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 $M$ ?

  • $\begingroup$ 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

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 $\dfrac12\left(\pm\sqrt{4 a b+c^2}+c\right)$.


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