My linear algebra professor proved that
Diagonalizable matrices with complex values are dense in set of $n \times n$ complex matrices.
He defined a metric (I believe) that was somehow related to the usual metric on $\mathbb{R}^{n^2}$.
Then somehow proved that diagonalizable matrices were dense because for any matrix $A$ if $\det(A – \lambda I) = 0$ on an open subset, then $\det(A – \lambda I)$ was the zero polynomial.
I Googled around a bit and found some stuff talking about the Zariski topology, and I am not sure this is what I want. Most of what I have found on this topology is much more general than what he was doing.
Does anyone know any resources that would explain how to go about proving this statement?
Edit: I found out how to prove this the way my professor did.
We define the norm of a matrix by
$$ |A| = \max\{|Ax| : |x| = 1 \}.$$
Now we have a distance $d(A, B) < \epsilon$. You can prove that if $(A – B)_{ij} < \epsilon/n$, then $|A – B| < \epsilon$.
Let $p(x)$ be the characteristic polynomial of $A$, an $n \times n$ matrix.
$$p(x) = \prod_1^n (x – x_i) = x^n + (-1)\sigma_1 x^{n – 1} + \cdots + (-1)^n\sigma_n$$
where $\sigma_1, \ldots, \sigma_n$ are the elementary symmetric polynomials of $x_1, \ldots, x_n$ which are the eigenvalues.
The discriminant of the characteristic polynomial is a symmetric polynomial, therefore it can be written in terms of the elementary symmetric polynomials, which in turn can be written in terms of the entries of the matrix.
But since the discriminant is a polynomial, it only has finitely many roots. Therefore for $\epsilon > 0$, by changing the entries of the matrix less than $\epsilon / n$ we can find a new matrix $B$ such that $|B – A| < \epsilon$ and the discriminant is not zero.
The discriminant being not zero means $B$ has distinct eigenvalues, thus has a basis of eigenvectors. Therefore, it is diagonalizable.
Thus the set of diagonalizable matrices is dense in the set of matrices with respect to that metric.
Best Answer
I cannot really follow the reasoning you are hinting in your question, but here's my take:
To talk about density you need a topology. Since $M_n(\mathbb{C})$, the space of complex $n\times n$ matrices is finite-dimensional, a very natural notion of convergence is entry-wise; so we can consider the metric $$ d(A,B)=\max\{ |A_{kj}-B_{kj}|\ : k,j=1,\ldots,n\}, \ \ \ A,B\in M_n(\mathbb{C}). $$ It is not hard to check that for any matrix $C$, $$ d(CA,CB)\leq d(A,B)\,\sum_{k,j=1}^n |C_{kj}|, $$ and the same inequality holds for multiplication on the right (this will be used in the last inequality below).
Now take any $A\in M_n(\mathbb{C})$. Let $J$ be its Jordan canonical form; then there exists a non-singular matrix $S$ such that $J=SAS^{-1}$. Fix $\varepsilon>0$. Let $$ m=\left(\sum_{k,j=1}^n |S_{kj}|\right)\,\left(\sum_{k,j=1}^n |(S^{-1})_{kj}|\right) $$
Now, the matrix $J$ is upper triangular, so its eigenvalues (which are those of $A$) are the diagonal entries. Let $J'$ be the matrix obtained from $J$ by perturbing the diagonal entries of $J$ by less than $\varepsilon/m$ in such a way that all the diagonal entries of $J'$ are distinct.
But now $J'$ is diagonalizable, since it has $n$ distinct eigenvalues. And $d(J,J')<\varepsilon/m$. Then $S^{-1}J'S$ is diagonalizable and $$ d(S^{-1}J'S,A)=d(S^{-1}J'S,S^{-1}JS)\leq m\,d(J',J)<\varepsilon. $$