt process preserves the eigenvector property. Thus repeated eigenvalues cause no obstruction, and $\boxed{Q=OD{O}^T\text{ for an orthogonal }O\text{ and real diagonal }D}$.

The quadratic form is ${\mathbf{x}}^{T}Q\mathbf{x}$ for the real symmetric matrixThe $y$ axis has eigenvalue $-3$. On the $xz$ plane, the characteristic polynomial isThus the orthogonal diagonalization from part (a) yieldsFor later use, one possible orthonormal basis isThen $u=(3x+z)/\sqrt{10}$, $v=y$, $w=(x-3z)/\sqrt{10}$. This explicit choice also confirms that the Euclidean norm is preserved: $x^2+y^2+z^2=u^2+v^2+w^2$.

In the orthonormal basis of part (b), the three quadric surfaces have the following forms. For $\alpha =1$,a two-sheeted hyperboloid with axis $u$, vertices $(\pm 1/\sqrt{6},0,0)$, and one sheet on each side of the plane $u=0$. For $\alpha =0$,an elliptic cone with vertex at the origin and the same axis. For $\alpha =-1$,a one-sheet hyperboloid with axis $u$ and a waist ellipse at $u=0$, whose semiaxes are $1/\sqrt{3}$ along $v$ and $1/2$ along $w$. The sections at fixed $u$ enlarge away from the waist.

Choose an orthonormal basis with $\mathbf{n}={\mathbf{e}}_3$. A half-turn about $\mathbf{n}$ kills all mixed axial-transverse components. Commutation with a quarter-turn makes the transverse block a scalar identity matrix plus an antisymmetric matrix. The resulting oriented-axis invariant Cartesian tensor, in coordinate-free form, iswith real constants $a,d,h$. The Kronecker delta, the outer product $n_i{n}_j$, and the displayed Levi-Civita symbol contraction are invariant under every proper rotation about $\mathbf{n}$, so this form is also sufficient. In the chosen basis,The antisymmetric part must be retained unless further symmetry is imposed.

Decompose the electric field as $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=(\mathbf{n}\cdot \mathbf{E})\mathbf{n}$. The electric current density isFor a nonzero real electric field, the alignment of current and field in an axisymmetric conductor gives the complete classification of collinearity:

For completeness, if ${\mathbf{E}}_{\perp }\ne 0$ and $\mathbf{J}=\lambda \mathbf{E}$, taking the dot product with $\mathbf{n}\times {\mathbf{E}}_{\perp }$ forces $h=0$. The transverse equation then gives $\lambda =a$, while a nonzero axial component gives $\lambda =d$. This proves there are no missing directions. Zero current is included as the zero scalar multiple; if “parallel” requires a direction for $\mathbf{J}$, exclude cases where that scalar is zero. For a strictly anisotropic crystal the last possibility is absent.

The matrix sending the standard basis vectors to the given columns isWith the Hermitian inner product, the squared column norms are $3/6+1/6+2/6=1$, $3/6+1/6+2/6=1$, and $4/6+2/6=1$. The off-diagonal products areThus the columns form an orthonormal basis, so $\boxed{U^{†}U=I}$ and $U$ is a unitary matrix. Here the operation maps the old basis vectors to the new vectors; the inverse coordinate change of basis uses $U^{†}$.

A real square matrix $Q$ is an orthogonal matrix when $Q^{T}Q=I$, equivalently its columns form an orthonormal basis. Here $Q^T$ is the transpose and $I$ the identity matrix. For every vector $x$,Taking nonnegative square roots provesThus an orthogonal matrix preserves the Euclidean norm.

All $n$ singular values are positive, so $\Sigma$ and $V$ are invertible matrices and $A$ has matrix rank $n$. The image of a linear map is thereforewith those columns furnishing an orthonormal basis. Also $Ax=0$ implies $\Sigma V^{T}x=0$ and hence $x=0$. ThusFinally $A^{T}z=V\Sigma U^{T}z=0$ exactly when $U^{T}z=0$. The remaining columns of the full orthogonal matrix span this orthogonal complement:If $m=n$, this last null space also equals $\{0\}$ and has the empty basis.

A real square orthogonal matrix $Q$ satisfies $\boxed{Q^{T}Q=Q{Q}^T=I}$. Equivalently its columns form an orthonormal basis. In particular it preserves the Euclidean norm, since $\Vert Qx{\Vert }_2^2=x^T{Q}^{T}Qx=\Vert x{\Vert }_2^2$.

