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I understand that given a linear operator $A$, its adjoint is define to be another linear operator $A^\ast$ such that $(Au,v)=(u,A^\ast v)$ $\forall\, u, v$ in the vector space. I'm wondering whether is that statement equivalent to $(Au,u)=(u, A^\ast u)$ $\forall\, u$?

Sirius Black
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    If the vector space is over $\mathbb C$, then this is true and can be proved by using the polarization formula. Note that in the real case $(Au,u) = (u,Au)$ holds for every linear $A$ because $(\cdot,\cdot)$ is symmetric. – Friedrich Philipp Mar 26 '16 at 03:08

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In the complex case, this is asking whether $(u,Bu)=(u,A^*u)$ for all $u$ implies $B=A^*$. That is true, as Friedrich Philipp commented, and one way to see this is linked here.

In the real case, having $(u,Bu)=(u,Cu)$ for all $u$ only implies that $B+B^*=C+C^*$. So for example, as Friedrich Philipp commented, you always have $(u,Au)=(u,A^* u)$ for all $u$ even when $A$ is not symmetric.

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Jonas Meyer
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