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My book, Linear Algebra Done Right, states that if we have:

$z=(z_1,...,z_n) \in \mathbb{C^n}$


$w=(w_1,...,w_n) \in \mathbb{C^n}$,

Then the inner product of $w$ with $z$ is: $w_1\overline{z_1}+...+w_n\overline{z_n}$.

Why is it that we must use the complex conjugate of $z$?


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up vote 3 down vote accepted

Because we require $\left\langle z,z\right\rangle \geq 0$ (in particular to be real); this doesn't make sense if $\left\langle z,z\right\rangle$ is not real (as we have no standard order on $\mathbf{C}$). For the same reason one defines the $L^2$ inner product (for complex valued continuous functions on $[a,b]$, say) by $\displaystyle\left\langle f,g\right\rangle:=\int_{[a,b]}f\bar{g}dx$.

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Great answer!!!!! – user123276 Mar 15 '14 at 11:34


  1. $\langle z,w\rangle=\overline{\langle w,z\rangle}$,

  2. $\langle z,z\rangle\ge 0$.

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