Let $A$ be the Hessian of some function $f,$ and $$A - LI \preceq 0.$$ (i.e. $A - LI$ negative semi-definite.) Does this mean that the largest eigenvalue of $A$ is upper-bounded by $L$? Why?
If $A - LI \preceq 0$ does the largest eigenvalue of $A$ upper bounded by $L$?
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linear-algebra
eigenvalues-eigenvectors
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0negative semidefinite@J.D. – 2012-08-03