I am currently studying linear algebra, specifically the orthogonality of a set of vectors. I learnt that the inner product of the 2 vectors in the field $\mathbb {C}$ can be expressed by the sum of multiplication of the one's counterpart with another's conjugated one.

$$ <u,v> = \sum a_i.\bar bi $$

This formula can apply to a set of $\mathbb {R}$ too. My question is why in the field $\mathbb{C}$ we need to use the conjugated vector instead of the normal one.

Thank you in advance

  • 3
    $\begingroup$ $a\overline a$ is real and non-negative but the same thing is not true for $a^{2}$. $\endgroup$ Jul 19, 2021 at 9:06
  • 1
    $\begingroup$ You do not want $(1,i)^T$ to be orthogonal to itself $\endgroup$ Jul 19, 2021 at 9:11

1 Answer 1


The reason why we use conjugated vectors in complex inner products is because there is a property we need verified with inner products : Hermitian symmetry.

Hermitian symmetry is defined as such (here, $E$ is an inner product space over $\mathbb{F}$):

For all $(x,y) \in E^2, \left\langle x,y \right\rangle = \overline{\left\langle y,x \right\rangle}$

This property makes $\left\langle x,y \right\rangle_? = \sum x_i y_i$ improper because it would be symmetrical. One could argue that it would be better for an inner product to be symmetrical, but Hermitian symmetry allows us to do this :

$ \left\langle x,x \right\rangle = \overline{\left\langle x,x \right\rangle}$

Therefore $ \left\langle x,x \right\rangle \in \mathbb{R} $

And on this consequence of Hermitian symmetry and another property of inner products relies the concept of canonical norm for an inner product, and on top of a norm, we can define the notion of distance between two vectors, of angle between two vectors, and then we can follow with the bases of topology on inner product spaces.

  • $\begingroup$ I see your point, but I feel that showing the reasoning through Hermitian symmetry also allows to see why the inner product is designed the way it is, especially when Hermitian symmetry isn't really intuitive. $\endgroup$ Jul 19, 2021 at 12:38
  • $\begingroup$ May you explain why supposing <x,y> = sum(x,y) symmetrical makes it imporer. I just dont know why we need to use conjugated vector for this formular in the first place. I mean, by using just normal <x,y> = sum(x,y) we can still ensure that <x,y> = <y,x>. I am so new to this field, so i am grateful to discover more $\endgroup$
    – John Pham
    Jul 20, 2021 at 5:08
  • $\begingroup$ We use the conjugated vector for two reasons : 1. Not using it would mean that, in $\mathbb{C}^2$, the vector $u = (i,1)$ would be orthogonal to itself : Indeed, $\left\langle u, u \right\rangle = 1 \times 1 + i \times i = 1 - 1 = 0$, and we don't want that (we want $\left\langle u, u \right\rangle = 0$ iff $ u = 0$) 2. Furthermore, the point of Hermitian symmetry (in complex spaces) is to ensure that, for any vector $x$ in a complex space $E$, we have $\left\langle u, u \right\rangle \in \mathbb{R}$, so we can derive a norm and all that jazz. $\endgroup$ Jul 20, 2021 at 8:35
  • $\begingroup$ I can see your point mentioning about a sole vector that does inner product with itself, which makes the rules <u,u> = 0 if and only if u = 0 improper. Excute me to give out 2 more questions: 1) Is there other reasons why we use <u,v> = u*v instead of u.v . 2) it seems to me that your answer was only about 1 vector, what about with two different vector <u,v> $\endgroup$
    – John Pham
    Jul 20, 2021 at 15:13
  • $\begingroup$ From what I have gathered, another benefit of that is that, for two vectors $u$ and $v$, and a complex $z$, we can have the following equality : $⟨zu,zv⟩ = z⟨u,zv⟩ = z\overline{z}⟨u,v⟩ = \left\vert z\right\vert^2⟨u,v⟩$, which is also something that happens in real spaces ($⟨\lambda u,\lambda v⟩ = \lambda ^2 ⟨u,v⟩$). The whole idea behind that is that we need all the useful little consequences that are true in real spaces to keep applying in complex spaces $\endgroup$ Jul 20, 2021 at 15:51

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