Samatix
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 Dec 12 awarded Caucus Aug 12 awarded Critic Jul 26 awarded Yearling Jul 26 answered $\forall v\in V:(Tv,v)=0\implies T^{\star}=-T$ Jul 26 revised True or false: If $||Tv+v ||=||Tv||+||v||$, then $1$ is eigenvalue of $T$ deleted 2 characters in body Jul 26 comment How can it be proved that the geometric mean function is concave? Did you try applying $ln()$ to the arithmetic mean function ? Jul 26 comment True or false: If $||Tv+v ||=||Tv||+||v||$, then $1$ is eigenvalue of $T$ What you have written are just implications and not equivalences. You can't say that $$||v+v ||=||v||+||v||$$ implies that $$||Tv+v ||=||Tv||+||v||$$ Jul 26 revised True or false: If $||Tv+v ||=||Tv||+||v||$, then $1$ is eigenvalue of $T$ added 133 characters in body Jul 26 answered True or false: If $||Tv+v ||=||Tv||+||v||$, then $1$ is eigenvalue of $T$ Jul 26 revised find the non constant polynomial so that P(x) δ_1' = δ_1' Added latex delimiters Jul 26 suggested approved edit on find the non constant polynomial so that P(x) δ_1' = δ_1' Jul 25 revised Duality and the Minimax Theorem added 335 characters in body Jul 25 revised Duality and the Minimax Theorem added 335 characters in body Jul 25 revised Duality and the Minimax Theorem added 335 characters in body Jul 25 revised Duality and the Minimax Theorem added 335 characters in body Jul 25 answered Duality and the Minimax Theorem Jul 25 comment Duality and the Minimax Theorem $v$ is not a constant. The variables of the problem are ($x_1$,$x_2$,...$x_{m_1}$,$v$). What's $A$? Jul 22 comment How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ @egreg, sorry I've written the answer before noticing that you added yours. Your answer was sufficient and complet Jul 22 revised How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ added 498 characters in body Jul 22 comment How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ Hey Simar, if you don't mind but $\sqrt(x^2) = |x|$ not $x$