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The plane spanned by $u_1,v_1$ is an invariant subspace for $M$, sinceIts orthogonal complement is also invariant: for $w$ orthogonal to both, $u_1^{T}Mw=-(M{u}_1{)}^{T}w=0$ and $v_1^{T}Mw=-(M{v}_1{)}^{T}w=0$. The restriction of the real symmetric matrix $M^2$ to this nonempty complement therefore has a unit eigenvector $u_2$, by the spectral theorem for real symmetric matrices.

If its eigenvalue is $-{\mu }_2^2<0$, the argument in part (b) shows that $v_2=M{u}_2/{\mu }_2$ is a unit eigenvector orthogonal to $u_2$. It also lies in the same complement; explicitly,Repeating the construction decomposes every negative eigenspace into orthogonal pairs. The zero eigenspace has dimension $n-2r$ for some number $r$ of pairs, and therefore has the same parity as $n$. This is the paired eigenspaces of the square of a real skew-symmetric matrix property.

For $n=2m$, there can be $m$ distinct negative eigenvalues and no zero, or at most $m-1$ negative ones plus zero, because a nonempty zero eigenspace has dimension at least two. For $n=2m+1$, there are at most $m$ distinct negative ones plus the mandatory zero. HenceThe bound is attained by skew two-by-two blocks with distinct positive parameters, adding one zero direction when $n$ is odd.

For a real symmetric matrix $A$, find its real eigenvalues from $\det (A-\lambda I)=0$, then solve $(A-\lambda I)v=0$ for the corresponding eigenspaces. Distinct eigenspaces are orthogonal. Within each repeated eigenspace, apply the Gram-Schmidt process and normalize to obtain an orthonormal basis of eigenvectors.

Put these vectors in the columns of $O$. The spectral theorem for real symmetric matrices guarantees a full basis, so $O^{T}O=I$ andThis is the orthogonal diagonalization of $A$ and the associated orthogonal change of coordinates.

The quadratic form has real symmetric matrixIts unit eigenvectors can be chosen as $e_u=(1,1,0)/\sqrt{2}$ with eigenvalue $1$, and $e_v=(1,-1,0)/\sqrt{2}$, $e_w=(0,0,1)$ with eigenvalue $9$. SetThe orthogonal diagonalization turns the surface equation intoIt is an ellipsoid, specifically a prolate spheroid centered at the origin. Its long principal axis is the line in direction $\boxed{(1,1,0)/\sqrt{2}}$, with semiaxis length $\boxed{3}$. Each direction in the plane perpendicular to that line has semiaxis length $\boxed{1}$; the displayed $e_v,e_w$ give one orthonormal choice of the two short principal axes. Their nonuniqueness reflects the repeated eigenvalue $9$ and rotational symmetry about the long axis.

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: $x^2+y^2+z^2=u^2+v^2+w^2$.

For an efficient QR algorithm, first reduce a general real matrix to upper Hessenberg matrix by orthogonal similarities. For a real symmetric matrix, use Householder tridiagonalization; the resulting symmetric tridiagonal matrix is the usual input in this setting. These structures are preserved by suitable QR algorithm implementations and greatly reduce the work per iteration.

The output is the eigenvalues, together with an orthogonal reduction when the factors are accumulated. In the symmetric case the target is a diagonal matrix, whose diagonal entries are the real eigenvalues; the accumulated orthogonal columns are eigenvectors. For a general real matrix, the target is the real form in a Schur decomposition: an upper block triangular matrix with diagonal blocks of size one or two. The two-by-two blocks represent complex conjugate eigenvalue pairs. An arbitrary unshifted iteration need not converge, so these describe the intended converged output, with appropriate shifts and block deflation in the QR algorithm.

Since every $u_{jj}>0$, the diagonal matrix $D$ is invertible. SetThe division is by the diagonal entry of row $i$. Thus $\hat{U}$ is an upper triangular matrix with diagonal entries one and $A=LD\hat{U}$.

