How to prove that, the set of all matrices $M_n{\mathbb R}$ with distinct eigen values is dense in $\mathbb R^n$? How to prove that the set of all matrices $M_n{\mathbb R}$  with distinct eigen values is dense in $\mathbb R^n$?
Is there any geometric interpretations behind this.. if it is so, then tell me how to construct a dense subset of $\mathbb R^n$ geometrically.
 A: It suffices to note that for any $A \in \Bbb R^{n \times n}$, we may construct a sequence $A_m \to A$ where each $A_m$ has distinct eigenvalues.
We begin by noting that $A$ is block upper-triangularizable, so that $A = STS^{-1}$, with
$$
T = \pmatrix{T_1&&* \\&\ddots\\&&T_k}
$$
Where $T_i$ is either an element of $\Bbb R$ or a block of the form 
$$
T_i = \pmatrix{a&-b\\b&a}
$$
with $a,b \in \Bbb R$.  Now define
$$
T^{(m)} = \pmatrix{\left(1 - \frac 1m\right) T_1&&* \\&\ddots\\&& \left( 1 - \frac km\right) T_k}
$$
We note that $T^{(m)}$ has distinct eigenvalues for all but finitely many $m$ (i.e. for all $m$ bigger than some $M \in \Bbb N$).  Now, define the sequence by $A_m = ST^{(m)}S^{-1}$.
A: Consider a map $f$ which takes an appropriately sized set of eigenvalues paired with their respective eigenvectors and outputs the (unique) matrix having those eigenvalues and eigenvectors. This function is bijective and continuous almost everywhere (since each entry of the outputed matrix would be a rational function of the input) and, clearly, in the input domain the set of sets of eigenvalues with no repeats would be dense, and by continuity, the image of this set would be dense too.
