I'm trying to solve a question (exam prep) which asks for Real eigenvalues of a matrix. My problem is that the Matrix is given over a field of Complex numbers (as shown on the picture) Matrix
Is there a special method for finding eigenvalues for matrices over a field of complex numbers? or can i use the normal method of finding Det(Matrix – IX)v =0 roots. If i use the normal method i get eigenvalue of -1.
Best Answer
The only why of finding eigenvalues is solving the characteristic polynomial, doesn't matter over what field your matrices are, they can be over anything. In your case is even easier because the matrix is effectively a real matrix, the fact that is over the complex field means that even if you have some complex eigenvalues you can still diagonalise the matrix. In anyway, you find the eigenvalues by the roots of the characteristic polynomial $$\det(A-\lambda\mathbb{I})=0$$ The problem asks you to take only the eigenvalues s.t. $\lambda\in\mathbb{R}$, so in this case, being the characteristic polynomial as follows $$-(\lambda+1)(\lambda^2+1)=0\implies\lambda_1 =-1\;\;\lambda_2=i\;\;\lambda_3=-i$$ the answer to your problem is $\lambda_1=-1$.
One way for a complex matrix to have all real eigenvalues is if the matrix is hermitian for the spectral theorem. See as a simple example the Puli matrices. But it's not the only way. As pointed out by user Lord Shark the Unknown even the matrix $$\left(\begin{matrix}1&i\\0&-1\end{matrix}\right)$$ has two real eigenvalues.