To use symmetry without assuming the conclusion, multiply by the matrix inverse of $L$ and its transpose:The matrix $M$ is symmetric because $A$ is a symmetric matrix. It is also an upper triangular matrix, since every factor on the right is an upper triangular matrix. A symmetric upper triangular matrix is diagonal. Its diagonal is exactly $D$, because $\hat{U}$ and $L^{-T}$ have diagonal entries one. Hence $M=D$, givingThis is the LDL decomposition.

Take the positive square roots of the diagonal entries:The matrix $\hat{L}$ is a lower triangular matrix with positive diagonal, andThis is a Cholesky decomposition. For a nonzero vector $x$, the vector ${\hat{L}}^{T}x$ is nonzero because ${\hat{L}}^T$ is invertible. ThereforeHere $\Vert \cdot {\Vert }_2$ is the Euclidean norm. Thus $A$ is a positive-definite matrix.

We use the spectral theorem to construct a sufficiently accurate approximate eigenpair. Complete $x$ to an orthonormal eigenbasis $x,x_2,\dots ,x_n$ of the symmetric matrix $B$, with corresponding eigenvalues $\lambda ,{\lambda }_2,\dots ,{\lambda }_n$. Simplicity of $\lambda$ means that every ${\lambda }_j-\lambda$ below is nonzero. PutThe orthogonal decomposition of $Ex$ shows that $x^{T}v=0$ andWith $u_{\epsilon }=x+\epsilon v$ and ${\mu }_{\epsilon }=\lambda +\epsilon a$, direct expansion cancels every first-order term:Also $\Vert u_{\epsilon }{\Vert }_2^2=1+{\epsilon }^2\Vert v{\Vert }_2^2\ge 1$. Thus its eigenpair residual has Euclidean norm $O({\epsilon }^2)$ in Big O notation.

For completeness, a real symmetric matrix $M$ always has an eigenvalue within $\Vert (M-\mu I)u{\Vert }_2/\Vert u{\Vert }_2$ of any real $\mu$ and nonzero $u$. Indeed, writing $u={\sum }_j{c}_j{q}_j$ in an orthonormal eigenbasis with eigenvalues ${\theta }_j$ givesApply this symmetric eigenvalue residual bound to $M=B+\epsilon E$. There is an eigenvalue ${\lambda }_{\epsilon }$ such thatwhich provesThis establishes the first-order perturbation of a simple symmetric eigenvalue without presupposing differentiability of the perturbed eigenvector. If $n=1$, the sum defining $v$ is empty and the displayed eigenvalue formula is exact.

Put $B=A_1^T{A}_1$. It is a symmetric matrix and a positive semidefinite matrix, while $A_2^T{A}_2=I_n-B$ is also a positive semidefinite matrix. By the spectral theorem, choose an orthogonal matrix $V=[v_1,\dots ,v_n]$ such thatThe bounds follow directly from $z^{T}Bz=\Vert A_{1}z{\Vert }_2^2\in [0,\Vert z{\Vert }_2^2]$. Define the diagonal matricesThe columns of $A_{1}V$ are mutually orthogonal, since $(A_1{v}_i{)}^T(A_1{v}_j)={\lambda }_j{\delta }_{ij}$. For each ${\lambda }_j>0$, set $u_{1j}=A_1{v}_j/\sqrt{{\lambda }_j}$. These columns are orthonormal. At a zero ${\lambda }_j$, one has $A_1{v}_j=0$; choose $u_{1j}$ by completing the already prescribed columns to an orthonormal set of $n$ columns in ${\mathbb{R}}^k$. Such a completion exists because $k\ge n$. This gives $A_{1}V=U_1{\Sigma }_1$, including the zero columns.

Likewise, the Gram matrix of $A_{2}V$ is $\mathrm{diag}(1-{\lambda }_j)$. Normalize its nonzero columns and complete at its zero columns in ${\mathbb{R}}^{m-k}$, using $m-k\ge n$. The resulting $U_2$ has orthonormal columns and $A_{2}V=U_2{\Sigma }_2$. Combining the identities givesThis is a form of the cosine-sine decomposition. Completing the zero columns is essential: neither block was assumed to have full column rank.

