Let $\lambda$ be a simple eigenvalue of $A \in L(C^n)$ and let $x$ be the corresponding eigenvector. Then for $E \in L(C^n)$, $A+E$ has an eigenvalue $\lambda(E)$ and an eigenvector $x(E)$ such that as $E\longrightarrow 0$, $\lambda(E) \longrightarrow \lambda$ and $x(E) \longrightarrow x$. This is 3.1.3 (Chapter 3, page 46) from Numerical Analysis: a second course by J. Ortega. I am using the book for self study.

The continuity of the eigenvalue is clear, but I do not understand the proof for the eigenvector bit. Could someone help me understand what is going on ? I looked at Continuity of a simple eigenvalue and its corresponding eigenvector but that does not answer my question. I would like to have a proof that does not invoke implicit/inverse function theorem.


1 Answer 1


Let us first note that this statement is not completely correct as it stands. This is mainly because eigenvectors are only unique up to a scalar multiple. So for a sequence of matrices $(E_n)_n$ with $E_n \rightarrow 0$, you could choose $x(E_n)$ in such a way that $\Vert x(E_n) \Vert = n$. In this case, your convergence certainly does not hold.

For simplicity note that by subtracting $\lambda \cdot \rm{id}$ from $A$, we can assume $\lambda = 0$.

The above means that you have to normalize the $x(E)$ in a suitable way.

For this, let us assume $\varphi(x) \neq 0$ for some linear functional $\varphi$ on $\Bbb{C}^n$. You could take e.g. $\varphi(z) = z_j$ for a suitable $j$.

Let us assume (by rescaling) that $x(E)$ is chosen so that $\varphi(x(E)) = \varphi(x)$ if $\varphi(x(E)) \neq 0$.

Now assume that your claim is false. Then there is a sequence $(E_n)_n$ with $E_n \rightarrow 0$ but $x(E_n) \not\rightarrow x$. Choosing a suitable subsequence (and denoting that by $(E_n)_n$ again), we can assume $\Vert x(E_n) - x(E)\Vert > \varepsilon$ for all $n$.

Let us now consider the "normalized" vectors $y(E_n) := \frac{x(E_n)}{\Vert x(E_n) \Vert}$. By restricting to a subsequence, we ca assume (by Bolzano Weierstraß) that $y(E_n) \rightarrow y$ for some $y \in \Bbb{C}^n$ with $\Vert y \Vert = 1$.

Note that $(A + E_n) y_n = \lambda_n y_n \rightarrow \lambda y = 0$ (because you already know that $\lambda_n \rightarrow \lambda$, as you say), i.e. $A y_n = (A + E_n) y_n - E_n y_n \rightarrow 0$, because $E_n \rightarrow 0$ and $(y_n)_n$ is bounded.

This shows $Ay = 0$, i.e. $y = c \cdot x$ for some $c \in \Bbb{C}$ (because $x$ is the only eigenvector for the eigenvalue $0$). We have $c \neq 0$ because $\Vert y \Vert = 1$.

This shows $\varphi(y_n) \rightarrow \varphi(y) = c \cdot \varphi(x) \neq 0$ and thus $\varphi(y_n) \neq 0$ and hence $\varphi(x(E_n)) \neq 0$ for $n$ large enough and thus (by our normalization of the $x(E_n)$) that $\varphi(x(E_n)) = \varphi(x)$ for $n$ large.

Finally, $$\Vert x(E_n)\Vert = \frac{\varphi(x(E_n))}{\varphi(y_n)} = \frac{\varphi(x)}{\varphi(y_n)} \rightarrow \frac{\varphi(x)}{\varphi(y)}.$$

We conclude $x(E_n) = \Vert x(E_n) \Vert \cdot y_n \rightarrow \frac{\varphi(x)}{\varphi(y)} \cdot y = \frac{\varphi(x)}{c \varphi(x)} \cdot c x = x$. This contradicts our assumption $\Vert x(E_n) - x\Vert > \varepsilon$ for all $n$.

  • $\begingroup$ I don't think you require $\varphi$ to be linear, just that it is continuous, and that it takes the value $\varphi(x)$ on every 1-dimensional subspace. In particular, you can take $\varphi(x) = \Vert x\Vert$. $\endgroup$
    – ronno
    May 24, 2014 at 8:58
  • $\begingroup$ You can not take $\varphi(x) = \Vert x \Vert$, because then you can again take something like $x'(E_n) = \alpha x(E_n)$ with $|\alpha| = 1$, $\alpha \neq 1$ and you cannot have $x'(E_n) \rightarrow x$ as well as $x(E_n) \rightarrow x$. The $\varphi$ has to be injective on $\rm{span}(x)$. $\endgroup$
    – PhoemueX
    May 24, 2014 at 9:47
  • $\begingroup$ I see. So it more or less needs to be linear. My mistake. $\endgroup$
    – ronno
    May 24, 2014 at 9:56
  • $\begingroup$ @PhoemueX : That is very neat, thanks. I was wondering though, could your proof be modified to make it constructive rather than use a contradiction argument ? $\endgroup$
    – me10240
    May 29, 2014 at 1:49
  • $\begingroup$ I am not sure about that at the moment. If I come up with something, I will let you know. $\endgroup$
    – PhoemueX
    May 30, 2014 at 10:07

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