Let $M_{n\times n}$ be the space of $n\times n$ matrices (over algebraically closed field). Then I know that the diagonalizable matrices are dense in $M_{n\times n}$ (because $M_{n\times n}$ is irreducible wrt. Zariski topology and the diagonalizable matrices are the open subset defined by the nonvanishing of the discriminant). If $(M_{n\times n})^k$ consists of $k$-tuples of matrices, do the tuples of simultaneously diagonalizable matrices form a dense subset?
Are simultaneously diagonalizable matrices dense in $(M_{n\times n})^k$
general-topologymatrices
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I cannot really follow the reasoning you are hinting in your question, but here's my take:
To talk about density you need a topology. Since $M_n(\mathbb{C})$, the space of complex $n\times n$ matrices is finite-dimensional, a very natural notion of convergence is entry-wise; so we can consider the metric $$ d(A,B)=\max\{ |A_{kj}-B_{kj}|\ : k,j=1,\ldots,n\}, \ \ \ A,B\in M_n(\mathbb{C}). $$ It is not hard to check that for any matrix $C$, $$ d(CA,CB)\leq d(A,B)\,\sum_{k,j=1}^n |C_{kj}|, $$ and the same inequality holds for multiplication on the right (this will be used in the last inequality below).
Now take any $A\in M_n(\mathbb{C})$. Let $J$ be its Jordan canonical form; then there exists a non-singular matrix $S$ such that $J=SAS^{-1}$. Fix $\varepsilon>0$. Let $$ m=\left(\sum_{k,j=1}^n |S_{kj}|\right)\,\left(\sum_{k,j=1}^n |(S^{-1})_{kj}|\right) $$
Now, the matrix $J$ is upper triangular, so its eigenvalues (which are those of $A$) are the diagonal entries. Let $J'$ be the matrix obtained from $J$ by perturbing the diagonal entries of $J$ by less than $\varepsilon/m$ in such a way that all the diagonal entries of $J'$ are distinct.
But now $J'$ is diagonalizable, since it has $n$ distinct eigenvalues. And $d(J,J')<\varepsilon/m$. Then $S^{-1}J'S$ is diagonalizable and $$ d(S^{-1}J'S,A)=d(S^{-1}J'S,S^{-1}JS)\leq m\,d(J',J)<\varepsilon. $$
Zariski density means that any polynomial identity in the entries of an $n\times n$ matrix which holds on all invertible matrices holds on all matrices.
If we furthermore are considering matrices with entries in $\mathbb R$ of $\mathbb C$, then we can also say that any identity between continuous functions on the space of all $n\times n$ matrices which holds on all invertible matrices holds on all matrices.
The proof for polynomial functions is not hard:
We can work over any infinite field $k$, which we can take to be $\mathbb R$ of $\mathbb C$ if you like.
A polynomial in the entries of an $n\times n$ matrix is just a polynomial in $n^2$ variables.
Check that for any non-zero polynomial in $n^2$ variables, there is at least one matrix on which it doesn't vanish. (This is where we use that $k$ is infinite, and it ultimately reduces to the fact that polynomials in one variable have only finitely many zeroes.)
The determinant, which I'll denote $\Delta$, is a non-zero polynomial in $n^2$-variable.
Suppose that $f$ is a polynomial which vanishes on all invertible matrices. Then the product $f \Delta$ vanishes on all matrices. By the first point, it must be the zero polynomial. Since $\Delta$ is non-zero, we see that $f$ must be the zero polynomial. That is, $f$ vanishes on all matrices.
The proof for continuous functions is similar, but involves some topology as well algebra: you have to check that any non-empty open subset of $n\times n$ matrices contains an invertible matrix. This is standard, but may not be clear to you if you're not used to making arguments in topology or manifold theory.
Added: Actually, Georges's comment below gives a nice proof of the statement in the preceding comment. The same argument can also be found in Pete Clark's answer here. (This is an answer to the question linked to by Jonas Meyer a comment above.)
Best Answer
No, quite the opposite in fact, even for $k=2$.
Two diagonalizable matrices are simultaneously diagonalizable iff they commute. This is a relatively rare situation, not a common one: it is defined by the simultaneous vanishing of all the entries in the matrix $AB-BA$. Instead of a Zariski-open condition, you have a Zariski-closed one, defining a variety of a fairly small dimension.