Multiplying the Arnoldi decomposition by $Q_k^T$ shows that the leading square block isIt is an upper Hessenberg matrix by the Arnoldi iteration. Because $A=A^T$, it is also a symmetric matrix. Entries below the first subdiagonal vanish by the Hessenberg structure, and their reflected entries above the first superdiagonal vanish by symmetry. ConsequentlyThe extra last row has only its final entry possibly nonzero. With the usual choice of positive normalization factors, the individual column relations are the three-term Lanczos algorithm recurrencewhere the first term is omitted for $j=1$. This reduction is an exact-arithmetic statement; round-off error can affect the orthogonality in an implementation.

Apply successive Householder reflections supported on smaller trailing coordinate sets. At step $k=1,\dots ,n-2$, let $x$ be the entries in rows $k+1,\dots ,n$ of column $k$ of the current symmetric matrix. If $x\ne 0$, part (c) gives an orthogonal reflection $H$ taking it to $c{e}_1$; if $x=0$, use the identity. Embed this as $Q_k=\mathrm{diag}(I_k,H)$ and replace the current matrix by $Q_k^{T}A{Q}_k$.

The first $k$ columns are unchanged by right multiplication. Left multiplication turns the tail of column $k$ into a vector supported only in row $k+1$. Previously treated columns have zero entries throughout this trailing coordinate set, so their zeros are preserved. Symmetry makes the matching row zeros appear as well. After the final step all entries below the first subdiagonal, and by symmetry above the first superdiagonal, vanish.

The product $Q=Q_1\cdots Q_{n-2}$ is orthogonal, andRestricting each transformation to its trailing coordinates is what preserves earlier zero patterns.

For $w\ne 0$, the Householder transformation isLet $P=w{w}^T/(w^{T}w)$. Then $P^T=P$ and $P^2=P$, since $P$ is the orthogonal projection matrix onto the span of $w$. Hence $H^T=H$ andThus $H$ is a symmetric matrix and an orthogonal matrix. Geometrically it reverses the component parallel to $w$ and fixes the perpendicular hyperplane.

Definition and equations. A normal mode is a motion in which every component of a linear system has the same time dependence with fixed amplitude ratios and phases. For a conservative system near equilibrium, the most useful form isThe vector $\mathbf{u}$ describes the mode shape, while $\omega$ is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.

Let $q_i$ be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix $M$ defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix $K$ is the Hessian of the potential energy. Substituting a mode gives the generalized eigenvalue problem for small oscillationsA practical calculation identifies the equilibrium, chooses independent coordinates, computes the quadratic kinetic and potential energies, solves for the eigenvalues and then obtains their eigenvectors. One can equivalently diagonalize the real symmetric matrix $M^{-1/2}K{M}^{-1/2}$. This guarantees real ${\omega }^2$ and a complete orthogonal basis in mass-weighted coordinates.

For distinct eigenvalues, the fact that $K$ and $M$ are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing ${\mathbf{u}}_i^{\mathsf{T}}M{\mathbf{u}}_j={\delta }_{ij}$ decouples the motion into ${\ddot{Q}}_i+{\omega }_i^2{Q}_i=0$, with total energy ${\sum }_i({\dot{Q}}_i^2+{\omega }_i^2{Q}_i^2)/2$. A positive-definite matrix $K$ gives positive eigenvalues and oscillatory stability. Zero modes can describe rigid translations or rotations without restoring force; a negative eigenvalue gives exponential instability rather than an oscillatory mode.

Symmetry. A symmetry preserving both quadratic energies maps modes into modes of the same frequency. Coordinates adapted to symmetry divide the eigenvalue problem into smaller independent blocks. For two identical subsystems, exchanging them gives symmetric and antisymmetric modes. The degeneracy of normal modes means that independent modes share one frequency. Higher-dimensional group representations of a symmetry can enforce such degeneracies; symmetry does not require every pair of modes to be degenerate, and accidental degeneracies can also occur. Within a degenerate eigenspace, orthogonal linear combinations are equally valid mode shapes.

