Here is the question I am trying to solve:
Show that the matrix \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 \end{pmatrix}
is diagonalizable over the field of complex numbers, but not over the field of real numbers.
I am wondering what does it mean in my calculations to show that a matrix is diagonalizable over the field of complex numbers, but not over the field of real numbers? does it only means that I have complex eigenvalues or what?
The linear transformation $\Bbb R^4\to\Bbb R^4$ which, given some basis, is represented by your matrix will not be diagonalizable. It does not have four eigenvalues from the scalar field (even counted with multiplicity), which means it doesn't yield the full basis of eigenvectors that a diagonalization requires.
The linear transformation $\Bbb C^4\to \Bbb C^4$ which, given some basis, is represented by your matrix will be diagonalizable. It has four distinct eigenvalues from the scalar field, each with a corresponding 1-dimensional eigenspace, which lets us create a full basis of eigenvectors.