Talk:Aufgaben:Problem 6
In my view, the statement \(Ker(A) \neq 0\) must be a typo in the textbook. I corrected it to \(Ker(A - \lambda 1)\neq 0\)
--Brynerm (talk) 22:36, 29 June 2015 (CEST)
The "alternative" solution has one typo in b) as \(\rho(g)\) is meant to be unitary (not self-adjoint), therefore there must be an \(\rho(g)^{-1}=\rho(g^{-1})\) in the equation with the scalar product. Apart from that, it's really all the same but more detailed and therefore it might be easier to follow.
--Brynerm (talk) 09:15, 30 June 2015 (CEST)
Thank you so much for noticing. I'll fix it tomorrow morning. I'll also add a source file, so everyone can edit it. Actually, that's a major mistake which could cost many points at the exam.
--Lilit (talk) 12:43, 30 June 2015 (CEST)
I think in part b) (short solution) one has to write that U is a \(\textbf{proper}\) non trivial invariant subspace.
Beni (talk) 16:31, 2 August 2015 (CEST)
What do you mean by proper subspace? I think "proper" only applies to sets, when talking about spaces "trivial" is used to say that the subspace not just the zero element and also not hole space.
Carl (talk) 08:22, 3 August 2015 (CEST)
Yes, that was exactly the point. But at least \(V\) is a vector space, so proper would make it unambiguous. But Wikipedia confirms you that it means also the whole space. I probably prefer to make it clear.
Beni (talk) 09:08, 3 August 2015 (CEST)
I think it should be fine as it is.
