---
tag: Y50G59S75Q
kind: definition
name: Eigenvalue perturbation
msc: [15A18, 47A55, 65F15]
defines: VR89YHC0TX
uses: [0BQAA2BK3Q, 0N6960W0BV, 03XVP8Q4GY]
sources:
  - url: https://en.wikipedia.org/wiki/Eigenvalue_perturbation
---

## Definition

Let $A$ and $B$ be $n \times n$ complex matrices, and let $A(\varepsilon) = A + \varepsilon B$ for $\varepsilon \in \C$.
Let $\lambda_0$ be a simple [[eigenvalue|0BQAA2BK3Q]] of $A$, that is, a root of [[multiplicity|03XVP8Q4GY]] $1$ of the characteristic polynomial.
Let $x$ be a right [[eigenvector|0N6960W0BV]] and $y$ a left eigenvector of $A$ for $\lambda_0$: $Ax = \lambda_0 x$ and $y^* A = \lambda_0 y^*$, with $x, y \ne 0$.
Then $y^* x \ne 0$.
There are a neighbourhood $D$ of $0$ in $\C$ and a unique analytic function $\lambda \colon D \to \C$ with $\lambda(0) = \lambda_0$ and the following property.
For each $\varepsilon \in D$, the number $\lambda(\varepsilon)$ is a simple eigenvalue of $A(\varepsilon)$.
The *eigenvalue perturbation* of $\lambda_0$ in the direction $B$ is this function $\lambda$.
Its *first-order coefficient* is
$$\lambda'(0) = \frac{y^* B x}{y^* x}.$$

## Remarks

So $\lambda(\varepsilon) = \lambda_0 + \varepsilon \, \dfrac{y^* B x}{y^* x} + O(\varepsilon^2)$.
To obtain the coefficient, differentiate $A(\varepsilon) x(\varepsilon) = \lambda(\varepsilon) x(\varepsilon)$ at $0$ and multiply on the left by $y^*$; the terms with $x'(0)$ cancel.

Example: for a Hermitian $A$ one may take $y = x$, so $\lambda'(0) = x^* B x / x^* x$.
For $A = \operatorname{diag}(1, 2)$ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, with $\lambda_0 = 1$ and $x = y = e_1$, this gives $\lambda'(0) = 0$.
Indeed the eigenvalues of $A + \varepsilon B$ are $\frac{3}{2} \pm \sqrt{\frac{1}{4} + \varepsilon^2}$, and the smaller one is $1 - \varepsilon^2 + O(\varepsilon^4)$.

Non-example: for $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, the eigenvalue $0$ is not simple.
With $B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, the eigenvalues of $A + \varepsilon B$ are $\pm \sqrt{\varepsilon}$, which are not analytic in $\varepsilon$ at $0$.
Here the right eigenvector $x = e_1$ and the left eigenvector $y = e_2$ satisfy $y^* x = 0$, so the formula for $\lambda'(0)$ has no meaning.

## Explanations

### L1 human

When a matrix depends smoothly on a small parameter, a simple eigenvalue moves smoothly too.
Its speed at the start is given by a short formula with the left and right eigenvectors.
A small value of $y^* x$ means that the eigenvalue is sensitive to changes.