Let $e_j=x-x^{(j)}$. The underlying Richardson iteration gives $e_j=(I-A{)}^j{e}_0$. The polynomials $1,(1-t),\dots ,(1-t{)}^k$ form a basis for polynomials of degree at most $k$. Thus writeDefine $y^{(k)}={\sum }_{j=0}^k{c}_j{x}^{(j)}$. Since the coefficients sum to one,The spectral theorem for real symmetric matrices supplies an orthonormal basis of eigenvectors. In that basis $p_k(A)$ has diagonal entries $p_k({\lambda }_i)$, soThe algebraic recombination is valid in exact arithmetic even if the original unaccelerated iterates themselves fail to converge.

The nullspace is the vector subspace $\boxed{\ker A=\{x\in {\mathbb{R}}^n:Ax=0\}}$. A full real singular value decomposition iswhere $U,V$ are orthogonal matrices and ${\Sigma }_{ij}=0$ for $i\ne j$, with ${\Sigma }_{jj}={\sigma }_j\ge 0$ for $1\le j\le n$. We may order ${\sigma }_1\ge \cdots \ge {\sigma }_n\ge 0$. The columns of $V$ form an orthonormal basis of the input space ${\mathbb{R}}^n$, and the columns of $U$ form one of the output space ${\mathbb{R}}^m$. The positive singular values number exactly the matrix rank.

Let $r$ be the matrix rank, and partition the right singular vectors as $V=(V_r{V}_0)$, with $V_0\in {\mathbb{R}}^{n\times (n-r)}$ consisting of the columns associated with zero singular values. From $Ax=0⟺\Sigma V^{T}x=0$, these columns form an orthonormal basis of the nullspace. Therefore its orthogonal projector isIt acts on the input space ${\mathbb{R}}^n$. Symmetry and idempotence follow from $V_0^T{V}_0=I_{n-r}$. At full column rank it is zero, and at rank zero it is the identity.

The intended orthogonal-polynomial convention is $\mathrm{deg}{p}_k=k$. An arbitrary orthonormal basis without this degree ordering does not satisfy the claim: permuting a standard basis can place the nonzero constant polynomial last, where it has no roots rather than $n+1$. The following proof uses the required degree ordering.

It makes $p_k$ orthogonal to every polynomial of degree less than $k$, since $p_0,\dots ,p_{k-1}$ form a basis of that lower-degree space. Let $c_1,\dots ,c_r$ be the distinct roots in $(a,b)$ at which $p_k$ changes sign, and put $q(x)={\prod }_{j=1}^r(x-c_j)$, with $q=1$ if $r=0$. The product $p_{k}q$ has constant sign on $(a,b)$ except at its finitely many zeros: each sign change of $p_k$ is canceled by that of $q$.

If $r<k$, then $q$ has degree below $k$, so orthogonality gives ${\int }_a^b{p}_{k}qw=0$. But a nonzero constant-sign continuous product against the positive weight has a nonzero integral, a contradiction. Thus $r\ge k$. Since $\mathrm{deg}{p}_k=k$, there are exactly $k$ such roots, all simple, with no additional endpoint roots. In particularThis also includes $k=0$, where the polynomial is nonzero constant and has no roots.

Let $E(t)$ be the matrix whose columns are the moving orthonormal basis vectors. Then $E{E}^T=I$. Differentiating givesso $W=\dot{E}{E}^T$ is skew-symmetric. Every real skew-symmetric $3\times 3$ matrix has the unique formwhich acts on a vector as $Wv=\mathbf{\omega }\times v$. Since $\dot{E}=WE$, each column satisfiesThe vector $\mathbf{\omega }$ is the instantaneous angular velocity. It may also be written $\mathbf{\omega }=\frac{1}{2}{\sum }_i{\mathbf{e}}_i\times {\dot{\mathbf{e}}}_i$.

Weak convergence means $\ell (u_j)\to \ell (u)$ for every bounded linear functional $\ell \in X^{\ast }$. Norm convergence implies this by $|\ell (u_j-u)|\le \Vert \ell \Vert \Vert u_j-u\Vert$. The converse fails for coordinate unit vectors in ${\ell }^2$: their norms are one, but pairing with any square-summable sequence tends to zero, so they are weakly null.

