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The hypothesis implies $v_1,v_2\ne 0$, since the zero vector is a scalar multiple of every vector. SetSince $v_1$ is not a scalar multiple of $v_2$, $w\ne 0$. Positive definiteness and bilinearity of the real inner product giveMultiplying by $\Vert v_2{\Vert }^2>0$ and taking square roots proves the strict Cauchy-Schwarz inequalitySince every real number $s$ satisfies $s\le |s|$, this proves, in particular,The argument derives the required inequality directly from the inner product axioms.

Symmetry of the real inner product gives $A^T=A$. Moreover, i with $y=x$ shows thatso the Gram matrix is a positive semidefinite matrix. To also check reality of every eigenvalue, allow a nonzero vector $z=a+ib\in {\mathbb{C}}^n$ with $Az=\lambda z$. Since $A$ is a real symmetric matrix,Here $z^{\ast }$ is the conjugate transpose, and $a^{T}Ab=b^{T}Aa$ cancels the imaginary part. Thusbecause $z^{\ast }z>0$. This proves both that every eigenvalue is real and that it is nonnegative.

Bilinearity of the real inner product gives, for the specified linear combinations,The last equality is the entrywise rule for matrix multiplication, with $x^T$ the transpose of the column vector $x$.

Apply the divergence theorem to $\mathbf{F}=u\nabla v-v\nabla u$. The product rule givesbecause the two gradient inner products cancel. Also $\mathbf{F}\cdot \mathbf{n}=u{\partial }_{n}v-v{\partial }_{n}u$. ThusThis is Green's second identity, under the usual regularity assumptions permitting these derivatives and boundary integrals.

The real spectral theorem states that a symmetric matrix has an orthonormal basis of real eigenvectors, and hence is a diagonalisable matrix. Here is a proof in this setting.

For $n\ge 1$, the quadratic form $q(x)=x^{T}Ax$ attains a maximum on the unit sphere, by compactness and the extreme value theorem. Choose a unit maximizer $u$ and put $\lambda =u^{T}Au$. For any $w\perp u$, the Rayleigh quotient along $u+tw$ has a maximum at $t=0$, soThe last equality uses $A=A^T$ and $u^{T}w=0$. Thus $Au$ is orthogonal to every $w\perp u$, so $Au=\lambda u$.

The orthogonal complement $u^{\perp }$ is an invariant subspace, because for $x\perp u$,The restriction of $A$ to $u^{\perp }$ is again symmetric with respect to the induced inner product. By mathematical induction on dimension, it has an orthonormal basis of eigenvectors; adjoining $u$ gives one for ${\mathbb{R}}^n$. If $Q$ has those basis vectors as columns, then $Q^{T}AQ$ is a real diagonal matrix and $Q^{-1}=Q^T$.

Adding and subtracting the two coordinates shows that $A(1,1{)}^T=(c+s)(1,1{)}^T$ and $A(1,-1{)}^T=(c-s)(1,-1{)}^T$. The normalized eigenvectors can therefore be chosen asTheir inner product is $(1-1)/2=0$, and their squared norms are one, proving orthonormality. If $s=0$, every nonzero vector is an eigenvector; this chosen orthonormal basis still works.

For $\mathbf{u}=(1,1{)}^T$ and $\mathbf{v}=(1,-1{)}^T$, the outer products areTheir inner product is zero and both squared norms are two. Thus $P_u=\mathbf{u}{\mathbf{u}}^T/2$ and $P_v=\mathbf{v}{\mathbf{v}}^T/2$ are orthogonal spectral projectors, with $P_u+P_v=I$. Since $A\mathbf{u}=5\mathbf{u}$ and $A\mathbf{v}=\mathbf{v}$, the eigenvalues are $\lambda =5$, $\mu =1$, andMultiplying these decompositions uses $P_u{P}_v=0$ and $P_u^2=P_u$, $P_v^2=P_v$, giving the identity matrix.

