0
$\begingroup$

If i have a matrix like \begin{align*} M= \begin{pmatrix} 0&1&0 \\ 0&0&1 \\ 4&-17&8 \end{pmatrix} \end{align*}

and merely exchange rows $R_1$ and $R_3$, so i have

\begin{align*} M'= \begin{pmatrix} 4&-17&8\\ 0&1&0 \\ 0&0&1 \end{pmatrix} \end{align*}

we find the eigenvectors of the row-exchanged matrix are really nice, i.e. $\Lambda' = (\lambda_1',\lambda_2',\lambda_3')=(1,1,4)$, whereas the eigenvalues of the original matrix are not nice.. My question is, is there a relation between the eigenvalues of matrix $M'$ and the eigenvalues of $M$ ?

$\endgroup$
  • $\begingroup$ You are replacing $M$ with $M'=PM$ where $P$ is a permutation matrix. If you also apply $P^T$ from the right and compute $M'' = PAP^T$, then you will also have permuted the columns and preserved the eigenvalues. This follows from the fact that $P^{-1} = P^T$ for a permutation matrix. $\endgroup$ – Carl Christian Apr 16 at 22:02
3
$\begingroup$

In general, there is no relation. For instance, if$$M=\begin{bmatrix}0&a&b\\0&0&c\\0&0&0\end{bmatrix},$$then its only eigenvalue is $0$. And indeed $0$ is also an eigenvalue of $M'$, but so are $\dfrac12\left(\pm\sqrt{4 a b+c^2}+c\right)$.

$\endgroup$

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.