For the subsequence assertion apply Gram-Schmidt process to a countable dense subset to choose a countable orthonormal basis $(e_m)$ of the separable Hilbert space and a bound $M$ for the sequence. Repeated subsequence selection and diagonalization make every coefficient $⟨u_{j_k},e_m⟩$ converge to a scalar $a_m$. Bessel inequality gives ${\sum }_{m=1}^N|a_m{|}^2\le M^2$ for every $N$, so $u={\sum }_m{a}_m{e}_m$ exists in $H$ with norm at most $M$. Pairings with finite linear combinations of the basis converge. Approximating any other test vector by such combinations and using the common norm bound proves $u_{j_k}\rightharpoonup u$. The finite-dimensional case uses the same argument with a finite basis.

Apply this to every subsequence of the bounded $L^2$ sequence. Any weakly convergent further subsequence has limit $v$ with ${\int }_a^{b}v=0$, since interval indicators are $L^2$ tests. The permitted fact gives $v=0$ almost everywhere. If the whole sequence failed to converge weakly to zero, some fixed test functional and subsequence would stay a positive distance from zero. Its weakly convergent further subsequence would then contradict the preceding conclusion. Thus the entire sequence converges weakly to zero in $L^2(0,1)$.

The spectral theorem for compact self-adjoint operators used here concerns a compact self-adjoint operator: its nonzero eigenvalues are real, have finite multiplicity and can accumulate only at zero; the orthogonal complement of its kernel is the closed span of its orthonormal eigenvectors. Self-adjointness is essential for this eigenbasis statement.

Let $Kf$ be the unique zero-parameter weak solution. Uniqueness gives linearity. Testing with $Kf$ and using coercivity gives $\Vert Kf{\Vert }_{H^1}\le {\nu }^{-1}\Vert f{\Vert }_2$. The Rellich-Kondrachov theorem makes the embedding $H_0^1(\Omega )\hookrightarrow L^2(\Omega )$ compact on this bounded $C^1$ domain, so $K$ is compact on $L^2$. Symmetry givesIt is self-adjoint and positive. It is injective: if $Kf=0$, the weak equation pairs $f$ to zero with all $H_0^1$ tests, dense in $L^2$, so $f=0$. Moreover $⟨Kf,f⟩=B[Kf,Kf]>0$ for $f\ne 0$.

The parameter equation is equivalent to $u=K(f+\lambda u)$, that isA solution of this operator equation in $L^2$ lies in $H_0^1$, because its right-side representation is $K(f+\lambda u)$, and the equivalence follows in both directions. For $\lambda \le 0$, the form $B[u,v]-\lambda ⟨u,v⟩$ remains coercive; the preceding minimization argument gives a unique weak solution.

The spectral theorem now supplies a complete orthonormal basis $({\varphi }_j)$ of $L^2$ with $K{\varphi }_j={\kappa }_j{\varphi }_j$, ${\kappa }_j>0$, and ${\kappa }_j\to 0$. The elliptic eigenvalues are ${\lambda }_j=1/{\kappa }_j>0$, with finite multiplicity and tending to infinity, and $L{\varphi }_j={\lambda }_j{\varphi }_j$ weakly. If $\lambda$ is not one of them, thenThis converges even in $H_0^1$: its squared energy norm is ${\sum }_j{\lambda }_j|⟨f,{\varphi }_j⟩{|}^2/|{\lambda }_j-\lambda {|}^2$, finite because the multiplier is bounded over $j$. Partial sums solve the weak equation with the corresponding $L^2$ projections of $f$. Passing to the limit proves the equation for every test; the same coefficients force uniqueness. Thus the expansion represents the unique weak solution away from the elliptic eigenvalues.

The Pusey-Barrett-Rudolph theorem excludes overlapping preparation distributions for distinct pure states in an ontological model of a quantum system satisfying preparation independence and the specified quantum predictions. It excludes neither all hidden variables nor every informational use of a quantum state. The precise overlap criterion is essential: a psi-epistemic model lets at least two distinct pure-state preparations assign positive common probability mass to the same underlying states; a psi-ontic model has mutually singular measures for each distinct pair.^[ref]

