I'm just a beginner in Linear Algebra, and I've proved myself the following:
A matrix $A^{n \times n}$ is block diagonalizable if and only if the base field $F^n$ can be divided into at least two subspaces, whose direct sum is $F^n$ and each being $A$ invariant.
Here, I removed the trivial case where every matrix can be seen as a single block diagonalized matrix. Surprising to me, I've learned that if the base field $F$ is algebraically closed and the given matrix has at least two distinct eigenvalues, then the matrix is block diagonalizable, using the theory of generalized eigenspaces.
I have two questions:
It follows that every algebraically closed field can be divided into at least two subspaces whose direct sum is $F^n$ and each being $A$ invariant if $A$ has at least two eigenvalues. What if the base field is not closed? Is there any simple counter-example(i.e. the base field is not closed and the matrix isn't block diagonalizable even though it has at least two distinct eigenvalues)?
What if the given matrix in the closed field has only one eigenvalue? I've conjectured that that matrix is block diagonalizable if and only if $F^n$=$N(A-\lambda I)+Col(A-\lambda I)$ where $N(A)$ is the nullspace, $Col(A)$ is the column space, $\lambda$ being the eigenvalue, $+$ here means direct sum. I counln't find any counter-example of this hypothesis. Any other elegant if and only if conditions for single eigenvalued matrix?