# Question on a linear algebra formula for **inner product**

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.

$$ = \sum a_i.\bar bi$$

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.

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

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.

• 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. Jul 19, 2021 at 12:38
• 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 Jul 20, 2021 at 5:08
• 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. Jul 20, 2021 at 8:35
• 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> Jul 20, 2021 at 15:13
• 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 Jul 20, 2021 at 15:51