2
$\begingroup$

Let's say I have a diagonalizable $3\times3$ matrix $$ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$$

with 3 distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$. Let's say I now add a small perturbation to $i$ of the form $-k^2$ so that the matrix becomes:

$$ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i - k^2 \end{pmatrix}$$

Is there are formula for how the eigenvalues will change? I'm not even sure where to start.

$\endgroup$
3
  • $\begingroup$ Are there any conditions on you $3\times 3$ matrix? $\endgroup$ – OgvRubin Oct 25 '18 at 21:15
  • $\begingroup$ @OlofRubin it is diagonalizable, has 3 distinct eigenvalues $\endgroup$ – Mike Flynn Oct 25 '18 at 21:17
  • $\begingroup$ Yes, there is a formula, i.e., a cubic equation (characteristic polynomial) and its roots. $\endgroup$ – Dietrich Burde Oct 25 '18 at 21:19
0
$\begingroup$

Write $A = \begin{bmatrix}a & b & c \\ d & e & f \\ g& h & i\end{bmatrix}.$

Say that $A$ is Hermitian that is $A = \overline{A}^T$ then you could use Weyl's perturbation theorem which says that if $B$ is another Hermitian matrix then

$$\max_{1\leq i \leq 3}|\lambda_i(A+B)-\lambda_i(A)|\leq \|B\|$$

Here $\lambda_i$ is the map which takes Hermitian matrices to their eigenvalues ordered non-increasingly.

In your case

$$B = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & k^2\end{bmatrix}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.