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Let A be a real $n \times n$ matrix such that $\langle Ax,x\rangle \geq 0$ Prove that $Au=0$ iff $A^tu=0$.

Now if we have $Au=0$ for some $u=0$ then clearly $A^tu=0$.

Hence, $Au=0$ for some $u\neq 0$ then on a contrary if $A^tu\neq 0$ then from $\langle A^tu,u \rangle = \langle u,Au \rangle = 0$ do we get any contradiction? Please help from here.

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    $\begingroup$ consider writing \langle and \rangle instead of < and > to denote an inner product. $\endgroup$
    – Masacroso
    Apr 21 '18 at 17:38
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    $\begingroup$ One method here is to show that $\langle Ax,x \rangle = 0 \iff (A + A^T)x = 0$ (and notably, $A + A^T$ is symmetric). $\endgroup$ Apr 21 '18 at 17:51
  • $\begingroup$ It is not at all clear what you're trying to say with the statement "now if we have $Au=0$ for some $u=0$...". All I can make sense of here is that you have assumed (for the purpose of contradiction) that there exists a $u$ such that $Au = 0$ but $A^tu \neq 0$, which seems like a reasonable way to start. $\endgroup$ Apr 21 '18 at 17:53
  • $\begingroup$ Could you prove your hint? $\endgroup$
    – user547341
    Apr 21 '18 at 17:53
  • $\begingroup$ @Willi perhaps, but how long the proof needs to be depends on how much you know. Do you know what "positive semidefinite" means? Have you heard of "Rayleigh's theorem" for symmetric matrices (or perhaps the "min-max theorem")? Have you heard of the "spectral theorem"? $\endgroup$ Apr 21 '18 at 17:56
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Let $v \in \mathbb R^n$ :

For every $t \in \mathbb R$, $$0 \le \langle A(u + tv), (u+tv)\rangle = \langle A^t (u + tv) , (u + tv)\rangle = \langle A^tu , u\rangle + t\langle A^t v,u\rangle + t \langle A^tu,v\rangle + t^2 \langle A^tv, v\rangle = t\left(\langle A^tu,v\rangle + t \langle A^tv, v\rangle\right)$$ taking $t > 0$ then dividing by $t$ we have $$\langle A^tu,v\rangle + t \langle A^tv, v\rangle \ge 0$$ Now $t\to 0$, $$\langle A^t u ,v \rangle \ge 0$$ taking $t < 0$ then dividing by $t$ we have $$\langle A^tu,v\rangle + t \langle A^tv, v\rangle \le 0$$ Now $t\to 0$, $$\langle A^t u ,v \rangle \le 0$$ So $$\langle A^tu, v\rangle = 0$$ This is true for every $v$ then by taking $v = A^tu$ we have $$\left\|A^tu\right\|^2 = \langle A^t u, A^t u \rangle = 0$$

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  • $\begingroup$ Thanks. But $A$ is not symmetric so why such $B$ exists ? $\endgroup$
    – Youem
    Apr 21 '18 at 18:32

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