If a matrix is upper-triangular, does its diagonal contain its eigenvalues? If yes, how can this be proven? My textbook and teacher just jumped over this statement (we are working over complex numbers, does the answer change if it's over reals?) and I was wondering if someone could provide a proof.
The following steps lead to a solution:
1)If a matrix $A$ is upper triangular, prove that $A$ is invertible iff none of the elements on the diagonal equals zero.
Suppose you have a matrix $A$ that is upper triangular. Consider $A - \lambda I$. Then for $A$ to have a non-zero eigenvector, the kernel of $A - \lambda I$ must not be trivial, in other words $A - \lambda I$ must not be invertible.
2) Hence prove that the eigenvalues of a matrix that is upper triangular all lie on its diagonal.
Hint 1: The determinant of a triangular matrix is the product of its diagonal elements.
Hint 2: To prove Hint 1 develop with respect to the first row resp. column and use induction.