TITLE: Why a non-diagonalizable matrix can be approximated by an infinite sequence of diagonalizable matrices?
QUESTION [13 upvotes]: It is known that any non-diagonalizable matrix, $A$, can be approximated by a set of diagonalizable matrices, e.g. $A \simeq \lim_{k \rightarrow \infty} A_k$. Why this is true?
Note: I was faced with it for the first time at a note about a simple proof for Cayley-Hamilton theorem, but I was not able to find it in my books nor in the internet to the extent I've googled.

REPLY [3 votes]: Let $B=[b_{i,j}]$ and $p((b_{i,j})_{i,j})$ be the discriminant of $\chi_B(x)=\det(B-xI)$. Since $p$ is not identically $0$, then $\{B|p(B)=0\}$ is Zariski-closed and  $U=\{B|p(B)\not= 0\}$ is Zariski-dense ($\star$). If $B\in U$, then $B$ has $n$ distinct complex eigenvalues and is diagonalizable over $\mathbb{C}$; according to ($\star$), every neighborhood of $A$ contains such a $B$ and we are done.
EDIT. @ tomasz , a proof as you want. Let $B_0\in U$ and let $D$ be the complex line that goes through $A,B_0$. $p_{|D}$ is not identically $0$ and consequently admits only a finite number of roots. Then there is a sequence $(B_i)_i$ in $D\cap U$ that converges to $A$. Moreover, the previous proof shows that $U$ is an arcwise connected set.