Are most matrices diagonalizable? More precisely, does the set of non-diagonalizable (over $\mathbb C$) matrices have Lebesgue measure zero in $\mathbb R^{n\times n}$ or $\mathbb C^{n\times n}$? 
Intuitively, I would think yes, since in order for a matrix to be non-diagonalizable its characteristic polynomial would have to have a multiple root. But most monic polynomials of degree $n$ have distinct roots. Can this argument be formalized? 
 A: Let $A$ be a real matrix with a non-real eigenvalue. It's rather easy to see that if you perturb $A$ a little bit $A$ still will have a non-real eigenvalue. For instance if $A$ is a rotation matrix (as in Georges answer), applying a perturbed version of $A$ will still come close to rotating the vectors by a fixed angle so this perturbed version can't have any real eigenvalues.
A: Yes. Here is a proof over $\mathbb{C} $.


*

*Matrices with repeated eigenvalues are cut out as the zero locus of the discriminant of the characteristic polynomial, thus are algebraic sets. 

*Some matrices have unique eigenvalues, so this algebraic set is proper.

*Proper closed algebraic sets have measure $0.$ (intuitively, a proper closed algebraic set is a locally finite union of embedded submanifolds of lower dimension)

*(over $\mathbb{C} $) The set of matrices that aren't diagonalizable is contained in this set, so it also has measure $0$. (not over $\mathbb{R}$, see this comment https://math.stackexchange.com/a/207785/565)

