Set of all pair $(A,E)$ for which $\det(A-\lambda E)\ne 0$ as a polynomial, forms a dense open set in $M_n(\mathbb{C})\times M_n(\mathbb{C})$?

I proved $GL_n(\mathbb{C})$ is dense and open set in $M_n(\mathbb{C})$ like this: $A$ be any complex matrix, found its eigenvalues $\lambda_1,\dots,\lambda_n$ and choose $\lambda<\min\{|\lambda_1|,\dots,|\lambda_n|\}$ Then the sequence $A_n=A-\frac{\lambda}{n}I\to A$ and since $\frac{\lambda}{n}\ne\lambda_i$, each entries of the sequence $\{A_n\}$ is an invertible matrix, converging to an arbitrary matrix $A$ in the set. I was trying to mimic this proof in my currently needed problem but did not succed. Thanks for helping.

  • 2
    $\begingroup$ Let $p(\lambda) = \det(A-\lambda E)$. If $(A,E)\in GL_{n}(\mathbb{C})\times M_n(\mathbb{C})$, then $p(0) = \det A \not=0 \implies p$ is not zero as a polynomial. Since $GL_{n}(\mathbb{C})\times M_n(\mathbb{C})$ is dense in $M_n(\mathbb{C})^{2}$, you're done. $\endgroup$ – Vinícius Novelli Sep 8 '17 at 18:04
  • $\begingroup$ I don't understand your proof. $\endgroup$ – Marso Sep 9 '17 at 23:40

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.