Analytical insight is obtained by considering the SVD $𝑺=𝑼𝚺{𝑽}^T$, and $𝑨={𝑺}^T𝑺=𝑽{𝚺}^2{𝑽}^T$ from which it results that singular values $\sigma$ of $𝑺$ are square roots of the eigenvalues of $𝑨$. The eigenvalues of a symmetric matrix are real-valued.

Recall (see L14) that eigenvalues of

can be obtained by the correspondence between continuum and discrete operators. The matrix $𝑫$ is a discrete version of the second-order differentiation operator ${\partial }_t^2$ obtained by centered finite differences

In the continuum case, the derivative at a point requires knowledge of values within a neighborhood, e.g., $u^{\text{'}}\left(t\right)={lim}_{h\to 0}(u(t+h)-u(t-h)/\left(2h\right))$. The corresponding feature of the discrete case is that the differentiation operator is not square. Consider the vector $\widehat{𝒖}\in {ℝ}^{m+2}$ obtained by sampling $u\left(t\right)$ at nodes $t_k=kh$, $h=\pi /(m+1)$, for $k=0,\dots ,m+1$, and the extended matrix

The finite difference approximation of the second derivative at $m$ interior points $u^{\text{'}\text{'}}\left(t_k\right),k=1,\dots ,m$ requires knowledge of $m+2$ function values, expressed in matrix form as

The eigensystem of the square matrix $𝑫\in {ℝ}^{m\times m}$ can however be obtained by noting that:

1. Eigenfunctions $\phi \left(t\right)$of ${\partial }_t^2$ are solutions of the ordinary differential equation ${\partial }_t^{2}u=\lambda u$, for $u:[0,\pi ]\to ℝ$.

2. Choosing boundary conditions on $u$ to recover the desired matrix $𝑫$.

3. Positing that eigenvectors of $𝑫$ are obtaining by evaluating $\phi \left(t\right)$ at $t_k$, $𝒛=\left[\begin{array}{lll}\phi \left(t_1\right)& \dots & \phi \left(t_m\right)\end{array}\right]$.

4. Verifying that $𝑫𝒛=\mu 𝒛$ is indeed verified with no dependence of $\mu$ on the component index $k$.

For the matrix $𝑫$, the appropriate boundary conditions are $u_0=u\left(t_0\right)=u\left(0\right)=0$ and $u_{m+1}=u\left(t_{m+1}\right)=u\left(1\right)=0_{}$, such that

At the interval endpoints

Since the boundary value problem

has eigenfunctions $\sin (\xi t)$, $\xi \in \mathbb{Z}$, posit that the eigenvectors $𝒛$ of the $𝑫$ matrix have components $z_k=\sin (\xi t_k)=\sin (\xi kh)$. Verify that $𝒛$ is indeed an eigenvector by computing the $k^{\mathrm{th}}$ component of $𝑫𝒛$

In the above calculation

does not depend on the component index $k$, hence indeed it is an eigenvalue since $𝑫𝒛=\lambda {𝒛}_{}$ is satisfied. Verify numerically:

∴ 

function D(m)
  spdiagm(0 => -2*ones(m), -1 => ones(m-1) , 1 => ones(m-1))
end;

∴ 

m=10; num=eigen(collect(D(m))).values;

∴ 

h=pi/(m+1); csi=m:-1:1; an=-(2*sin.(csi*h/2)).^2;

∴ 

[num an]

∴ 

From the above, the eigenvalues of $𝑩=h^2𝑫+4𝑰$ can readily be found as

∴ 

The fact that the matrix $𝑨={𝑺}^T𝑺$ differs from $𝑩$ just through a rank-one update at the right endpoint,

suggests one of two approaches:

1. Define different boundary conditions for the ordinary differential equation ${\partial }_t^{2}u=\lambda u$.

2. Find the effect of a rank-one perturbation upon the eigenvalues of $𝑩$.

Consider the first approach, and require

The left end-condition remains the same $u_0-2u_1+u_2=-2u_1+u_2\Rightarrow u_0=0$. At the right obtain

In the continuum case the above is satisfied when

Carrying out a Taylor series expansion and truncating at first order gives the Robin boundary condition

This leads to consideration of the Sturm-Liouville problem

Let $\lambda ={\xi }^2$ and obtain that a nontrivial solution $u\left(t\right)=\sin (\xi t)$ is obtained for $\xi$ that satisfies the right end-condition

A numerical root-finding procedure determines solutions $\{{\xi }_1,\dots ,{\xi }_m\}$ of the above equation.

∴ 

using Roots

∴ 

f(x,m)=x+2*(m+1)*tan(x*pi)/pi; f10(x)=f(x,10);

∴ 

csi=find_zero.(f10,1:10);

∴ 

function C(m)
  d = -2*ones(m); d[m]=-3; h=pi/(m+1); s=1/h^2
  spdiagm(0 => s*d, -1 => s*ones(m-1) , 1 => s*ones(m-1))
end;

∴ 

num=eigen(collect(C(10))).values

∴ 

collect(C(10))

∴ 

Verify whether $𝒖=\left[\sin (\xi t_j)\right]$ is indeed an eigenvector of $𝑪$.

∴ 

∴ 

m=10; num=eigen(collect(D(m))).values;

∴ 

h=pi/(m+1); csi=m:-1:1; an=-(2*sin.(csi*h/2)).^2;

∴ 

[num an]

∴ 

∴ 

C10=collect(C(10))

∴ 

For $m=10$,

are the roots. Since $𝑨=h^2𝑪+4𝑰$ the smallest and largest eigenvalues of $𝑨$ are

and the smallest, largest singular values of $𝑺$ are

For $m=10,20,\dots ,100$ the root-finding procedure gives

The extremal singular values are

However the matrix $𝑨$ is a rank-one perturbation of $𝑩=𝑸\Psi {𝑸}^T$, $𝑨=𝑸\Psi {𝑸}^T-{𝒆}_m{𝒆}_m^T$, $\Psi =\mathrm{diag}({\mu }_1,\dots ,{\mu }_m)$. Numerical experimentation shows that eigenvalues of $𝑩$

From here there are two ways to proceed:

1) Eigenvalues of $𝑨$ are roots of the characteristic polynomial

Use the factorization

and $\det \left(𝑸\right)=1$, $\det (𝑭𝑮)=\det \left(𝑭\right)\det \left(𝑮\right)$, to obtain

From the SVD $𝑺=𝑼𝚺{𝑽}^T$, $𝑨={𝑺}^T𝑺=𝑽{𝚺}^2{𝑽}^T$ deduce that singular values $\sigma$ of $𝑺$ are square roots of the eigenvalues of $𝑨$

The largest, smallest singular values ${\sigma }_1$, ${\sigma }_m$ are obtained for $\xi =1$, $\xi =m$, respectively. This leads to

Verify the above numerical experiment.

∴ 

kappa(m)=cos(pi/2/(m+1))/cos(m*pi/2/(m+1));

∴ 

r=250:250:1000; num=cond.(Matrix.(S.(r))) ./ r; an=kappa.(r) ./ r; [num an]