Choked flow

28 year-old Painting Investments Worker Truman from Regina, usually spends time with pastimes for instance interior design, property developers in new launch ec Singapore and writing. Last month just traveled to City of the Renaissance. In mathematics, the Bauer–Fike theorem is a standard result in the perturbation theory of the eigenvalue of a complex-valued diagonalizable matrix. In its substance, it states an absolute upper bound for the deviation of one perturbed matrix eigenvalue from a properly chosen eigenvalue of the exact matrix. Informally speaking, what it says is that the sensitivity of the eigenvalues is estimated by the condition number of the matrix of eigenvectors.

Theorem (Friedrich L. Bauer, C.T.Fike – 1960)

Let $A\in {ℂ}^{n,n}$ be a diagonalizable matrix, and $V\in {ℂ}^{n,n}$ be the non-singular eigenvector matrix such that $A=V\Lambda V^{-1}$. Moreover, let $\mu$ be an eigenvalue of the matrix $A+\delta A$; then an eigenvalue $\lambda \in \sigma (A)$ exists such that:

$|\lambda -\mu |\le {\kappa }_p(V)\Vert \delta A{\Vert }_p$

where ${\kappa }_p(V)=\Vert V{\Vert }_p\Vert V^{-1}{\Vert }_p$ is the usual condition number in p-norm.

Proof

If $\mu \in \sigma (A)$, we can choose $\lambda =\mu$ and the thesis is trivially verified (since ${\kappa }_p(V)\ge 1$).

So, be $\mu \notin \sigma (A)$. Then $det(\Lambda -\mu I)\ne 0$. $\mu$ being an eigenvalue of $A+\delta A$, we have $det(A+\delta A-\mu I)=0$ and so

$0=det(V^{-1})det(A+\delta A-\mu I)det(V)=det(\Lambda +V^{-1}\delta AV-\mu I)$

$=det(\Lambda -\mu I)det[(\Lambda -\mu I{)}^{-1}{V}^{-1}\delta AV+I]$

and, since $det(\Lambda -\mu I)\ne 0$ as stated above, we must have

$det[(\Lambda -\mu I{)}^{-1}{V}^{-1}\delta AV+I]=0$

which reveals the value −1 to be an eigenvalue of the matrix $(\Lambda -\mu I{)}^{-1}{V}^{-1}\delta AV$.

For each consistent matrix norm, we have $|\lambda |\le \Vert A\Vert$, so, all p-norms being consistent, we can write:

$1\le \Vert (\Lambda -\mu I{)}^{-1}{V}^{-1}\delta AV{\Vert }_p\le \Vert (\Lambda -\mu I{)}^{-1}{\Vert }_p\Vert V^{-1}{\Vert }_p\Vert V{\Vert }_p\Vert \delta A{\Vert }_p$

$=\Vert (\Lambda -\mu I{)}^{-1}{\Vert }_p{\kappa }_p(V)\Vert \delta A{\Vert }_p$

But $(\Lambda -\mu I{)}^{-1}$ being a diagonal matrix, the p-norm is easily computed, and yields:

$\Vert (\Lambda -\mu I{)}^{-1}{\Vert }_p=\underset{\Vert 𝐱{\Vert }_p\ne 0}{max}\frac{\Vert (\Lambda -\mu I{)}^{-1}𝐱{\Vert }_p}{\Vert 𝐱{\Vert }_p}$

$=\underset{\lambda \in \sigma (A)}{max}\frac{1}{|\lambda -\mu |}=\frac{1}{\underset{\lambda \in \sigma (A)}{min}|\lambda -\mu |}$

whence:

$\underset{\lambda \in \sigma (A)}{min}|\lambda -\mu |\le {\kappa }_p(V)\Vert \delta A{\Vert }_p.$

The theorem can also be reformulated to better suit numerical methods. In fact, dealing with real eigensystem problems, one often has an exact matrix $A$, but knows only an approximate eigenvalue-eigenvector couple, ($\tilde{\lambda }$,$\widetilde{𝐯}$), and needs to bound the error. The following version comes in help.

Theorem (Friedrich L. Bauer, C.T.Fike – 1960) (alternative statement)

Let $A\in {ℂ}^{n,n}$ be a diagonalizable matrix, and be $V\in {ℂ}^{n,n}$ the non singular eigenvector matrix such as $A=V\Lambda V^{-1}$. Be moreover ($\tilde{\lambda }$,$\widetilde{𝐯}$) an approximate eigenvalue-eigenvector couple, and $𝐫=A\widetilde{𝐯}-\tilde{\lambda }\widetilde{𝐯}$; then an eigenvalue $\lambda \in \sigma (A)$ exists such that:

$|\lambda -\tilde{\lambda }|\le {\kappa }_p(V)\frac{\Vert 𝐫{\Vert }_p}{\Vert \widetilde{𝐯}{\Vert }_p}$

where ${\kappa }_p(V)=\Vert V{\Vert }_p\Vert V^{-1}{\Vert }_p$ is the usual condition number in p-norm.