Expanding the determinant gives $\det (\lambda I-M)=(\lambda -1)[(\lambda -1{)}^2-25]$. Solving each eigenvalue equation yieldsTheir pairwise inner products are $16-25+9=0$, $12-12=0$ and $12-12=0$. Their squared Euclidean norms are $50,50,25$, respectively, so division by $\sqrt{50},\sqrt{50},5$ gives an orthonormal basis. This also follows from the spectral theorem for real symmetric matrices: symmetry makes eigenvectors for distinct eigenvalues orthogonal.

By the vector triple product identity,The last inner product vanishes by perpendicularity. Taking the inner product with $d$ givesThis is the identity for cross products with a common vector, and holds without nondegeneracy assumptions.

The standard Fourier series iswith Fourier coefficientsThese follow by taking inner products with the individual basis vectors and using their orthogonality and squared norms.

With inner product $⟨u,v⟩={\int }_{-\pi }^{\pi }u(x)v(x)dx$, orthogonality meansThe constant function has squared norm $2\pi$, and each sine or cosine has squared norm $\pi$. Thus the displayed family is orthogonal but not normalized; these different lengths determine the coefficient factors.

With the standard unweighted inner product on real square-integrable functions, orthogonality on $[-L,L]$ meansMultiplying this inner product by a common nonzero normalising constant does not change orthogonality.

Use the real inner productOrthogonality means this integral is zero for any two distinct listed functions. The individual squared norms are positive: the constant function has squared norm $2L$, while every listed sine or cosine has squared norm $L$. Thus they are orthogonal but not normalized to unit length in this inner product.

Take the inner product of the Fourier series with $\sin (n\pi x/L)$. Fourier orthogonality eliminates the constant, all cosine terms, and all sine terms except the $n$th. The remaining squared norm issoFor a square-integrable function, the calculation is justified by taking inner products of finite Fourier partial sums and passing to their mean-square limit.

General definitions. The Euler-Lagrange equation gives $m_i{\ddot{q}}_i+{\sum }_j{V}_{ij}{q}_j=0$. Write $M=\mathrm{diag}(m_i)$ and $K=(V_{ij})$. A normal mode has fixed displacement pattern $v$ and common time factor, $q(t)=vQ(t)$; its normal frequency satisfiesHere $\omega$ is angular frequency. A normal coordinate is the coefficient of one pattern in a complete modal expansion, so these coordinates decouple the equations. Since $M$ is positive definite, the change $z=M^{1/2}q$ turns the mode equation into the ordinary symmetric eigenvalue problem for $M^{-1/2}K{M}^{-1/2}$.

Three-mass equations. Let $x_1,x_2,x_3$ be horizontal displacements from the unstretched equilibrium, ordered left to right. The kinetic energy and potential energy areDifferentiating the potential energy givesThus $M=m\mathrm{diag}(1,2,3)$ and

Frequencies and patterns. Put $\lambda =m{\omega }^2/k$. Expansion of the generalized eigenvalue problem for small oscillations givesSolving at each root givesThe zero normal mode is uniform translation. In the lower nonzero mode, the central mass stays fixed, the two end masses move oppositely, and their unequal displacements balance both the end-mass momenta and the forces on the central mass. In the upper mode the end masses move together and the centre oppositely. Both vibrating modes leave the centre of mass fixed.

Orthogonality. Because $K$ is symmetric, two mode equations implyDistinct eigenvalues therefore give the mass-matrix orthogonality of normal modes:The Euclidean dot product need not vanish for unequal masses; for example $v_0^T{v}_1=-2$, while $v_0^{T}M{v}_1=m(-3+3)=0$. The modal squared norms are $6m,12m,12m$.

Normal coordinates. Expand $x=Q_0{v}_0+Q_1{v}_1+Q_3{v}_3$. Project using the mass-weighted inner product and divide by each modal norm to obtainFor example, the inverse equations are $x_1=Q_0-3Q_1+Q_3$, $x_2=Q_0-2Q_3$, $x_3=Q_0+Q_1+Q_3$. They reduce the energies toand henceThe first coordinate is the centre of mass and has free linear motion. Multiplying $Q_0,Q_1,Q_3$ by $\sqrt{6m},\sqrt{12m},\sqrt{12m}$ respectively gives mass-normalized normal coordinates if desired.

