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Well, I know the fact that $GL_n(\mathbb{R})$ is open set in $M_n(\mathbb{R})$, how to show that it is dense also? Well I thought like this: If $A\in M_n(\mathbb{R})$ and If $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $A$ and if I take $\lambda<\min\{|\lambda_1|,\dots,|\lambda_n|\}$ then the sequence $$B_n= A + \frac{\lambda}{n}I \rightarrow A$$ Where each $A + \frac{\lambda}{n}I$ is invertible matrix. Is my approach is correct? Is the set of all diagonalizable matrices also dense?

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Your approach is correct (in fact, you just have to take $A$ non-invertible).

The set of diagonalizable matrices over $\Bbb R$ is not dense, for example the matrix $\pmatrix{0&1\\ -1&0}$ is not diagonalizable, and by continuity of the roots of a polynomial, cannot be approached by diagonalizable matrices.

However, it's true when the field is $\Bbb C$. Take $A\in\mathcal M_n(\Bbb C)$, then after changing the basis if necessary, we can assume $A$ upper-triangular. Either all the eigenvalues of $A$ are distinct, and in this case $A$ is diagonalizable, or not. In this case, let $A_j$ the diagonal matrix $\operatorname{diag}(j^{-1},(2j)^{—1},\ldots,(nj)^{-1})$. Then the eigenvalues of $A+A_j$ are distinct by construction and $A_j\to 0$.

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you said my approach is correct but what if some $\lambda_i=0$ for some $i$? then what will be my $\lambda$? – Un Chien Andalou Jul 17 '13 at 21:44

Your approach is right, and you don't need to think about the diagonalizable matrices.

You simply observe that the eigenvalues of $A+\lambda I$ (counting multiplicities) are exactly the eigenvalues of $A$ plus $\lambda$. So, with your choice of $\lambda$ (or you don't even need $\lambda$, you just say "for $n$ big enough"), $A+\lambda I$ has all nonzero eigenvalues, and so it is invertible.

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