# Talk:Nilpotent matrix

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Field:  Algebra

## Algebraically closed?

In the article it is mentioned that `The same theorem holds true for matrices over any algebraically closed field.' The theorem about which this is mentioned, is one that already holds over the reals, though, which of course aren't algebraically closed.

The reason why you need algebraic closure, it seems to me, is the fact that you're looking at eigenvalues of the matrix. If you take a matrix over an arbitrary field and consider the eigenvalues in its algebraic closure (for matrices over the real numbers you're also looking at complex eigenvalues, so this shouldn't cause any problems), I'd say the theorem holds for all fields. This deserves some attention, but the current version of the theorem seems to require to strong conditions at the moment. 21:07, 3 August 2009 (UTC)

I agree, I think Algebraic Closure is too strong. It is true all eigenvalues (whether in the algebraic closure, or not) will be zero. But this just means that THE eigenvalue is in the base field, regardless of whether or not the field happens to be algebraically closed. 06:13, 5 December 2009 (UTC)Eulerianpath (talk)

It just not true! Take for example the algebraic closure of ${\displaystyle \mathbb {F} _{2}}$ and matrix ${\displaystyle A={\begin{bmatrix}1&0\\0&1\end{bmatrix}}}$. It should state any field with characteristic 0. —Preceding unsigned comment added by 131.174.41.58 (talk) 15:52, 3 November 2010 (UTC)