User:BRousselet/abouteigenvalues

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x=3 x=5

x=3 x=5

x=3 x=5

About eigenvalue perturbation

$\mathbf {K} _{0}\delta \mathbf {x} _{i}+\delta \mathbf {K} \mathbf {x} _{0i}=\lambda _{0i}\mathbf {M} _{0}\delta \mathbf {x} _{i}+\lambda _{0i}\delta \mathbf {M} \mathrm {x} _{0i}+\delta \lambda _{i}\mathbf {M} _{0}\mathbf {x} _{0i}.\qquad (3)$

we left multiply with $\mathbf {x} _{0i}^{T}$ and use (2) as well as its first order variation $\delta \mathbf {x} _{j}\mathbf {M} \mathbf {x} _{0i}+\mathbf {x} _{0j}\mathbf {M} \delta \mathbf {x} _{i}+\mathbf {x} _{0j}\delta \mathbf {M} \mathbf {x} _{0i}=0$ get

$\mathbf {x} _{0i}^{T}\delta \mathbf {K} \mathbf {x} _{0i}=\lambda _{0i}\mathbf {x} _{0i}^{T}\delta \mathbf {M} \mathrm {x} _{0i}+\delta \lambda _{i}$

or

$\delta \lambda _{i}=\mathbf {x} _{0i}^{T}\delta \mathbf {K} \mathbf {x} _{0i}-\lambda _{0i}\mathbf {x} _{0i}^{T}\delta \mathbf {M} \mathrm {x} _{0i}$

We notice that it is the first order perturbation of the generalized Rayleigh quotient :$R(K,M;x_{0i})=x_{0i}^{T}Kx_{0i}/x_{0i}^{T}Mx_{0i},{\text{ with }}x_{0i}^{T}Mx_{0i}=1$

Eigenvector perturbation

We left multiply (3) with $x_{0j}^{T}$ for $j\neq i$ and get

$\mathbf {x} _{0j}^{T}\mathbf {K} _{0}\delta \mathbf {x} _{i}+\mathbf {x} _{0j}^{T}\delta \mathbf {K} \mathbf {x} _{0i}=\lambda _{0i}\mathbf {x} _{0j}^{T}\mathbf {M} _{0}\delta \mathbf {x} _{i}+\lambda _{0i}\mathbf {x} _{0j}^{T}\delta \mathbf {M} \mathrm {x} _{0i}+\delta \lambda _{i}\mathbf {x} _{0j}^{T}\mathbf {M} _{0}\mathbf {x} _{0i}.$

We use $\mathbf {x} _{0j}^{T}K=\lambda _{0j}\mathbf {x} _{0j}^{T}M{\text{ and }}\mathbf {x} _{0j}^{T}\mathbf {M} _{0}\mathbf {x} _{0i}=0.$ for $j\neq i$.

$\lambda _{0j}\mathbf {x} _{0j}^{T}\mathbf {M} _{0}\delta \mathbf {x} _{i}+\mathbf {x} _{0j}^{T}\delta \mathbf {K} \mathbf {x} _{0i}=\lambda _{0i}\mathbf {x} _{0j}^{T}\mathbf {M} _{0}\delta \mathbf {x} _{i}+\lambda _{0i}\mathbf {x} _{0j}^{T}\delta \mathbf {M} \mathrm {x} _{0i}.$

or

$(\lambda _{0j}-\lambda _{0i})\mathbf {x} _{0j}^{T}\mathbf {M} _{0}\delta \mathbf {x} _{i}+\mathbf {x} _{0j}^{T}\delta \mathbf {K} \mathbf {x} _{0i}=\lambda _{0i}\mathbf {x} _{0j}^{T}\delta \mathbf {M} \mathrm {x} _{0i}.$

As the eigenvalues are simple, for $j\neq i$

$\epsilon _{ij}=\mathbf {x} _{0j}^{T}\mathbf {M} _{0}\delta \mathbf {x} _{i}={\frac {-\mathbf {x} _{0j}^{T}\delta \mathbf {K} \mathbf {x} _{0i}+\lambda _{0i}\mathbf {x} _{0j}^{T}\delta \mathbf {M} \mathrm {x} _{0i}}{(\lambda _{0j}-\lambda _{0i})}},i=1,\dots n;j=1,\dots n;j\neq i.$

Moreover the firs order variation of ... yields $2\epsilon _{ii}=2\mathbf {x} _{0i}^{T}\mathbf {M} \delta x_{i}=-\mathbf {x} _{0i}^{T}\delta M\mathbf {x} _{0i}.$ We have obtained all the components of $\delta x$ .

