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?

Rodrigo de Azevedo
  • 18,977
  • 5
  • 36
  • 95
  • 12,539
  • 1
  • 24
  • 46
  • 5
    A matrix can have a characteristic polynomial with a multiple root and still be diagonalizable (e.g. the identity matrix). – Dave Jun 20 '17 at 02:32
  • 5
    Right, but if the set of matrices with repeated eigenvalues has measure zero, then so does the set of non-diagonalizable matrices. – florence Jun 20 '17 at 02:33
  • 5
    Sure, having a repeated eigenvalue is necessary for non-diagonalizability. However, I was just noting that it's not sufficient. – Dave Jun 20 '17 at 02:37
  • 1
    see https://math.stackexchange.com/questions/207723/my-proof-of-the-set-of-diagonalizable-matrices-is-zariski-dense-in-m-n-mathbb and https://en.wikipedia.org/wiki/Discriminant – AnonymousCoward Jun 20 '17 at 02:44
  • 1
    Related: https://mathoverflow.net/questions/12657/proving-almost-all-matrices-over-c-are-diagonalizable – symplectomorphic Jun 20 '17 at 02:54
  • Hey OP, it wouldn't hurt to accept an answer by now. – AnonymousCoward Dec 28 '17 at 14:50

2 Answers2


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)
  • 5,162
  • 30
  • 39

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.

  • 8,246
  • 1
  • 24
  • 50