Proof

We solve this problem with Tarık's method: m$\tilde{\lambda }\notin \sigma (A)$ (otherwise, we can choose $\lambda =\tilde{\lambda }$ and theorem is proven, since ${\kappa }_p(V)\ge 1$). Then $(A-\tilde{\lambda }I{)}^{-1}$ exists, so we can write:

$\widetilde{𝐯}=(A-\tilde{\lambda }I{)}^{-1}𝐫=V(D-\tilde{\lambda }I{)}^{-1}{V}^{-1}𝐫$

since $A$ is diagonalizable; taking the p-norm of both sides, we obtain:

$\Vert \widetilde{𝐯}{\Vert }_p=\Vert V(D-\tilde{\lambda }I{)}^{-1}{V}^{-1}𝐫{\Vert }_p\le \Vert V{\Vert }_p\Vert (D-\tilde{\lambda }I{)}^{-1}{\Vert }_p\Vert V^{-1}{\Vert }_p\Vert 𝐫{\Vert }_p$

$={\kappa }_p(V)\Vert (D-\tilde{\lambda }I{)}^{-1}{\Vert }_p\Vert 𝐫{\Vert }_p.$

But, since $(D-\tilde{\lambda }I{)}^{-1}$ is a diagonal matrix, the p-norm is easily computed, and yields:

$\Vert (D-\tilde{\lambda }I{)}^{-1}{\Vert }_p=\underset{\Vert 𝐱{\Vert }_p\ne 0}{max}\frac{\Vert (D-\tilde{\lambda }I{)}^{-1}𝐱{\Vert }_p}{\Vert 𝐱{\Vert }_p}$

$=\underset{\lambda \in \sigma (A)}{max}\frac{1}{|\lambda -\tilde{\lambda }|}=\frac{1}{\underset{\lambda \in \sigma (A)}{min}|\lambda -\tilde{\lambda }|}$

whence:

$\underset{\lambda \in \sigma (A)}{min}|\lambda -\tilde{\lambda }|\le {\kappa }_p(V)\frac{\Vert 𝐫{\Vert }_p}{\Vert \widetilde{𝐯}{\Vert }_p}.$

The Bauer–Fike theorem, in both versions, yields an absolute bound. The following corollary, which, besides all the hypothesis of Bauer–Fike theorem, requires also the non-singularity of A, turns out to be useful whenever a relative bound is needed.

Corollary

Be $A\in {ℂ}^{n,n}$ a non-singular, diagonalizable matrix, and be $V\in {ℂ}^{n,n}$ the non singular eigenvector matrix such as $A=V\Lambda V^{-1}$. Be moreover $\mu$ an eigenvalue of the matrix $A+\delta A$; then an eigenvalue $\lambda \in \sigma (A)$ exists such that:

$\frac{|\lambda -\mu |}{|\lambda |}\le {\kappa }_p(V)\Vert A^{-1}\delta A{\Vert }_p$

(Note: $\Vert A^{-1}\delta A\Vert$can be formally viewed as the "relative variation of A", just as $|\lambda -\mu ||\lambda {|}^{-1}$ is the relative variation of λ.)

Proof

Since μ is an eigenvalue of (A+δA) and $det(A)\ne 0$, we have, left-multiplying by $-A^{-1}$:

$-A^{-1}(A+\delta A)𝐯=-\mu A^{-1}𝐯$

that is, putting$\tilde{A}=\mu A^{-1}$ and $\widetilde{\delta A}=-A^{-1}\delta A$:

$(\tilde{A}+\widetilde{\delta A}-I)𝐯=\mathrm{𝟎}$

which means that$\tilde{\mu }=1$is an eigenvalue of$(\tilde{A}+\widetilde{\delta A})$, with $𝐯$eigenvector. Now, the eigenvalues of $\tilde{A}$are $\frac{\mu }{{\lambda }_i}$, while its eigenvector matrix is the same as A. Applying the Bauer–Fike theorem to the matrix$\tilde{A}+\widetilde{\delta A}$ and to its eigenvalue$\tilde{\mu }=1$, we obtain:

$\underset{\lambda \in \sigma (A)}{min}|\frac{\mu }{\lambda }-1|=\underset{\lambda \in \sigma (A)}{min}\frac{|\lambda -\mu |}{|\lambda |}\le {\kappa }_p(V)\Vert A^{-1}\delta A{\Vert }_p$

Remark

If A is normal, V is a unitary matrix, and $\Vert V{\Vert }_2=\Vert V^{-1}{\Vert }_2=1$, so that ${\kappa }_2(V)=1$.

The Bauer–Fike theorem then becomes:

$\mathrm{\exists }\lambda \in \sigma (A):|\lambda -\mu |\le \Vert \delta A{\Vert }_2$

( $\mathrm{\exists }\lambda \in \sigma (A):|\lambda -\tilde{\lambda }|\le \frac{\Vert 𝐫{\Vert }_2}{\Vert \widetilde{𝐯}{\Vert }_2}$ in the alternative formulation)

which obviously remains true if A is a Hermitian matrix. In this case, however, a much stronger result holds, known as the Weyl's theorem on eigenvalues.

References

1. F. L. Bauer and C. T. Fike. Norms and exclusion theorems. Numer. Math. 2 (1960), 137–141.
2. S. C. Eisenstat and I. C. F. Ipsen. Three absolute perturbation bounds for matrix eigenvalues imply relative bounds. SIAM Journal on Matrix Analysis and Applications Vol. 20, N. 1 (1998), 149–158