Suppose a real, dense and asymmetric square matrix $A\in\mathbb{R}^{n\times n}$, all its eigenvalues $\lambda_i \in \mathbb R$

Is it possible to construct a symmetric matrix $B\in\mathbb{R}^{n\times n}$, the eigenvalues of which are $\lambda_i^2$ to compute absolute values of $\lambda_i$ without the need of solving eigenvalues of $A$?

  • 1
    $\begingroup$ I don't understand what you want the relationship between $A$ and $B$ to be? $\endgroup$ – Wolfgang Bangerth Feb 13 '14 at 3:13
  • $\begingroup$ Originally, I would expect $B=A^TA$, but it is not the case unless $A$ is symmetric. $\endgroup$ – LCFactorization Feb 13 '14 at 4:35
  • $\begingroup$ Aren't you just looking for $A^2$? $\endgroup$ – Stefano M Feb 13 '14 at 10:59
  • 1
    $\begingroup$ Sorry, I didn't notice you are asking for a symmetric matrix. So we can restate your question as how to construct a symmetric matrix $B$ with the same eigenvalues of $A^2$, without first computing the eigenvalues of $A$? $\endgroup$ – Stefano M Feb 13 '14 at 12:40
  • 1
    $\begingroup$ But $B_{3\times 3}$ is only of size 3, right? You can basically write down the eigenvalues of this matrix. The entire cost is going to be to form the product $JJ^T$ which is still far cheaper than computing the eigenvalues of $A$. Or what am I missing? $\endgroup$ – Wolfgang Bangerth Feb 15 '14 at 13:30

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.