For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and coupled to the other by a spring of stiffness $k_c$,The symmetric mode ${\mathbf{u}}_{+}\propto (1,1)$ leaves the coupling spring unstretched and has ${\omega }_{+}^2=k/m$. The antisymmetric mode ${\mathbf{u}}_{-}\propto (1,-1)$ stretches it and has ${\omega }_{-}^2=(k+2k_c)/m$. Superposing the two gives beats when their frequencies are close.

Continuous systems. The same idea replaces matrices by differential operators and mode vectors by functions satisfying boundary conditions. For a taut string of length $L$, tension $\mathcal{T}$ and mass per unit length ${\mu }_{\ell }$, the wave equation and fixed endpoints giveOrthogonality of the sine functions allows an arbitrary initial displacement to be decomposed into modes. Changing the endpoints changes the spectrum: boundary conditions are part of the physical mode problem.

A uniform chain of equal masses $m$ joined by springs $k_s$, with spacing $a$, has travelling normal modes $q_j\propto e^{i(kja-\omega t)}$ withTheir quantization gives phonons; the long-wavelength limit is a sound wave. Rectangular electromagnetic cavity modes satisfy Maxwell's equations and conductor boundary conditions, with $\omega =c|\mathbf{k}|$ in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion $\omega \propto k^2$. These examples show how restoring forces and geometry determine the frequency spectrum.

Real systems. Small-amplitude motion is often approximately linear even when the full equations are nonlinear. Large amplitudes produce frequency shifts, mode coupling and transfer of energy between modes. Friction broadens resonances and makes free oscillations decay. A damping matrix proportional to $M$ or $K$ can retain the undamped mode shapes; general damping need not, and the appropriate eigenfrequencies and modes can be complex. External periodic forcing predominantly excites modes near resonance, with amplitude set by coupling to the forcing and dissipation. Imperfections split ideal degeneracies, while radiation or coupling to a surrounding fluid can remove energy from the chosen subsystem. Normal modes remain useful when linearization, the boundary conditions and the treatment of damping accurately describe the motion being measured.

Modes, stability and symmetry. A normal mode of the linearized system has a fixed displacement pattern multiplied by one temporal oscillation. For $n$ independent generalized coordinates, a conservative small-disturbance system hasWhen the mass matrix $\mathsf{M}$ is positive definite and the stiffness $\mathsf{K}$ is a symmetric matrix, the generalized eigenvalue problem for small oscillations has $n$ independent mode directions, counting degeneracy. Positive ${\omega }^2$ gives stable oscillation, negative ${\omega }^2$ an exponentially growing and decaying pair, and zero ${\omega }^2$ a neutral direction whose nonlinear stability needs further analysis. An arbitrary linear motion is a superposition of these modes.

If a symmetry preserves both kinetic and potential energies, it maps modes to modes of the same frequency. Mode patterns can be classified by that symmetry, for example reflection-even and reflection-odd patterns. Degenerate eigenspaces admit different bases, so a mode shape need not be uniquely selected inside such a subspace.

Two pendulums free to swing in any direction. Let $x_j$ be the small horizontal displacement along the line joining the supports and $y_j$ the perpendicular horizontal displacement. Put ${\omega }_0^2=g/L$ and $\kappa =k/m$. To quadratic order,The spring is unstretched at equilibrium. Its first-order length change is only $x_2-x_1$; a transverse difference changes its length at second order and produces no quadratic transverse restoring energy. Consequently the four modes areThis is the transverse degeneracy of unstretched spring-coupled pendulums: ${\omega }_0$ has a three-dimensional eigenspace. Independent transverse motions form an equally valid basis for two of those directions. A pre-tensioned spring would change the transverse result.

