Difference between revisions of "Talk:Aufgaben:Problem 6"
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--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 22:36, 29 June 2015 (CEST) | --[[User:Brynerm|Brynerm]] ([[User talk: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 \(\rho(g)^{-1}\) 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. | + | 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. |
--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 09:15, 30 June 2015 (CEST) | --[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 09:15, 30 June 2015 (CEST) | ||
Revision as of 07:27, 30 June 2015
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.
