How can I show that each smooth conic in $\Bbb P _2\Bbb C $ is isomorph to the conic given by the equation $x^2+y^2+z^2=0$? I thought about considering the quadratic form $A$ associated to my conic, A is symmetric so I can say that $D=P^{-1}AP$ where D is diagonal with coefficients $d_1, d_2, d_3$. Know I probably have to define a maps from $\Bbb P _2\Bbb C $ to $\Bbb P _2\Bbb C $ using $P$, but I can't understand which is the correct way to do that! Can anyone help me? Thanks in advance!
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See as well here – Jean Marie Mar 20 '21 at 18:55
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@JeanMarie isn't it possible to go on from my reasoning? Here is quite different... – MJane Mar 20 '21 at 21:21
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A small suggestion. Go the other way. Start with the conic $,x^2+y^2+z^2=0,$ and use any invertable linear map on $,\mathbb{C}^3,$ (which also induces a map $,P_2\mathbb{C}\to P_2\mathbb{C}$) and apply it to the conic. – Somos Mar 20 '21 at 22:35