A Hermitian matrix obeys $H^{†}=H$, where the dagger is conjugate transpose. If $Hv=\lambda v$ with $v\ne 0$, then $v^{†}Hv=\lambda v^{†}v$. This scalar is real because its complex conjugate equals $v^{†}{H}^{†}v=v^{†}Hv$. Since $v^{†}v>0$, the eigenvalue $\lambda$ is real.

For the Gram matrix, conjugate symmetry of the inner product givesThus $G$ is Hermitian. The assumption that the vectors form a basis additionally makes $G$ positive definite, though only Hermiticity is needed here.

A real orthogonal matrix satisfies $Q^{T}Q=I$; a complex unitary matrix satisfies $U^{†}U=I$. Their columns are orthonormal in the corresponding real or complex inner product.

Adopt the convention that a complex inner product is linear in its second argument, conjugate symmetric, and positive definite: $⟨u,u⟩\ge 0$, with equality exactly when $u=0$. Conjugate symmetry and second-argument linearity then imply antilinearity in the first:For the specified real symmetric matrix use $b=a^2$. Its upper block is the outer product $(1,a{)}^T(1,a)$, so it has one zero eigenvalue and one eigenvalue $1+a^2$. A complete orthonormal eigenbasis isIf $c=1+b$, the last two eigenvalues coincide and their eigenspace is the span of $v_{+},v_c$; every nonzero vector in that span is an eigenvector. Since $c>0$, the zero eigenspace remains one-dimensional.

By the spectral theorem, the Hermitian matrix $A$ has a complete orthonormal eigenbasis. If $Av=\lambda v$, idempotence gives ${\lambda }^2=\lambda$, so every eigenvalue is zero or one. Since $\det A\ne 1$, at least one is zero. ThereforeThe trace of an idempotent equals its rank; here Hermiticity further identifies $A$ as an orthogonal projection matrix.

A complex inner product $⟨u,v⟩$ is conjugate-linear in its first argument, linear in its second, conjugate symmetric and positive definite: $⟨v,u⟩=\overline{⟨u,v⟩}$ and $⟨u,u⟩>0$ for nonzero $u$. An orthonormal basis is a spanning linearly independent family $e_i$ with $⟨e_i,e_j⟩={\delta }_{ij}$. The associated norm is $\Vert u\Vert =\sqrt{⟨u,u⟩}$.

For two distinct basis elements $x,y$, their orthogonality and unit norms give $\Vert x-y{\Vert }^2=⟨x,x⟩+⟨y,y⟩-⟨x,y⟩-⟨y,x⟩=2$. Hence their distance is $\sqrt{2}$; if the same element is selected twice, it is zero.

Finally, for $B^{†}=B$, the scalar quadratic form obeys $(x^{†}Bx{)}^{\ast }=x^{†}{B}^{†}x=x^{†}Bx$, proving $x^{†}Bx$ is real.

A Hermitian matrix satisfies $A^{†}={\overline{A}}^T=A$. If $Av=\lambda v$ with $v\ne 0$, then $v^{†}Av=\lambda v^{†}v$ is real because its conjugate equals $v^{†}{A}^{†}v=v^{†}Av$. Since $v^{†}v>0$, its eigenvalue $\lambda$ is real. If $Aw=\mu w$ with $\mu \ne \lambda$, thenso $v^{†}w=0$: distinct eigenspaces are orthogonal for the standard complex inner product.

The converse is false. For example $\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$ has the two real eigenvalues one and two but is not a Hermitian matrix. It can be embedded as a block in a $3\times 3$ example if desired.

For Hermitian $A,B$, $(AB{)}^{†}=B^{†}{A}^{†}=BA$. Thus $AB$ is Hermitian exactly when $AB=BA$, equivalently when their commutator vanishes. Also $(AB+BA{)}^{†}=AB+BA$, so a value that works for every such pair isThis sum is their anticommutator.

Use the inner product $⟨u,v⟩=u^{†}v$ on ${\mathbb{C}}^n$. The distinct real eigenvalues permit a full orthonormal eigenvector expansion; the two child solutions establish that expansion and apply it to the proposed quadratic form.