subtle confusion about the definition of Jordan normal form

Question

When I tried to comprehend the definition of Jordan normal form, I noticed that the Field should contains all eigenvalues of the linear operator. I feel confused that if eigenvalue of linear transformation is defined in the scalar-field of the vector space, then the requirement have to be self-evident. I know my consideration is definitely wrong, but what is the true meaning of "all eigenvalue" here ?

Answer 1

For example if you consider the field $\mathbb{R}$, there are many operators which do not have real eigenvalues (for example rotations). Thus we cannot form a real Jordan normal form, since on the diagonal of the JNF, the eigenvalues of the operator in question occur. But if all the eigenvalues are real, you can of course calculate the JNF. I think, the theorem I have in mind states, that the field has to be algebraically closed, then you do not run into such problems.

Answer 2

"All eigenvalues" here means all eigenvalues of the operator considered over the algebraic closure of the field (not just over the original field). In other words, all the roots of the characteristic polynomial should lie in the field.

Put another way: Starting from a matrix $A$ over any field $F$, in order to put $A$ in Jordan normal form, you need to extend $F$ to contain all the roots of the characteristic polynomial of $A$.