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If yes in general, then yes for $L=I$, the identity. And also yes for symmetric $n-by-n$ n$-by-$nH$with$H^2=K$and the same eigenvectors as$K$. It seems to me that if finding the eigenvectors of a pair$(H,H^2)$turned out easier than finding the eigenvectors of$H^2=K$, then$H$would have to enjoy some special structure that a generic$K$doesn't enjoy, enjoy--- because one easily completes any given$H$to a pair$(H,H^2)$. That means first Thus having$K${\em given} doesn't help and the problem turns out as hard as finding eigenvectors of$H$. But nothing generally distinguishes operators that happen to turnout turn out equal to the square-root of given generic operators (with 1-dim eigenspaces and distinct eigenvalues). 1 If yes in general, then yes for$L=I$, the identity. And also yes for symmetric$n-by-nH$with$H^2=K$and the same eigenvectors as$K$. It seems to me that if finding the eigenvectors of a pair$(H,H^2)$turned out easier than finding the eigenvectors of$H^2=K$, then$H$would have to enjoy some special structure$K$doesn't enjoy, because one easily completes$H$to a pair$(H,H^2)$. That means first having$K$doesn't help and the problem turns out as hard as finding eigenvectors of$H\$. But nothing generally distinguishes operators that happen to turnout equal to the square-root of given generic operators (with 1-dim eigenspaces and distinct eigenvalues).