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Diagonalizable Matrices.

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Today, during a lecture, we were posed the question whether D_n(K), the set of diagonalizable n \times n matrices over an algebraically closed field K, is Zariski-open, i.e. open in the Zariski topology. This would imply that in case K = \C, the set D_n(M) would be open and dense in M_n(K) = \R^{n \times n} in the standard (Euclidean) topolgy.

Unfortunately, the answer turns out to be “no” for the case K = \C (as well as K = \R):

Example.

Let n \ge 2. Consider the matrix

A := \Matrix{ 0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 0 } \in D_n(\C),

as well as the sequence

A_m := \Matrix{ 0 & 1/m & 0 & \cdots & 0 \\ \vdots & \ddots & 0 & \ddots & \vdots \\ \vdots & & \ddots & \ddots & 0 \\ \vdots & & & \ddots & 0 \\ 0 & \cdots & \cdots & \cdots & 0 } \in M_n(\C).

Clearly, \lim_{m\to\infty} A_m = A. Assume that D_n(\C) is open in M_n(\C); then we must have A_m \in D_n(\C) for almost all m \in \N. But m A_m is in Jordan canonical form, and clearly not diagonalizable; but this means that A_m \not\in D_n(\C) for all m \in \N. Therefore, D_n(\C) is not open in M_n(\C).

But nonetheless, D_n(K) contains a Zariski-open subset of M_n(K) in case K is algebraically closed (which implies that D_n(\C) lies dense in M_n(\C)). For that recall that \chi_A = \det(x E_n - A) \in K[x] is the characteristic polynomial of A \in M_n(K).

Proposition.

Consider the set

V_n(K) := \{ A \in M_n(K) \mid \chi_A \text{ is squarefree } \}.

Then V_n(K) \subseteq M_n(K) and V_n(K) is Zariski-open in M_n(K). In fact, V_n(K) is the complement of a hypersurface in M_n(K).

Proof.

Note that in case \chi_A is squarefree, \chi_A splits into distinct linear factors since K is algebraically closed. Hence, A has n distinct eigenvalues in K and therefore one obtains n linearly independent eigenvectors of A; i.e., A is diagonalizable over K. Therefore, V_n(K) \subseteq M_n(K).

Now we show that M_n(K) \setminus V_n(K) is a hypersurface in M_n(K), i.e. there exists a polynomial f \in K[x_{ij} \mid 1 \le i, j \le n] such that V_n(K) = \{ A \in M_n(K) \mid f(A) \neq 0 \}. For that, consider the maps f_0, \dots, f_{n-1} : M_n(K) \to K defined by \chi_A = x^n + \sum_{i=0}^{n-1} f_i(A) x^i for all A \in M_n(K). Obviously, these f_i must be polynomials. Next, consider the discriminant D(\chi_A) of \chi_A; this is a polynomial expression in the coefficients of \chi_A, i.e. in 1, f_0(A), \dots, f_{n-1}(A), whose value is zero if, and only if, \chi_A is squarefree. Therefore, A \in V_n(K) \Leftrightarrow D(\chi_A) \neq 0. Finally, f := D(\chi_A) is a polynomial, whence V_n(K) = \{ A \in M_n(K) \mid f(A) \neq 0 \} is Zariski-open in M_n(K).

Note that the situation is different over \R:

Proposition.

In the standard topology,

& \overline{D_n(\R)} = \overline{D_n(\R) \cap V_n(\R)} \\ {}={} & \{ A \in M_n(\R) \mid A \text{ has only real eigenvalues } \} \\ {}={} & \{ A \in M_n(\R) \mid A \text{ has a Jordan canonical form over } \R \}.
Proof.

Assume that A has at least one eigenvalue \lambda \in \C with imaginary part \Im \lambda \neq 0. If (A_m)_m is a sequence of matrices with \lim_{m\to\infty} A_m = A, each A_m must have an eigenvalue \lambda_m \in \C with \lim_{m\to\infty} \lambda_m = \lambda. But then, for infinitely many m, we must have \lambda_m \not\in \R (since \C \setminus \R is open), whence we cannot have A_m \in D_n(\R) for infinitely many m. Hence, A is not in the closure of D_n(\R) \cap V_n(\R).

