Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When calculating the inner product$^1$ of two complex vectors $u$ and $v$, why is the complex conjugate of $v$ used? Why not just compute the inner product as with real vectors?

1:Where the inner product of two vectors is defined as the summation of the product of corresponding elements.

share|cite|improve this question
You want the associated quadratic form to be real valued in order to impose the positive-definitess condition and so obtain what are called hermitian forms. Note that these forms are not bilinear but just sesquilinear. – Giuseppe May 26 '12 at 16:53
@Farhad: Great question. +1 – Babak S. May 26 '12 at 17:01
A requirement is $\langle z,z \rangle \geq 0$. For $z = i$, without cnjugates you get $\langle i, i \rangle = -1 < 0$. – TMM May 26 '12 at 19:50
@Babak Thank You! – user26649 May 26 '12 at 20:14
up vote 0 down vote accepted

A quick answer is that if $z$ is a complex vector, $$\sum_{k=1}^n z_k^2$$ will not be real. In this case we do not have $$\|z\|^2= \langle z, z\rangle.$$

share|cite|improve this answer
But wouldn't multiplying by the the original vector still provide a complex answer? For example if $v= e^{i\theta_1}$ and $u=e^{i\theta_2}$, $vu=e^{i(\theta_1 + \theta_2)}$, but multiplying $v\overline{u} = e^{i(\theta_1 - \theta_2)}$. Both products are still complex, so how does this change anything? – user26649 May 26 '12 at 17:12
$\langle z,z \rangle$ is supposed to be nonnegative for all vectors $z$. As ncmathsadist showed, if you do not use the complex cojnjugate, it is not. – Stefan Smith May 26 '12 at 19:10
Ah! I see, thanks! – user26649 May 26 '12 at 20:14
@ncmathsadist One last question: If vector $a$ is at $70$ degrees and vector $b$ is at $80$ degrees, the angle between them is $10$. However the complex conjugate of $b$ is at $80+180 = 260$ degrees. Therefore the angle between vector $a$ and the complex conjugate of be vector $b$ is $260-70 = 190$. Therefore won't using the dot product of the negative conjugate give you a "wrong" angle? (for example in this case $190$ instead of $10$). I've added 180 degrees to find the complex conjugate because $i$ determines the y axis and multiplying $i$ by $-1$ would rotate the vector 180 degrees. – user26649 May 28 '12 at 13:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.