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.
$N \times N$ matrices with distinct eigen-values
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linear-algebra
matrices
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0@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..... – 2012-10-22
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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.
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0took a while to understand your argument considering the fact that I am a novice to all that stuff!! :):) – 2012-10-22