Diagonalizable Matrices. « Felix' Math Place

Diagonalizable Matrices.

Abstract.
We consider the property of an n times n matrix of being diagonalizable. Is this property open in the standard topology, or the Zariski topology? The emphasis lies on the real and complex numbers, as well as on arbitrary algebraically closed fields.

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 $\displaystyle A := \Matrix{ 0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 0 } \in D_n(\C),$ as well as the sequence $\displaystyle A_m := \Matrix{ 0 & 1/m & 0 & \cdots & 0 \\ \vdots & \ddots & 0 & \ddots & \vdots \\ \vdots & & \ddots & \ddots & 0 \\ \vdots & & & \ddots & 0 \\ 0 & \cdots & \cdots & \cdots & 0 } \in M_n(\C).$ 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 $Formula does not parse: m \in \IN$ . 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 $Formula does not parse: m \in \IN$ . 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 $\displaystyle V_n(K) := \{ A \in M_n(K) \mid \chi_A \text{ is squarefree } \}.$ 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 $M_n(K)$ .
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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.
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Also note that $V_n(\R) \not\subseteq D_n(\R)$ for $n > 1$ , as the example $\displaystyle \Matrix{ 0 & 1 \\ -1 & 0 }$ (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 $\displaystyle \overline{V_n(\R)} = M_n(\R).$

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 $f \in \overline{S}$ .
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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 $\displaystyle T^{-1} A T = \Matrix{ C_{p_1} & & 0 \\ & \ddots & \\ 0 & & C_{p_t} },$ 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 $\displaystyle A_m := T \Matrix{ C_{p_1^{(m)}} & & 0 \\ & \ddots & \\ 0 & & C_{p_t^{(m)}} } T^{-1} \in M_n(\R);$ 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 $A \in \overline{V_n(\R)}$ .
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Tags: diagonalizable matrices, topological argument, Zariski topology

This entry was posted on friday, january 29th, 2010 at 04:47 utc and is filed under Algebraic Geometry, Linear Algebra. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Diagonalizable Matrices.

Abstract.

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