Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
2  
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

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer
    
took a while to understand your argument considering the fact that I am a novice to all that stuff!! :):) –  dineshdileep Oct 22 '12 at 12:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.