Hi guys.

I have the following matrix:

$\displaystyle \begin{bmatrix}1 & 1 & 0 & 0\\ -1 & -1 & 0 & 0\\ -2 & -2 & 2 & 1\\ 1 & 1 & -1 & 0\end{bmatrix}$

its characteristic polynomial is: $\displaystyle f(\lambda)=\lambda^2(\lambda-1)^2$

I've been asked if the above matrix has diagonal matrix over $\displaystyle \mathbb{R}$, I wrote "no, since the geometric multiplicity of the eigenvalue 1 is 1 (and thus smaller then the algebraic multiplicity).

and then, I've been asked if the above matrix has diagonal matrix over $\displaystyle \mathbb{C}$.

I'm guessing no, but I came here to ask, just to make sure I'm not missing anything here:

why does it make a difference if we're over $\displaystyle \mathbb{R}$ or $\displaystyle \mathbb{C}$ in this case?

Is it possible to get a different geometric multiplicity in different fields, for the same roots?