Given the eigenvalues and eigenvectors of a matrix $R^{n\times n}$ is that possible to recover the same matrix from smaller matrices $R^{(n-1) \times (n-1)}$ where one of its eigenvalues and eigenvectors had been eliminated?. I have a Matrix $A$ of $3 \times 3$ dimension and doing a process i have obtained three diferent matrices of dimension $2 \times 2$ where each one conserve two of the three original eigenvectors and eigenvalues.

The process what i have done is: Matrix $A$, has n eigenvalues, $n=3$ for explaining. Is possible to obtain three new matrices through this $A_k=(SMP^{k}S')S(MP^k)^{-1}A(MP^k)S'(SMP^kS')^{-1}$ and as well with $A_k=SMEP^kS^T(SMP^kS')^{-1}$ where: $S$ is a selection matrix to eliminate a row and a column, in the case of $n=3$ is the following:


$P$ is a permutation matrix for eigenvectors $$P=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}$$

$E$ is a diagonal Matrix which contains the eigenvalues $$E=\begin{pmatrix}\lambda_{1}&0&0\\0&\lambda_2&0\\0&0&\lambda_3\end{pmatrix}$$ M is the eigenvectors matrix, $$M=\begin{pmatrix}v_{1}&v_{2}&v_{3}\end{pmatrix}$$ and $k$ let us to find the $n+1$ solutions i mean $A_1$, $A_2$ and $A_3$ . Another way to express that solutions is $A_k=SAMP^kS^T(SMP^kS^T)^{-1}$ . What I have done is like a projection from $R^3$ over $R^2$ or $C^2$ eliminating one eigenvalue in each case to obtain $A_1$, $A_2$, $A_3$.

Thanks for your help!


  • $\begingroup$ It would save everyone a lot of time and effort comparing them if you pointed out the relationship between this question and this other question of yours. At first sight this appears to be a duplicate, but it's hard to tell -- please clarify. $\endgroup$ – joriki Dec 14 '12 at 12:08
  • $\begingroup$ @joriki is the same question but maybe in other words, please tell me what dont you understand to clarify?, I am agree with you that it would save a lot of time having just one clear question sorry for that. But please let me know what is not clear to clarify in one of both. Thanks $\endgroup$ – Gina Torres Jan 8 '13 at 11:49

Your Answer

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

Browse other questions tagged or ask your own question.