If yes in general, then yes for $L=I$, the identity. And also yes for symmetric $n-by-n$ n$-by-$n$ $H$ 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).
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If yes in general, then yes for $L=I$, the identity. And also yes for symmetric $n-by-n$ $H$ 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). |
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