Now assume that A has only real eigenvalues. Then there exist some T \in GL_n(\R) with T^{-1} A T in Jordan canonical form. By pertubing the diagonal elements of T^{-1} A T slightly, we can obtain a sequence of matrices B_m \in V_n(\R) \cap D_n(\R) with \lim_{m \to \infty} B_m \to T^{-1} A T. But then, \lim_{m\to\infty} T B_m T^{-1} = A and T B_m T^{-1} \in V_n(\R) \cap D_n(\R) for every m \in \N.

Note that this implies A \in \overline{V_n(\R) \cap D_n(\R)}; moreover, this also implies D_n(\R) \subseteq \overline{D_n(\R) \cap V_n(\R)}. Hence, the first two equalities hold. The third equality is standard.

Also note that V_n(\R) \not\subseteq D_n(\R) for n > 1, as the example

\Matrix{ 0 & 1 \\ -1 & 0 }

(which is diagonalizable over \C, with eigenvalues \pm i) shows. So what about \overline{V_n(\R)}? In fact, as in the case of K = \C, it turns out that \overline{V_n(\R)} is M_n(\R).

Proposition.

We have

\overline{V_n(\R)} = M_n(\R).

For the proof, we need a little lemma.

Lemma.

Let S := \{ f \in \R[x] \mid f \text{ is squarefree } \}. Then \overline{S} = \R[x].

Proof.

Let f \in \R[x] be an arbitrary polynomial. Write f = \lambda \prod_{i=1}^n p_i, where \lambda \in \R^* and p_i \in \R[x] is irreducible and monic, 1 \le i \le n. Now the coefficients of all p_i's (except the highest coefficients) are a finite set in \R of d := \sum_{i=1}^n \deg p_i elements, whence there exists sequences (a_1^{(m)}, \dots, a_d^{(m)}) with pairwise distinct a_1^{(m)}, \dots, a_d^{(m)} such that \lim a_i^{(m)} converges to one coefficent of one p_j. In particular, we can construct monic polynomials p_i^{(m)} \in \R[x] with \deg p_i^{(m)} = \deg p_i, \lim_{m\to\infty} p_i^{(m)} = p_i and p_i^{(m)} \neq p_j^{(m)} for every i \neq j. Even more, we can make sure that every p_i^{(m)} is irreducible; this enforces that f_m := \prod_{i=1}^n p_i^{(m)} is squarefree, i.e. f_m \in S. Therefore, we found a sequence in S converging to f, whence f \in \overline{S}.

Proof (Proof of the Proposition).

Let A \in M_n(\R) whose characteristic polynomial \chi_A can be written as \prod_{i=1}^t p_i, with not necessarily distinct, but monic and irreducible polynomials p_1, \dots, p_n \in \R[x]. There exists a matrix T \in GL_n(\R) with

T^{-1} A T = \Matrix{ C_{p_1} & & 0 \\ & \ddots & \\ 0 & & C_{p_t} },

where C_{p_i} is the companion matrix of p_i; this is a Frobenius normal form of A. Now we can find a sequence of squarefree polynomials p_i^{(m)} \in \R[x] such that for every m, p_1^{(m)}, \dots, p_t^{(m)} are pairwise coprime, and that \lim_{m\to\infty} p_i^{(m)} = p_i. Then set

A_m := T \Matrix{ C_{p_1^{(m)}} & & 0 \\ & \ddots & \\ 0 & & C_{p_t^{(m)}} } T^{-1} \in M_n(\R);

clearly, \lim_{m\to\infty} A_m = A. Moreover, the characteristic polynomial of A_m is given by \prod_{i=1}^t p_i^{(m)}, i.e. it is squarefree by choice of the p_i^{(m)}. Therefore, A_m \in V_n(\R) and A \in \overline{V_n(\R)}.

Categories: Algebraic Geometry Linear Algebra
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