Spectrum theorem for p-adic matrix analysis

Recently, I met a problem related to p-adic matrices in my research, the key of the problem can be summarized in the following way: 1: whether there exist spectrum theorem for p-adic matrix.\

2: whether invertible p-adic matrix is dense in $$C_{p}^{n\times n}$$

3:whether diagonalizable p-adic matrix is dense in $$C_{p}^{n \times n}$$

Any reference and comments will be appreciated.

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If you mean finite-dimensional matrices in part 1, then yes, because there is such a theorem over any field. For infinite-dimensional matrices, not sure what you want. In part 2, yes, take one entry whose adjoint is nontrivial and add a very small number to it. This works in any non-discrete topological field. –  Will Sawin Mar 4 '13 at 14:29
@Will Sawin thanks for your comments. –  yaoxiao Mar 4 '13 at 15:03

All three results are true.

Beginning with the third, it suffices to show that the set of $n\times n$-matrices in $\mathbb{C}_p$ whose characteristic polynomials have distinct roots is dense since these will certainly be diagonalizable.

Now the characteristic polynomial of a matrix may be described as a polynomial of degree $n$ whose coefficients are themselves polynomial in the entries of the matrix (using the usual definition of the characteristic polynomial of $A$ as the determinant of $tI-A$). It follows easily that the (formal) derivative of the characteristic polynomial also has coefficients that are polynomial in the entries of the matrix. Finally using the theory of resultants http://en.wikipedia.org/wiki/Resultant one may find a single polynomial $f$ in the matrix coefficients so that $f(a_{11},a_{12},\ldots,a_{nn})\neq 0$ for $A=(a_{ij})$ if and only if the characteristic polynomial of $A$ and its derivative have no roots in common. This last happens precisely if the characteristic polynomial has distinct roots.

To summarise the previous two paragraphs we have seen that there exists a single polynomial $f$ in $n^2$ variables such that a matrix $(a_{ij})$ is diagonalizable if and only if $f(a_{ij})\neq 0$. In fact, this works over any algebraically closed field since $f$ actually has coefficients in the integers.

It is now easy to see that $f^{-1}(\mathbb{C}_p\backslash 0)$ is an open dense subset of the $n\times n$ matrices with coefficients in $\mathbb{C}_p$ (it is even Zariski dense) as required.

The second follows easily from the third since any neighbourhood of a diagonal matrix will contain an invertible (and diagonal) matrix or from the comments.