The open three-pendulum chain. Restrict motion to the vertical plane containing the supports, with horizontal coordinates $x_1,x_2,x_3$. There are springs between neighbours and no wall springs. The equations areThe spring matrix has eigenvalues $0,1,3$, with respective eigenvectors $(1,1,1)$, $(1,0,-1)$ and $(1,-2,1)$. The normal modes of three freely ended coupled pendulums are thereforeThe lowest mode stretches neither spring. In the middle mode the outer masses oppose each other and the central mass is stationary. The highest mode has the central mass moving oppositely with twice the outer amplitude. Reflection about the middle support makes the first and last patterns even and the middle pattern odd.

An even ring. For a large ring, use tangential displacement $x_j$ and periodic indexing $x_{j+N}=x_j$. The locally straight nearest-neighbour equations areCyclic translation symmetry selects Fourier modes $x_j\propto e^{i(qj-\omega t)}$, with $q=2\pi s/N$. HenceFor even $N$, the lowest pattern has equal tangential displacements, while the highest alternates signs around the ring. Thus

These are the even and odd pendulum-ring modes. For exact finite regular-polygon geometry with straight chord springs, the first-order spring extension includes a factor $\cos (\pi /N)$, so its squared contribution is multiplied by ${\cos }^2(\pi /N)$. That correction tends to one in the requested large-$N$ limit.

An odd ring. Perfect alternation cannot close after an odd number of steps. The two highest wave indices are instead $s=(N-1)/2$ and $(N+1)/2$. They give a degenerate pair, represented by real sine and cosine standing patterns, withThe maximum is slightly below the even-ring limiting value, and the sign pattern must contain a gradual phase mismatch rather than exact alternation. The uniform lowest mode is unchanged, and there are still $N$ independent tangential mode directions.

Expand the vector field about the reference trajectory:Over a sufficiently short interval, the tangent-linear propagator isConsequentlyIf $S=J+J^T$ has $S\mathbf{e}=\lambda \mathbf{e}$ and the perturbation is initially parallel to $\mathbf{e}$, its squared Euclidean norm obeysPositive $\lambda$ means instantaneous norm-squared growth, negative $\lambda$ decay, and zero $\lambda$ no first-order change. The corresponding norm growth rate is $\lambda /2$, not $\lambda$. These are eigenvalues of the symmetric part governing the chosen error metric, not generally the eigenvalues of the nonsymmetric Jacobian matrix or the long-time Lyapunov exponents. The direction and the reference Jacobian matrix evolve during a finite forecast.

For the instantaneous Lorenz error growth at a zero-y crossing, direct differentiation of the Lorenz system givesAt a crossing of $y=0$ its symmetric part is particularly simple:Thus a perturbation initially in the $z$ direction has norm-squared rate $-2b$ and initially decays in norm at rate $b$. The $x$-dependent couplings between $y$ and $z$ are skew-symmetric at the crossing and do not contribute to the instantaneous total norm derivative. They can rotate the perturbation later: a pure $\delta z$ immediately induces $\dot{\delta y}=-x\delta z$. It is therefore incorrect to claim that such an error remains an independently decaying $z$ mode along the complete trajectory.

For the $(x,y)$ block, the trace is $-2(\sigma +1)<0$ and the determinant is $4\sigma -(\sigma +r-z{)}^2$. A negative determinant gives one positive and one negative eigenvalue. Equivalently,In particular the requested sufficient lower-height condition isThere is also a second growth range $z>\sigma +r+2\sqrt{\sigma }$; it must not be lost by removing the absolute value. At either endpoint the largest rate is zero. Between them both block eigenvalues are negative. At a crossing in either growing range, some planar error directions have positive instantaneous exponential rate. This statement is local in time: a single instantaneous positive eigenvalue of $S$ does not prove sustained exponential error growth over an evolving Lorenz trajectory.