Perturbation of an implicit function.

In the next paragraph, we shall use the Implicit function theorem (Statement of the theorem ); we notice that for a continuously differentiable function $f:\mathbb {R} ^{n+m}\to \mathbb {R} ^{m},\;f:(x,y)\mapsto f(x,y)$, with an invertible Jacobian $J_{f,b}(x_{0},y_{0})$, from a point $(x_{0},y_{0})$ solution of $f(x_{0},y_{0})=0$, we get solutions of $f(x,y)=0$ with $x$ close to $x_{0}$ in the form $y=g(x)$ where $g$ is a continuously differentiable function ; moreover the Jacobian of $g$ is provided by the linear system

$J_{f,y}(x,g(x))J_{g,x}(x)+J_{f,x}(x,g(x))=0\quad (1)$.

As soon as the hypothesis of the theorem is satisfied, the Jacobian of $g$ may be computed with a first order expansion of $f(x_{0}+\delta x,y_{0}+\delta y)=0$, we get

$J_{f,x}(x,g(x))\delta x+J_{f,y}(x,g(x))\delta y=0$; as $\delta y=J_{g,x}(x)\delta x$, it is equivalent to equation $(1)$.

Eigenvalue perturbation: theoretical basis

we use the previous paragraph (Perturbation of an implicit function) with somewhat different notations suited to eigenvalue perturbation; we introduce ${\tilde {f}}:\mathbb {R} ^{2n^{2}}\times \mathbb {R} ^{n+1}\to \mathbb {R} ^{n+1}$, with

• ${\tilde {f}}(K,M,\lambda ,x)={\binom {f(K,M,\lambda ,x)}{f_{n+1}(x)}}$ with

$f(K,M,\lambda ,x)=Kx-\lambda x,f_{n+1}(M,x)=x^{T}Mx-1$. In order to use the Implicit function theorem, we study the invertibility of the Jacobian $J_{{\tilde {f}};\lambda ,x}(K,M;\lambda _{0i},x_{0i})$ with

$J_{{\tilde {f}};\lambda ,x}(K,M;\lambda _{i},x_{i})(\delta \lambda ,\delta x)={\binom {-Mx_{i}}{0}}\delta \lambda +{\binom {K-\lambda M}{2x_{i}^{T}M}}\delta x_{i}$. Indeed, the solution of

$J_{{\tilde {f}};\lambda _{0i},x_{0i}}(K,M;\lambda _{0i},x_{0i})(\delta \lambda _{i},\delta x_{i})=$${\binom {y}{y_{n+1}}}$ is

$$\delta \lambda _{i}=-x_{0i}^{T}y,\;{\text{ and }}(\lambda _{0i}-\lambda _{0j})x_{0j}^{T}M\delta x_{i}=x_{j}^{T}y,j=1,\dots ,n,j\neq i\;;$$ ${\text{ or }}x_{0j}^{T}M\delta x_{i}=x_{j}^{T}y/(\lambda _{0i}-\lambda _{0j}),{\text{ and }}\;2x_{0i}^{T}M\delta x_{i}=y_{n+1}$

When $\lambda _{i}$ is a simple eigenvalue, as the eigenvectors $x_{0j},j=1,\dots ,n$ form an orthonormal basis, for any right-hand side, we have obtained one solution therefore, the Jacobian is invertible.

