The following is my proof of the assertion that the set of diagonalizable matrices is Zariski-dense in $M_n(\mathbb F)$. Is this right ?

Let $\mathbb F$ be an infinite field (not necessarily algebraically closed)and $M_n(\mathbb F)$ the set of all $n \times n$ matrices with entries in $\mathbb F$.

We denote by $D_n(\mathbb F)$ the set of $n \times n$ diagonalizable matrices with entries in $\mathbb F$.

For each $A \in M_n(\mathbb F)$, we denote by $d(A)$ the discriminant of the characteristic polynomial of $A$.

Since $d(A)$ is a polynomial in the entries of $A$ with coefficients in $\mathbb F$, the set $U := \{ X \in M_n(\mathbb F) : d(X) \not = 0 \}$ is Zariski-open. (Here, we are identifying $M_n(\mathbb F)$ with ${\mathbb A}^{n^2}$.)

It follows from the fact that ${\mathbb A}^{n^2}$ is irreducible that $U$ is Zariski-dense in ${\mathbb A}^{n^2}$. As $U$ is contained in $D_n(\mathbb F)$, $D_n(\mathbb F)$ is also Zariski-dense in ${\mathbb A}^{n^2}$.

Thanks in advance.

  • 3
    $\begingroup$ Yes, your proof is right. $\endgroup$
    – Stephen
    Commented Oct 5, 2012 at 12:45
  • 2
    $\begingroup$ For completeness, you should also point out why $U$ is nonempty. (In an irreducible variety, every NONEMPTY Zariski open is Zariski dense.) Other than that, looks fine to me. $\endgroup$ Commented Oct 5, 2012 at 13:07
  • $\begingroup$ @DavidSpeyer, thank you for your comment. I think it is enough to point out that any diagonal matrix with distinct entries is an element of $U$. Is this correct ? $\endgroup$
    – Aki
    Commented Oct 5, 2012 at 13:45
  • $\begingroup$ @Steve, thank you for your super quick answer. Recently I came across the proof just for the case that $F$ is algebraically closed. So, I lost confidense in my proof. Is there something special in that case ? $\endgroup$
    – Aki
    Commented Oct 5, 2012 at 13:46
  • 1
    $\begingroup$ @Aki Yup, that's right. $\endgroup$ Commented Oct 5, 2012 at 13:56

1 Answer 1


Your proof is only correct if by "diagonalizable" you mean "diagonalizable" over an extension field of $\mathbb F$.
However, in my experience this is not the most usual interpretation of diagonalizable.
The rotation $\begin {pmatrix} 0&-1\\1&0\end {pmatrix}$ by $\pi/2$ in the plane over $\mathbb R$ for example is not diagonalizable over $\mathbb R$, even though its characteristic polynomial is $X^2+1$ has nonzero discriminant.
In your proof however it counts as diagonalizable, and that is the controversial point.

I have just checked that Hoffman-Kunze explicitly write on page 185 of their Linear Algebra that the above matrix is not diagonalizable.

  • $\begingroup$ Thank you for your answer. I've just looked at the page of Hoffman-Kunze. I think this is a crucial point for me, because I've been trying to understand a proof of "the ring of invariants for the conjugation action of $GL(V)$ on $End(V)$ is generated by the elementary symmetric polynomials of eigenvalues", where $V$ is a finite dimensional $\mathbb F$-vector space. In the proof,the lemma saying that every invariant function on $M_n(\mathbb F)$ is completely determined by its restriction to the set of diagonal matrices is used. To understand this, I attempted to apply the fact above. $\endgroup$
    – Aki
    Commented Oct 5, 2012 at 15:57

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