2
$\begingroup$

Question is to prove that the set of all diagonalizable matrices are dense in $M_n(\mathbb{C})$.

I am sure this question is discussed in this site previously but i am looking for a more constructive proof..

The very first example of an element in $M_2(\mathbb{C})$ that is not diagonalizable is $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$..

Now, I want a sequence of matrices $A_n$ converging to $A$ such that all $A_n$ are diagonalizable...

I see matrix convergence as convergence of each element...

So, atleast we need a sequence converging to $0$.. The very first sequence that we come across is $\frac{1}{n}$

So, i was considering $A_n=\begin{bmatrix}1&1\\\frac{1}{n}&1\end{bmatrix}$ converging to $A$..

Incidentally all these $A_n$ are diagonalizable with eigenvalues $\frac{n-\sqrt{n}}{n}$ and $\frac{n+\sqrt{n}}{n}$..

$A_n$ for me was just a random choice..

I just want to know if some thing like this works in case of $n\times n$ matrices..

$\endgroup$
4
  • $\begingroup$ Hint: If you replace the bottom right with any number different from $1$, it will be diagonalizable. $\endgroup$ Commented Mar 4, 2016 at 9:33
  • $\begingroup$ @TobiasKildetoft For general n matrix?/ i have checked it using $\frac{1}{n}$ in place of zeros and ended up having complicated eigenvalues $\endgroup$ Commented Mar 4, 2016 at 9:37
  • $\begingroup$ For the specific matrix in question. The idea is that having distinct eigenvalues ensured it being diagonalizable. $\endgroup$ Commented Mar 4, 2016 at 9:38
  • $\begingroup$ @TobiasKildetoft Ok... $\endgroup$ Commented Mar 4, 2016 at 9:42

1 Answer 1

Reset to default
0
$\begingroup$

The denseness follows easily from two properties: eigenvalues depend continuously on the entries of a matrix and matrices with distinct eigenvalues are diagonalizable.

A nondiagonalizable matrix $A$ is conjugate to a Jordan canonical form $J=S^{-1}AS$. Simply change slightly the entries of the main diagonal of $J$ to obtain a matrix $J'$ and consider $SJ'S^{-1}$. You have just obtained a diagonalizable matrix that can be made arbitrarily close to $A$.

$\endgroup$
6
  • $\begingroup$ The proof in the second paragraph is correct. The first paragraph isn't quite correct; the eigenvalues could theoretically depend continuously on the entries in such a way that they remain non-distinct in a neighbourhood of some matrix. $\endgroup$ Commented Mar 4, 2016 at 10:04
  • $\begingroup$ That's not very helpful. I did read the argument carefully. I can't give an example because I'm concerned with a logical possibility that, as we agree, isn't in fact realized. Perhaps you could point out what's wrong about my argument? $\endgroup$ Commented Mar 4, 2016 at 10:06
  • 1
    $\begingroup$ OK, I'll try to flesh it out. Your first paragraph states that the denseness follows from the continuous dependence of the eigenvalues on the entries and the diagonalizability of matrices with distinct eigenvalues. I don't think it does. As an extreme case, all eigenvalues of all matrices could be the same, and no matrices could be diagonalizable. Then your two premises would hold, but diagonalizable matrices wouldn't be dense. $\endgroup$ Commented Mar 4, 2016 at 10:11
  • $\begingroup$ Could we please take this away from the ad hominem level and exchange ideas about the subject matter? Where does my argument go wrong? $\endgroup$ Commented Mar 4, 2016 at 10:13
  • $\begingroup$ I did. You replied with ad hominem attacks instead of engaging with my arguments. $\endgroup$ Commented Mar 4, 2016 at 10:36

You must log in to answer this question.