$N \times N$ matrices with distinct eigen-values

I am using Rajendra Bhatia's "Matrix Analysis" for a self-study. I came across this problem where he asks to prove "Set of all $N \times N$ matrices with distinct eigen values is dense in the space of $N \times N$ matrices". I am not able to prove it. Any help or solution would be appreciated.

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Do you know what the Jordan Canonical form is? –  N. S. Oct 22 '12 at 5:47
Not much on it. But does learning it help to solve this problem? –  dineshdileep Oct 22 '12 at 6:12
a matrix with distinct eigenvalues is diagonalizable hence proving the above statement is same as proving "diagonalizable matrices with complex values are dense in set of n×n complex matrices" which you can find here math.stackexchange.com/questions/107945/… –  jim Oct 22 '12 at 7:05
@jim that helped!! –  dineshdileep Oct 22 '12 at 12:25
@dineshdileep Yes. Any matrix can be written as $PJP^{-1}$ where $J$ is a very special upper triangular matrix. Then, it is easy to find a diagonal matrix $D$, so that $J+ \epsilon D$ has distinct diagonal entries, and since is upper triangular, it has distinct eigenvalues..... –  N. S. Oct 22 '12 at 15:06

Let $A$ be a matrix. Whether or not $A$ has distinct eigenvalues can be checked by the discriminant $D$ of the characteristic polynomial $\chi_A(X)$. All in all, the map $A\mapsto D$ is polynomial in each matrix entry $a_{i,j}$. If this $n^2$-variate polynomial were identical to $0$ on an open neighbourhood of $A$, then it would be the zero polynomial and identically $0$, i.e. there would not even exist matrices with $n$ distinct eigenvalues - contradiction.