The implicit function theorem provides a continuously differentiable function $(K,M)\mapsto (\lambda _{i}(K,M),x_{i}(K,M))$ hence the expansion with little o notation: $\lambda _{i}=\lambda _{0i}+\delta \lambda _{i}+o(\|\delta K\|+\|\delta M\|)$ $x_{i}=x_{0i}+\delta x_{i}+o(\|\delta K\|+\|\delta M\|)$. with

$\delta \lambda _{i}=\mathbf {x} _{0i}^{T}\delta \mathbf {K} \mathbf {x} _{0i}-\lambda _{0i}\mathbf {x} _{0i}^{T}\delta \mathbf {M} \mathrm {x} _{0i};$ $\delta x_{i}=\mathbf {x} _{0j}^{T}\mathbf {M} _{0}\delta \mathbf {x} _{i}\mathbf {x} _{0j}{\text{ with}}$$\mathbf {x} _{0j}^{T}\mathbf {M} _{0}\delta \mathbf {x} _{i}={\frac {-\mathbf {x} _{0j}^{T}\delta \mathbf {K} \mathbf {x} _{0i}+\lambda _{0i}\mathbf {x} _{0j}^{T}\delta \mathbf {M} \mathrm {x} _{0i}}{(\lambda _{0j}-\lambda _{0i})}},i=1,\dots n;j=1,\dots n;j\neq i.$

Eigenvalue sensitivity, a small example

A simple case is $K={\begin{bmatrix}2&b\\b&0\end{bmatrix}}$; however you can compute eigenvalues and eigenvectors with the help of online tools such as [1] (see introduction in Wikipedia WIMS) or using Sage SageMath. You get the smallest eigenvalue $\lambda =-\left[{\sqrt {b^{2}+1}}+1\right]$ and an explicit computation ${\frac {\partial \lambda }{\partial b}}={\frac {-x}{\sqrt {x^{2}+1}}}$; more over, an associated eigenvector is ${\tilde {x}}_{0}=[x,-({\sqrt {x^{2}+1}}+1))]^{T}$; it is not an unitary vector; so $x_{01}x_{02}={\tilde {x}}_{01}{\tilde {x}}_{02}/\|{\tilde {x}}_{0}\|^{2}$; we get $\|{\tilde {x}}_{0}\|^{2}=2{\sqrt {x^{2}+1}}({\sqrt {x^{2}+1}}+1)$ and ${\tilde {x}}_{01}{\tilde {x}}_{02}=-x({\sqrt {x^{2}+1}}+1)$ ; hence $x_{01}x_{02}=-{\frac {x}{2{\sqrt {x^{2}+1}}}}$; for this example , we have checked that ${\frac {\partial \lambda }{\partial b}}=2x_{01}x_{02}$ or $\delta \lambda =2x_{01}x_{02}\delta b$.

Divers

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${\begin{aligned}\mathbf {K} _{0}\mathbf {x} _{0i}&+\delta \mathbf {K} \mathbf {x} _{0i}+\mathbf {K} _{0}\delta \mathbf {x} _{i}+\delta \mathbf {K} \delta \mathbf {x} _{i}=\\[6pt]&\lambda _{0i}\mathbf {M} _{0}\mathbf {x} _{0i}+\lambda _{0i}\mathbf {M} _{0}\delta \mathbf {x} _{i}+\lambda _{0i}\delta \mathbf {M} \mathbf {x} _{0i}+\delta \lambda _{i}\mathbf {M} _{0}\mathbf {x} _{0i}\\[12pt]&+\lambda _{0i}\delta \mathbf {M} \delta \mathbf {x} _{i}+\delta \lambda _{i}\delta \mathbf {M} \mathbf {x} _{0i}+\delta \lambda _{i}\mathbf {M} _{0}\delta \mathbf {x} _{i}+\delta \lambda _{i}\delta \mathbf {M} \delta \mathbf {x} _{i}.\end{aligned}}$

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Example 2

Suppose $[K_{0}]$ is the 2x2 identity matrix, any vector is an eigenvector; then $u_{0}=[1,1]^{T}/{\sqrt {2}}$ is one possible eigenvector. But if one makes a small perturbation, such as

$[K]=[K_{0}]+{\begin{bmatrix}\epsilon &0\\0&0\end{bmatrix}}$

Then the eigenvectors are $v_{1}=[1,0]^{T}$ and $v_{2}=[0,1]^{T}$; they are constant with respect to $\epsilon$ so that $\|u_{0}-v_{1}\|$ is constant and does not go to zero.