Talk:Nilpotent matrix

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. — Preceding unsigned comment added by HSNie (talk • contribs) 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. Eulerianpath (talk) 06:13, 5 December 2009 (UTC)


 * It just not true! Take for example the algebraic closure of $$\mathbb{F}_2$$ and matrix $$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$. It should state any field with characteristic 0. —Preceding unsigned comment added by 131.174.41.58 (talk) 15:52, 3 November 2010 (UTC)

False characterization
The characterization of nilpotent matrices in the article is false.

N nilpotent implies the spectrum is {0}. However, the converse is generally false which can be seen by looking at the the following matrix over R

0 0 0

0 0 1

0 -1 0.

It would be sufficient to further assume, that N is triangularizable (e.g. use Cayley-Hamilton), which is always the case in C (that is probably what the author was thinking off). — Preceding unsigned comment added by 2001:638:1558:F73A:D883:A216:F19:FCA0 (talk) 13:18, 11 May 2018 (UTC)

"Nilpotent endomorphism" listed at Redirects for discussion
A discussion is taking place to address the redirect Nilpotent endomorphism. The discussion will occur at Redirects for discussion/Log/2021 April 11 until a consensus is reached, and readers of this page are welcome to contribute to the discussion. 𝟙𝟤𝟯𝟺𝐪𝑤𝒆𝓇𝟷𝟮𝟥𝟜𝓺𝔴𝕖𝖗𝟰 (𝗍𝗮𝘭𝙠) 15:50, 11 April 2021 (UTC)

In Nilpotent every eigen values of matrix are zero
Am I right 2402:E280:230A:4F5:7544:1B28:E108:23E8 (talk) 05:21, 15 March 2022 (UTC)


 * Yes. This is clearly stated in the Characterization section.—Anita5192 (talk) 06:10, 15 March 2022 (UTC)