Bauer–Fike Theorem

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''.

The theorem was proved by Friedrich L. Bauer and C. T. Fike in 1960.

The setup

In what follows we assume that:
* is a diagonalizable matrix;
* is the non-singular eigenvector matrix such that , where is a diagonal matrix.
* If is invertible, its condition number in -norm is denoted by and defined by:
::$\kappa_p(X)=\, X\, _p \left \, X^ \right \, _p.$

The Bauer–Fike Theorem

:Bauer–Fike Theorem. Let be an eigenvalue of . Then there exists such that:
::$, \lambda-\mu, \leq \kappa_p (V) \, \delta A \, _p$

Proof. We can suppose , otherwise take and the result is trivially true since . Since is an eigenvalue of , we have and so

:$\begin 0 &= \det(A+\delta A-\mu I) \\ &= \det(V^)\det(A+\delta A-\mu I)\det(V) \\ &= \det \left ( V^ (A+\delta A-\mu I) V \right ) \\ &= \det \left ( V^AV + V^\delta AV- V^ \mu I V \right ) \\ &= \det \left (\Lambda+V^\delta AV-\mu I \right ) \\ &= \det(\Lambda-\mu I)\det \left ((\Lambda-\mu I)^V^\delta AV +I \right ) \\ \end$

However our assumption, , implies that: and therefore we can write:

:$\det \left ((\Lambda-\mu I)^V^\delta AV +I \right ) = 0.$

This reveals to be an eigenvalue of

:$(\Lambda-\mu I)^V^\delta AV.$

Since all -norms are consistent matrix norms we have where is an eigenvalue of . In this instance this gives us:

:$1 = , -1, \leq \left \, (\Lambda-\mu I)^V^\delta AV \right \, _p \leq \left \, (\Lambda-\mu I)^ \right \, _p \left \, V^\right \, _p \, V\, _p\, \delta A\, _p = \left \, (\Lambda-\mu I)^ \right \, _p\ \kappa_p(V)\, \delta A\, _p$

But is a diagonal matrix, the -norm of which is easily computed:

:$\left \, \left (\Lambda-\mu I \right )^ \right \, _p\ =\max_ \frac =\max_\frac\ = \frac$

whence:

:$\min_, \lambda-\mu, \leq\ \kappa_p(V)\, \delta A\, _p.$

An Alternate Formulation

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

:Bauer–Fike Theorem (Alternate Formulation). Let be an approximate eigenvalue-eigenvector couple, and . Then there exists such that:
::$\left , \lambda-\lambda^a \right , \leq \kappa_p (V)\frac$

Proof. We can suppose , otherwise take and the result is trivially true since . So exists, so we can write:

:$\boldsymbol^a = \left (A-\lambda^a I \right )^ \boldsymbol= V \left (D-\lambda^a I \right )^V^\boldsymbol$

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

:$\left \, \boldsymbol^a \right \, _p= \left \, V \left (D-\lambda^a I \right )^V^\boldsymbol \right \, _p \leq \, V\, _p \left \, \left (D-\lambda^a I \right )^ \right \, _p \left \, V^ \right \, _p \, \boldsymbol\, _p =\kappa_p(V) \left \, \left (D-\lambda^a I \right )^ \right \, _p \, \boldsymbol\, _p.$

However

:$\left (D- \lambda^a I \right )^$

is a diagonal matrix and its -norm is easily computed:

:$\left \, \left (D-\lambda^a I \right )^ \right \, _p = \max_\frac =\max_ \frac=\frac$

whence:

:$\min_ \left , \lambda-\lambda^a \right , \leq\kappa_p(V) \frac.$

A Relative Bound

Both formulations of Bauer–Fike theorem yield an absolute bound. The following corollary is useful whenever a relative bound is needed:

:Corollary. Suppose is invertible and that is an eigenvalue of . Then there exists such that:
::$\frac\leq\kappa_p (V) \left \, A^\delta A \right \, _p$

Note. can be formally viewed as the ''relative variation of'' , just as is the relative variation of .

Proof. Since is an eigenvalue of and , by multiplying by from left we have:

:$-A^(A+\delta A)\boldsymbol=-\mu A^\boldsymbol.$

If we set:

:$A^a =\mu A^, \qquad (\delta A)^a = -A^\delta A$

then we have:

:$\left (A^a + (\delta A)^a - I \right )\boldsymbol = \boldsymbol$

which means that is an eigenvalue of , with as an eigenvector. Now, the eigenvalues of are , while it has the same eigenvector matrix as . Applying the Bauer–Fike theorem to with eigenvalue , gives us:

:$\min_ \left, \frac-1\ = \min_\frac \leq \kappa_p (V) \left \, A^\delta A \right \, _p$

The Case of Normal Matrices

If is normal, is a unitary matrix, therefore:

:$\, V\, _2= \left \, V^ \right \, _2 = 1,$

so that . The Bauer–Fike theorem then becomes:

:$\exists\lambda\in\Lambda(A): \quad , \lambda-\mu, \leq\, \delta A\, _2$

Or in alternate formulation:

:$\exists\lambda\in\Lambda(A): \quad \left , \lambda-\lambda^a \right , \leq\frac$

which obviously remains true if is a Hermitian matrix. In this case, however, a much stronger result holds, known as the Weyl's theorem on eigenvalues. In the hermitian case one can also restate the Bauer–Fike theorem in the form that the map that maps a matrix to its spectrum is a non-expansive function with respect to the Hausdorff distance on the set of compact subsets of .

References

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{{DEFAULTSORT:Bauer-Fike theorem
Spectral theory
Theorems in analysis
Articles containing proofs