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Complex symmetric does not imply normal

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The taxonomy diagram indicates that complex symmetric matrices are normal. It is not the case, see: https://math.stackexchange.com/questions/3319576/complex-symmetric-matrices-are-normal for a counter-example. — Preceding unsigned comment added by Donvinzk (talkcontribs) 23:26, 10 August 2019 (UTC)[reply]

Thanks for noticing it. I've removed the corresponding arrow. Saung Tadashi (talk) 01:39, 30 August 2019 (UTC)[reply]
Thanks for the notification; I'll also change it in the original file and other version that are around on the web. Cheers–Jérôme (talk) 05:42, 26 May 2020 (UTC)[reply]
It's also updated in the source.–Jérôme (talk) 16:00, 2 June 2020 (UTC)[reply]

Layout

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I have renamed and moved some sections for clarification and for a more natural order. Nevertheless some entries are misplaced, sometimes with a wrong definition. For example Bézout matrix, Sylvester matrix and companion matrix must be in the same section, as they are all generated from polynomials. I will try to fix this in further edits. D.Lazard (talk) 11:07, 19 April 2021 (UTC)[reply]

Complex skew symmetric does not imply normal

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There is an additional issue in the taxonomy diagram, which is closely related to the one already posted above. Basically, complex skew symmetric matrices also need not be normal. For example, consider the following matrix

C = ((0, -i, -1-i),(i, 0 ,0), (1+i, 0, 0))

It is skew symmetric since C^T = -C (where ^T denotes the transpose). However, it is not normal, since C(C^H) does not give the same as (C^H)C (where ^H denotes the conjugate transpose).

Therefore, I think the arrow from skew symmetric to normal should be erased. Eseabrook (talk) 22:36, 10 February 2023 (UTC)[reply]