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I'm trying to prove that for a certain matrix $A$, and its conjugate transpose $A^*$, we have $⟨Ax,y⟩=⟨x,A^*y⟩$, where $⟨⟩$ represent the inner product. So here it's simply the dot product in $R^n$.

Does anyone have an idea of how to prove this?

Thank you!

tremblay
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1 Answers1

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The inner product (dot product on $\mathbb{R}^n$ or $\mathbb{C}^n$) is defined as $$ \langle x,y \rangle = x^* y = \sum_{k=0}^{n-1} \overline{x_k} y_k $$ Some definitions switch them around ($x^\intercal \overline{y}$) but they are essentially the same. $$ \langle Ax,y \rangle = (Ax)^* y = x^* A^* y = \langle x, A^* y \rangle $$ Since you are using $\mathbb{R}^n$, the conjugation can be dropped without any problem.

Henricus V.
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    Can you explain the $(Ax)^y =x^A^*y$ step? – MJP May 20 '20 at 19:40
  • how can $x^\intercal \overline{y}$ be the same as $x^* y$? I think $\overline{x^\intercal}y$ is the same as $x^*y$ by definition, but $\overline{x^\intercal}y$ is a complex conjugate of $x^\intercal \overline{y}$, so they are not the same, are they? – starriet 차주녕 Oct 17 '22 at 01:17