TITLE: Given a perturbation of a symmetric matrix, find an expansion for the eigenvalues
QUESTION [11 upvotes]: Let $A$ be a real, symmetrix $n\times n$ matrix with $n$ distinct,
  non-zero eigenvalues, and let $V$ be a real, symmetric $n\times n$
  matrix. 
Consider $A_{\varepsilon}=A+\varepsilon V$, a perturbation of $A$, where
  $\varepsilon$ is a small number. 

Find an expansion as an $\varepsilon$-series for the eigenvalues and
  eigenvectors of $A_{\varepsilon}$ around the eigenvalues and
  eigenvectors of the original matrix, $A$.
Assume that $\epsilon$ is sufficiently small that one can neglect
  all terms of order $\varepsilon^2$ and higher in the expansion. How
  close are the eigenvalues of $A_{\varepsilon}$ to those of $A$? What
  about the eigenvectors?
What can we say about the eigenvalues and eigenvectors of
  $A_{\varepsilon} $ if $A$ and $V$ are arbitrary $n\times n$ matrices?
  You may assume that $\det A \neq 0$.


Ideas: If $A$ and $V$ commute (thus are simultaneously diagonalizable), there are just the linear term $\varepsilon v_{ii}$, when we look with the right basis. I'm getting the feeling I really don't know what they want me to do with this problem. Any ideas?

REPLY [15 votes]: $\newcommand{\+}{^{\dagger}}%
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Let's consider a particular eigenvector $v_{0}$ of $A$ where $\lambda_{0}$ is the correspondent eigenvalue. Namely, $Av_{0} = \lambda_{0}v_{0}$. 'Under the action' of $A + \epsilon V$, the eigenvalue problem becomes
$\pars{A + \epsilon V}v = \lambda v$. We expand $v$ and $\lambda$ in powers of $\epsilon$ as:
$$
v = v_{0} + v_{1}\epsilon + v_{2}\epsilon^{2} + \cdots\,,
\qquad
\lambda = \lambda_{0} + \lambda_{1}\epsilon + \lambda_{2}\epsilon^{2} + \cdots
$$ 
In order to derive the contributions to the eigenvector and eigenvalue
$\ul{\mbox{up to order}\ \epsilon}$, it's sufficient to write:
$$
\pars{A + \epsilon V}\pars{v_{0} + \epsilon v_{1}}
=
\pars{\lambda_{0} + \lambda_{1}\epsilon}\pars{v_{0} + \epsilon v_{1}}\,,
\quad
\left\vert%
\begin{array}{rcl}
Av_{0} & = & \lambda_{0}v_{0}
\\
Av_{1} + Vv_{0} & = & \lambda_{0}v_{1} + \lambda_{1}v_{0}
\end{array}\right.
$$
where we equated terms which correspond to the same power of $\epsilon$
$\pars{~\ul{\mbox{up to order}\ \epsilon}~}$. 
Applying $v_{0}^{\sf T}$, by the left, to both sides of the last expression:
$$
v_{0}^{\sf T}Av_{1} + v_{0}^{\sf T}Vv_{0} = \lambda_{0}v_{0}^{\sf T}v_{1} + \lambda_{1}v_{0}^{\sf T}v_{0}
$$
Since $A$ is a symmetric matrix, we'll have
$$
\color{#ff0000}{v_{0}^{\sf T}Av_{1}} = \pars{v_{1}^{\sf T}Av_{0}}^{\sf T}
=
\pars{v_{1}^{\sf T}\lambda_{0}v_{0}}^{\sf T}
=
\color{#ff0000}{\lambda_{0}v_{0}^{\sf T}v_{1}}
\quad\mbox{such that}\quad
\lambda_{1} = {v_{0}^{\sf T}Vv_{0} \over v_{0}^{\sf T}v_{0}}
$$

Then, $\pars{~\ul{\mbox{up to order}\ \epsilon}~}$:
$$\color{#0000ff}{\large%
\lambda
\approx
\lambda_{0} + {v_{0}^{\sf T}Vv_{0} \over v_{0}^{\sf T}v_{0}}\,\epsilon}
$$