Let preparation $x$ sample $\lambda \in \Lambda$ from ${\mu }_x$. A fixed quantum measurement $M$ has nonnegative response functions ${\xi }_M(j|\lambda )$ with ${\sum }_j{\xi }_M(j|\lambda )=1$. Averaging over the preparation must reproduce the Born rule. The response is fixed by $M$ and the complete underlying state, not by additional knowledge of the preparation label. No outcome determinism or Bell local causality is assumed; the joint measurement response need not factor into local responses. For independently operated preparations, preparation independence further identifies the joint state space with the appropriate product and assignsOperationally independent devices do not logically entail this ontic factorization; it is an additional physical premise.^[ref]

The two-qubit example can be checked directly. Set $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. In computational basis order $00,01,10,11$, take the following normalized columns as an orthonormal basis:Their pairwise inner products vanish. Direct multiplication shows that the respective preparations $00,0+,+0,++$ have zero probability amplitude for respective outcomes $1,2,3,4$. This is the PBR exclusion measurement for zero and plus: the measurement does not perfectly distinguish the preparations, but each outcome rules one out.

Suppose a common submeasure $\nu$ of ${\mu }_0$ and ${\mu }_{+}$ has mass $\omega >0$; with densities one can take $\nu (\lambda )=\min \{{\mu }_0(\lambda ),{\mu }_{+}(\lambda )\}$. Preparation independence makes $\nu \otimes \nu$ a common submeasure of all four product preparations, of mass ${\omega }^2$. The zero probability for outcome $j$ under its excluded preparation and nonnegativity forceSumming and using normalization yields ${\omega }^2=0$, a contradiction. This measure formulation avoids treating a set with merely nonempty topological support as a positive-probability overlap.

For arbitrary distinct nonorthogonal pure states, write $|{\psi }_{0,1}⟩=\cos (\theta /2)|0⟩\pm \sin (\theta /2)|1⟩$, with overlap $\cos \theta$ and $0<\theta <\pi /2$. The general proof constructs an exclusion measurement on $n$ independently prepared copies when $2^{1/n}-1\le \tan (\theta /2)$. A finite $n$ always exists. The same common-submeasure contradiction then excludes positive overlap. Orthogonal preparations are already disjoint if their perfectly distinguishing measurement is reproduced. The original paper also bounds residual overlap for imperfect exclusion data; exact disjointness concerns the ideal theorem, not a finite experiment's proof of exact zero.^[ref]

The scope is consequently sharp. Bohmian mechanics is not excluded: its underlying state includes the wave function, making distinct pure preparations disjoint even though extra configuration variables remain. A psi-ontic model need not be complete with the wave function alone, and the theorem does not identify the wave function with a three-dimensional material field. Models abandoning preparation independence, or accounts not representing preparations as uncertainty over observer-independent states of this specified kind, are outside this conclusion. Epistemic uncertainty between mixed preparations is not banned. Nor does the theorem establish a collapse mechanism, choose between Bohmian and Everettian theories, or solve the measurement problem. Its achievement is a conditional impossibility theorem about a precise form of pure-state epistemic overlap.

A Hermitian operator satisfies the adjoint symmetry conditionfor vectors in its specified domain. For a finite matrix this means $A=A^{†}$. For an unbounded differential operator, boundary conditions and domains matter: symmetry on a chosen domain is not alone sufficient for self-adjointness. Physical observables require the corresponding self-adjoint operator and its spectral measure. This distinction prevents purely formal integration by parts from being mistaken for a complete operator definition.

Real eigenvalues. If $A|a⟩=a|a⟩$ with nonzero norm, Hermiticity impliesso $a=a^{\ast }$. Thus eigenvalues representing measurement outcomes are real.

Orthogonality. For eigenvectors $|a⟩,|b⟩$,Hence $(a-b)⟨b|a⟩=0$, and distinct eigenvalues have orthogonal eigenvectors. Within a degenerate eigenspace, Gram-Schmidt orthogonalization can choose an orthonormal basis. In finite dimensions the spectral theorem gives a complete orthonormal eigenbasis andwhere $P_a$ projects onto each eigenspace. In infinite dimensions a continuous spectrum may require a spectral integral rather than a normalizable eigenvector for every possible outcome; a position operator on the full line is a familiar example.

Measurements and probabilities. In the discrete case the Born rule assigns probability $⟨\psi |P_a|\psi ⟩$ to the outcome $a$. For a nondegenerate eigenbasis, $|\psi ⟩={\sum }_a{c}_a|a⟩$ gives probabilities $|c_a{|}^2$, withThe variance vanishes precisely when the state lies in an eigenspace with a single sharp outcome. A repeatable ideal projective measurement prepares the corresponding eigenspace; superpositions of different eigenvalues do not have a preassigned sharp value of that observable in this description.

