What does algebraically closed field play the role in Schur's unitary triangulation theorem?

Question

Schur's unitary triangulation theorem said that

Theorem (Schur’s Triangularization Theorem) Every square complex matrix A is unitarily similar to an upper-triangular matrix, i.e., there exists a unitary matrix $U$ such that $T = U^∗AU$ is triangular.

Here, the theorem is about square complex matrix, but in the lecture, professor mentioned that any algebraically closed field matrices can fit the theorem [with similarity instead of unitary similarity] (if there are mistake on the expression, it must be me misunderstanding).

My questions is, why algebraically closed field can work, how to connect the proof with the original square complex matrix proof?

Answer 1

suppose you are working over some algebraically closed field $\mathbb K$. Just mimic the Schur Triangularization process, but at each stage when you complete to a basis (invertible matrix) you ignore the requirement for orthonormal columns. The end result is
$A=STS^{-1}$
(Exactly this is walked through step by step in chapter 4 of Artin's Algebra.)

how to connect the proof with the original square complex matrix proof?

once you have the above, if you are working over $\mathbb C$, you can run QR factorization,
$S=QR\implies A=QRTR^{-1}Q^{-1}=Q\big(RTR^{-1}\big)Q^*$
which recovers standard Schur Triangularization-- recalling that upper triangular matrices form a (semi)group.