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In the solution of problem 2.10(b) of Stephen Boyd & Lieven Vandenberghe's Convex Optimization, it is mentioned that if

$$g^Tv = 0, \qquad v^TAv \geq 0 \qquad \forall v$$

where $A$ is a positive semidefinite matrix and $g$ is a vector with real elements), then there must exist $\lambda$ such that $A+\lambda gg^T$ is positive semidefinite. How to obtain this?

Here is the solution image which I am talking about:


enter image description here


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    $\begingroup$ You need to provide more details. What is given here? $A$ and $g$? What about $v$? For all $v$? etc. $\endgroup$
    – max_zorn
    May 15 '18 at 3:34
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    $\begingroup$ If $g^T v=0$ for all $v$ then $g=0$. $\endgroup$
    – NicNic8
    May 15 '18 at 3:41
  • $\begingroup$ @NicNic8 no $g^Tv$ is not zero for all $v$ but $v^TAv\geq 0$ for all $v$ $\endgroup$ May 15 '18 at 3:44
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    $\begingroup$ @FrankMoses: If $v^T A v \geq 0$ for all $v$ then you can choose $\lambda =0$ (as $A$ is already semidefinite). $\endgroup$
    – Fabian
    May 15 '18 at 3:48
  • $\begingroup$ @Fabian can you please explain why I cannot chose any value I like because $g^Tv=0$? $\endgroup$ May 15 '18 at 3:52
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From what's given, $A$ is positive semidefinite. Therefore,

\begin{align} v^T (A + \lambda g g^T ) v &= v^T A v + \lambda v^T g g^T v \\ &= v^T A v + \lambda (v^T g)(g^Tv) \\ &= v^T A v + 0 \\ &= v^T A v \\ &\geq 0. \end{align}

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