$[[sqrt2/2,-sqrt2/2,0,0]][[2,1,0,0],[1,2,0,0],[0,0,2,-1],[0,0,-1,2]][[sqrt2/2],[-sqrt2/2],[0],[0]]=1$
$[[0,0,sqrt2/2,sqrt2/2]][[2,1,0,0],[1,2,0,0],[0,0,2,-1],[0,0,-1,2]][[0],[0],[sqrt2/2],[sqrt2/2]]=1$
$[[sqrt6/6,sqrt6/6,0,0]][[2,1,0,0],[1,2,0,0],[0,0,2,-1],[0,0,-1,2]][[sqrt6/6],[sqrt6/6],[0],[0]]=1$
$[[0,0,sqrt6/6,-sqrt6/6]][[2,1,0,0],[1,2,0,0],[0,0,2,-1],[0,0,-1,2]][[0],[0],[sqrt6/6],[-sqrt6/6]]=1$
una base ortonormale è quella del mio messaggio precedente:
$[[sqrt2/2],[-sqrt2/2],[0],[0]] ^^ [[0],[0],[sqrt2/2],[sqrt2/2]] ^^ [[sqrt6/6],[sqrt6/6],[0],[0]] ^^ [[0],[0],[sqrt6/6],[-sqrt6/6]]$
Infine, poichè:
$[[0,0,sqrt2/2,sqrt2/2]][[2,1,0,0],[1,2,0,0],[0,0,2,-1],[0,0,-1,2]][[0],[0],[sqrt2/2],[sqrt2/2]]=1$
$[[-sqrt2/2,sqrt2/2,0,0]][[2,1,0,0],[1,2,0,0],[0,0,2,-1],[0,0,-1,2]][[-sqrt2/2],[sqrt2/2],[0],[0]]=1$
$[[0,0,-sqrt2/2,sqrt2/2]][[2,1,0,0],[1,2,0,0],[0,0,2,-1],[0,0,-1,2]][[0],[0],[-sqrt2/2],[sqrt2/2]]=3$
$[[sqrt2/2,sqrt2/2,0,0]][[2,1,0,0],[1,2,0,0],[0,0,2,-1],[0,0,-1,2]][[sqrt2/2],[sqrt2/2],[0],[0]]=3$
la soluzione riportata dall'autore:
essendo manifestamente sbagliata, rischia di ingenerare gravi misconcezioni.
salvesalvino ha scritto:... in quanto ho 4 autovalori distinti ...
Veramente, gli autovalori sono 2:
$\lambda=1 ^^ \lambda=3$
di molteplicità algebrica 2.