Compatibility and dynamics. Commuting Hermitian matrices admit simultaneous diagonalization, permitting a common set of sharp labels. For example $L^2,L_z$ commute, whereas $L_x,L_y$ do not. In the differential-operator setting the corresponding spectral compatibility requires appropriate domain conditions. Noncommuting observables obey the Robertson uncertainty principleThe canonical commutation relation then gives $\Delta x\Delta p\ge \hslash /2$.

The Hamiltonian operator represents energy. A self-adjoint $H$ generates unitary time evolution, $U(t)=e^{-iHt/\hslash }$, preserving norm and total probability; its eigenvalues determine stationary-state phases and transition energies. Hermiticity supplies real outcomes and orthogonal eigenspaces, while self-adjointness and the spectral theorem make these measurement and evolution rules mathematically complete.

The specified energy requires $n_x+n_y=2$, so the relevant energy eigenspace has the orthonormal basisApplying the creation and annihilation operators givesThus its matrix in this ordered basis isWriting $q=\hslash \lambda$, the characteristic polynomial gives $\lambda ({\lambda }^2-4)=0$. The possible angular momentum eigenvalues are thereforeFor completeness, normalized simultaneous energy and angular momentum eigenstates areEach value occurs once in this energy eigenspace. Equivalently, circular ladder operators of a two-dimensional oscillator give $\hat{H}=\hslash \omega ({\hat{N}}_{+}+{\hat{N}}_{-}+1)$ and ${\hat{L}}_z=\hslash ({\hat{N}}_{+}-{\hat{N}}_{-})$. With $N_{+}+N_{-}=2$, the difference can only be $2,0,-2$.

Orthogonality means zero inner product, not merely that two objects look different. For complex vectors, $⟨u,v⟩={\sum }_i{u}_i^{\ast }{v}_i$; for functions a common choice is $⟨u,v⟩=\int w(x)u(x{)}^{\ast }v(x)dx$ with positive weight $w$. An orthonormal set has $⟨u_m,u_n⟩={\delta }_{mn}$.

For an orthogonal set, the component of $f$ along $u_n$ has coefficientA complete orthonormal set gives $f={\sum }_n{c}_n{u}_n$ in the appropriate norm and Parseval's identity $\Vert f{\Vert }^2={\sum }_n|c_n{|}^2$. Without completeness there can be a nonzero residual orthogonal to the entire chosen set. Orthogonal nonzero vectors are linearly independent, but independence alone does not imply orthogonality.

Fourier coefficients are projections becauseLikewise, for a self-adjoint operator, $A{u}_n={\lambda }_n{u}_n$ gives $({\lambda }_n-{\lambda }_m)⟨u_m,u_n⟩=0$, so distinct eigenvalues have orthogonal eigenvectors. Degeneracy does not itself force a chosen pair to be orthogonal, but an orthonormal basis can be selected within each eigenspace. The differential version uses the correct Sturm-Liouville weight and boundary conditions.

For mechanical normal modes with unequal masses the relevant inner product is $⟨u,v{⟩}_{\mathsf{M}}=u^{†}\mathsf{M}v$. This makes the modes in D14 orthogonal even when an ordinary unweighted dot product does not. Expanding in such coordinates diagonalizes the kinetic and potential quadratic forms, explaining why modal motions evolve independently and their energies add.

The principal moments of inertia are the eigenvalues of the symmetric inertia tensor. In units of $\mathrm{kg}{\mathrm{m}}^2$, its characteristic determinant isTherefore the requested algebraic answer isAn orthonormal basis of principal axes is $(1,1,0)/\sqrt{2}$, $(1,-1,0)/\sqrt{2}$, and $(0,0,1)$; the last two span the degenerate eigenspace and may be replaced by any orthonormal basis of it.

There is a physical inconsistency in calling the supplied matrix a rigid body's inertia tensor. The moment-of-inertia triangle inequalities require each principal moment to be no greater than the sum of the other two. Here $3>1+1$. In principal coordinates, $I_2+I_3-I_1=2\int x_1^{2}dm$ cannot be negative for a nonnegative mass distribution. Thus the eigenvalue calculation is well defined, but no ordinary rigid mass distribution realizes this tensor. The positive second-moment matrix test for an inertia tensor makes the same obstruction explicit.