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..
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 $A$.