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
1 Answers
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Take an eigenvector $v$ as a column vector with eigenvalue $\lambda$ and transpose (row vector) $v'.$ I am assuming that $L$ is a real number. We get $ 0 \geq v' (A - LI) v = v' (\lambda v - Lv) = (\lambda - L) (v' v). $ Now, $v'v$ is positive as it is the ordinary dot product of $v$ with itself, so $ 0 \geq \lambda - L, $ $ L \geq \lambda .$