At the specified state, the symmetric matrix isThe planar characteristic equation is ${\lambda }^2+22\lambda -104=0$, giving $4$ and $-26$, while the third eigenvalue is $-6$. An orthonormal set of eigenvectors isWith comparable prior uncertainty in these dimensionless coordinates and equally accurate scalar observations, prioritize the growing combinationAn observation of this projection removes error in the direction that grows most strongly over a sufficiently short forecast; its norm initially grows at rate $2$ in Lorenz time units. The strongly contracting direction is a lower priority under those assumptions. This is not an observation of the absolute state amplitude alone: the relevant error projection is $(\delta x+2\delta y)/\sqrt{5}$. Anisotropic prior covariance, observation noise or a different forecast norm would modify the optimal choice; for a finite forecast the leading right singular vector of the actual propagator replaces this instantaneous direction.

If one retains the Euler approximation $A=I+J\delta t$, its exact norm matrix isThe additional term can mix the instantaneous symmetric-part eigenvectors, so they no longer generally diagonalize finite-time error amplification. However, when $\delta t^2$ is material, a reliable physical propagator must itself be calculated beyond first-order Euler: a large Euler-step amplification can be a numerical artifact. For constant $J$, the exact map is $\Phi =e^{J\Delta t}$; for a changing reference trajectory it solves $\dot{\Phi }=J(t)\Phi$, $\Phi (0)=I$.

Use its singular value decomposition, $\Phi =U\Sigma V^T$. The initial orthonormal right singular vectors ${\mathbf{v}}_i$ diagonalize ${\Phi }^T\Phi$ and satisfyA singular value $s_i>1$ gives finite-time norm growth, $s_i<1$ decay, and the leading right singular vector is the initial perturbation with greatest amplification in the stated norm. The left singular vector describes the resulting final error pattern. Non-normal dynamics can give transient growth even if every Jacobian matrix eigenvalue has negative real part, so selecting observations from those eigenvalues alone is insufficient. Over a genuinely short interval, $s_i^2=1+{\lambda }_i\delta t+O(\delta t^2)$, recovering the symmetric-part result. Initial observations should target the relevant growing right singular projections, with prior covariance and observational precision included when they are not isotropic.

Taking the transpose gives $M^T=(X^T+X)/\sqrt{2}=M$, so $M$ is a symmetric matrix. For $i<j$, $M_{ij}=(X_{ij}+X_{ji})/\sqrt{2}$ has the standard normal distribution, whereas $M_{ii}=\sqrt{2}{X}_{ii}$ has the normal distribution $N(0,2)$. Distinct upper-triangular entries depend on disjoint sets of the original independent random variables, so they remain independent.

Consequently the joint probability density function, relative to ${\prod }_{i\le j}d{m}_{ij}$, isHere $\mathrm{tr}(m^2)={\sum }_i{m}_{ii}^2+2{\sum }_{i<j}{m}_{ij}^2$ because $m$ is a symmetric matrix. This is the Gaussian orthogonal ensemble, with off-diagonal variance one and diagonal variance two; the rescaled matrix $M/\sqrt{n}$ has the corresponding $1/n$ off-diagonal variance.

We will also use its entry covariance. Expanding in the independent entries of $X$, for which $\mathbb{E}[X_{ij}{X}_{kl}]={\delta }_{ik}{\delta }_{jl}$, givesThe symbols ${\delta }_{ab}$ are Kronecker deltas. This formula includes the diagonal case, where the two terms both contribute.

A projective conic in ${\mathbb{P}}^2(\mathbb{C})$ is the zero locus of a nonzero homogeneous quadratic polynomial, up to nonzero scaling of that polynomial. Under an invertible linear change of homogeneous coordinates, its symmetric matrix has rank $3$, $2$, or $1$. Over $\mathbb{C}$, the corresponding forms are a nonsingular conic such as $XZ-Y^2=0$, two distinct lines such as $XY=0$, or a double line $X^2=0$. In the last case the repeated factor records multiplicity, even though the underlying point set is one line.

A projective line is the projectivization of a two-dimensional linear subspace of ${\mathbb{C}}^3$, equivalently the zero locus of one nonzero linear form. For two such subspaces $U,W$, the dimension formula for a sum of subspaces gives $\dim (U\cap W)\ge 2+2-3=1$. A nonzero vector in the intersection therefore supplies a common projective point. Distinct lines have exactly one such point.