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A fixed circular elastic membrane of radius $a$ has radial normal modes satisfying the displayed Sturm-Liouville problem, with weight $r$, $R(a)=0$ and regularity at the origin. The regular solutions are Bessel functions of the first kind $J_0(\sqrt{\lambda }r)$. Integration by parts gives the positive Rayleigh quotient $\lambda ={\int }_0^{a}r|R^{\prime }{|}^{2}dr/{\int }_0^{a}r|R{|}^{2}dr$. Distinct modes are orthogonal in the weighted inner product ${\int }_0^{a}r\overline{R}Sdr$.

For noninteger real $\gamma$, expand a smooth forcing as $F={\sum }_{n\ge 0}{F}_n{T}_n$ in Chebyshev polynomials. Because $(1-x^2)T_n^{\prime \prime }-x{T}_n^{\prime }=-n^2{T}_n$, division of each coefficient by ${\gamma }^2-n^2$ gives the regular solution. The coefficients use the weighted inner product with weight $(1-x^2{)}^{-1/2}$. Choosing the same endpoint domain as the regular eigenfunctions of the Sturm-Liouville problem is essential: finite values alone also allow homogeneous solutions $\cos (\gamma \arccos x)$ and $\sin (\gamma \arccos x)$, which generally have divergent endpoint derivatives. Setting $x=\cos \theta$ converts the regular endpoint condition into the Neumann boundary conditions $Z_{\theta }(0)=Z_{\theta }(\pi )=0$. The homogeneous equation then has no nonzero solution when $\gamma$ is noninteger, proving uniqueness in this domain.

For a uniform string with tension $T>0$, linear mass density $\rho >0$, left end fixed, and a freely transverse-moving point mass $M\ge 0$ at $x=L$, the wave equation and endpoint law areThe last equation is Newton's second law: the string pulls the mass back toward the interior, giving the minus sign. With wave speed $c=\sqrt{T/\rho }$, a nonzero normal mode has spatial part $X(x)=\sin (\omega x/c)$ and satisfiesFor $M>0$, its dimensionless roots $q=\omega L/c$ occur once in each $(n\pi ,n\pi +\pi /2)$, $n\ge 0$, because $\cot q$ decreases from infinity to zero while $Mq/(\rho L)$ increases. For $M=0$, $q=(n+1/2)\pi$: the right end has a homogeneous Neumann boundary condition. There is no nonzero zero-frequency mode, since a linear spatial profile vanishing at the left end and with zero right slope must vanish everywhere.

Multiplying the spatial equation by $X$ and using integration by parts givesFor two different modes, the same calculation gives the weighted inner product orthogonality $\rho {\int }_0^L{X}_m{X}_{n}dx+M{X}_m(L)X_n(L)=0$. The endpoint mass therefore contributes to both inertia and mode orthogonality.

Let $A$ and $M$ be real symmetric positive-definite matrices. The principal square root of a positive semidefinite matrix construction gives an invertible matrix for each. SetThen $Ax=b$ is equivalent to $By=c$. This is a matrix congruence, and $B$ is a symmetric positive-definite matrix: $B^T=B$, and for every nonzero vector $v$,since $S^{-1}v\ne 0$. This short proof establishes both requirements for applying the conjugate gradient method to the transformed linear system.

The left transformed matrix has the same eigenvalues becauseIts correct geometry uses the weighted inner product $⟨u,v{⟩}_M=u^{T}Mv$: $⟨u,M^{-1}Av{⟩}_M=u^{T}Av=⟨M^{-1}Au,v{⟩}_M$. Thus it is symmetric in this inner product, even when it is not a symmetric matrix in ordinary coordinates. The spectral condition number governing the transformed symmetric problem is ${\kappa }_2(B)$, not generally ${\kappa }_2(M^{-1}A)$.

The Preconditioned conjugate gradient method realizes this transformation using auxiliary solves $M{z}_k=r_k$, without constructing $S$ or its matrix inverse. With fixed $M$, initialize $r_0=b-A{x}_0$ and $p_0=z_0$; while $r_k\ne 0$, usethen solve $M{z}_{k+1}=r_{k+1}$ and setStop at a zero linear-system residual before forming another ratio. Positive definiteness makes the denominators positive before termination.^[ref]

An invertible matrix alone is insufficient: with $A=I$ and $M=\mathrm{diag}(-1,1)$, the left transformed matrix has eigenvalues $-1,1$, and $M$ has no real symmetric positive-definite matrix square root. Even a valid symmetric positive-definite matrix can worsen conditioning if its scaling is unsuitable, as the next worked example shows.

Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.

A symmetric positive-definite matrix $M$ defines an inner product on real coordinate vectors byFor diagonal $M$, positive diagonal entries weight the squared coordinate errors. This is the finite-coordinate version of a weighted inner product; the positive-definite matrix example appears in Francis Narcowich, Inner Products and Norms, standard inner products, example 8.

For $M=\mathrm{diag}(4,1)$, the squared norm of $(a,b{)}^T$ is $4a^2+b^2$. A first-coordinate error of one costs as much as a second-coordinate error of two. The unit sphere is the ellipse $4a^2+b^2=1$, and orthogonality means $4u_1{v}_1+u_2{v}_2=0$. Thus “closest” and “perpendicular” depend on the chosen inner product. The orthogonal projection residual example shows how this changes a best approximation.

In the weighted inner product $⟨f,g⟩={\int }_0^{1}xf(x)g(x)dx$, the best approximation to $x^2$ in the span of $1,x$ is the displayed polynomial. The normal equations are $1/4-A/2-B/3=0$ and $1/5-A/3-B/4=0$. Their solution gives $A=-3/10$, $B=6/5$. The residual is orthogonal to that span, so the Pythagorean theorem proves minimality and uniqueness. Its squared norm is $1/600$. A weight vanishing at a single endpoint still gives a positive-definite inner product on continuous functions: a nonzero continuous function remains nonzero on an interval where the weight is positive.

On $(0,L)$, the differential operator $\mathcal{L}={\partial }_x^2+\beta {\partial }_x=e^{-\beta x}{\partial }_x(e^{\beta x}{\partial }_x)$ with homogeneous Neumann boundary conditions is self-adjoint in the weighted inner product with weight $e^{\beta x}$. Its eigenfunctions and nonnegative decay rates areTheir squared norms are $N_0=(e^{\beta L}-1)/\beta$ and $N_k=L[1+{\beta }^2/(4q_k^2)]/2$. At $\beta =0$, interpret the continuous limit as $N_0=L$. The Sturm-Liouville eigenfunction expansion givesThis kernel acts against the measure $e^{\beta y}dy$, not Lebesgue measure alone. It is symmetric in $x,y$; the transition density against Lebesgue measure is $e^{\beta y}{K}_{\beta }(x,y,t)$.

A positive measurable weight $w$ defines the displayed norm on the L2 space for the measure $w(x)dx$, with functions identified when equal almost everywhere. This is the norm induced by the weighted inner product. Changing $w$ changes the contribution of each region to approximation error. If $p_0,\dots ,p_N$ are nonzero mutually orthogonal real functions, completing squares shows that the unique best approximation to real $f$ in their linear span has coefficients $a_j=⟨f,p_j{⟩}_w/\Vert p_j{\Vert }_w^2$.