# Why does the inner product of two vectors have to be positive definite?

I'm studying an undergraduate level geometry book and was studying about inner products when I got a bit confused. I've tried to find other answers here and elsewhere, but none of the answers were exactly intuitive and so it was hard for me to understand, and so decided to ask my own question.

According to the book, one of the properties of the inner product between two vectors is that it must be positive definite. To borrow the exact words:

An inner product on $$\Bbb{R}^n$$ is a function $$\langle\ \cdot\ ,\ \cdot\ \rangle: \Bbb{R}^n \times \Bbb{R}^n \rightarrow \Bbb{R}$$ on two vector variables that satisfies the following properties:

1. Positive definiteness: The necessary and sufficient condition for $$\langle\mathbf{a}, \mathbf{a} \rangle \ge 0$$ and $$\langle\mathbf{a}, \mathbf{a}\rangle = 0$$ is $$\mathbf{a} = \mathbf{0}$$.

2. Commutativity: $$\langle \mathbf{a}, \mathbf{b} \rangle = \langle \mathbf{b}, \mathbf{a} \rangle$$

3. Linear on the first argument: $$\langle \mathbf{a}_1 + \mathbf{a}_2, \mathbf{b} \rangle = \langle \mathbf{a}_1, \mathbf{b} \rangle + \langle \mathbf{a}_2, \mathbf{b} \rangle$$ and $$\langle \alpha \mathbf{a}, \mathbf{b} \rangle = \alpha \langle \mathbf{a}, \mathbf{b} \rangle = \langle \mathbf{a}, \alpha \mathbf{b} \rangle$$

I'm having trouble understanding the positive definiteness. Why is that so? What is the geometric meaning of an inner product having to be positive definite? In fact, I've never even heard of this before when I was studying linear algebra. I merely learned that the inner product of two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ is:

$$\mathbf{a} \cdot \mathbf{b} = \sum_{i = 1}^n a_ib_i$$

One Reddit answer brought up the concept of "distance" and that if an inner product is not positive definite then we cannot define distance between two vectors, but I'm having trouble understanding that as well.

Also, I thought that positive definiteness did not include equality (i.e. $$\ge$$) and rather positive semi-definiteness is the one that included equality.

Would anyone be able to shed some light on this concept? Thanks in advance.

• There are situations where a product is defined that does not require $\langle a, a\rangle \ge 0$ for all $a$ - some might have a negative product (you also have to loosen the restriction that $\langle a, a\rangle = 0$ if and only if $a = 0$). Special relativity is such a case. That said, it is not hard to show that an anti-inner product space (one where $\langle a,a\rangle \le 0$ for all $a$) has only one vector ($0$). – Cameron Williams Mar 19 '20 at 12:50

Yes, it is part of the definition of inner product that we always have $$\langle v,v\rangle\geqslant0$$. That's because that allows us to define a norm $$\lVert v\rVert=\sqrt{\langle v,v\rangle}$$ and from that norm we get a distance: the distance from $$v$$ to $$w$$ is $$\lVert v-w\rVert$$.

But I don't think I've ever seen “Positive definiteness” as a name for this property. It has nothing to do with positive definite matrices.

• If $A$ is a positive definite matrix, then $⟨x,y⟩:=x^TAy$ defines an inner product. – Dietrich Burde Mar 19 '20 at 12:38
• I have frequently seen the term positive definite applied to inner products to mean exactly this. – Matthew Leingang Mar 19 '20 at 12:38
• I have always seen this called positive definiteness. – shalop Mar 19 '20 at 13:03
• Thanks for the answer and comments everyone. Why isn't it called positive semi-definiteness? I'm bringing this point up because of the equality sign. – Seankala Mar 19 '20 at 13:03
• Because the inequality is strict unless v=0. – shalop Mar 19 '20 at 13:05

I merely learned that the inner product of two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ is:

$$\mathbf{a} \cdot \mathbf{b} = \sum_{i = 1}^n a_ib_i$$

This is the usual definition of inner product in $$\Bbb R^n$$. In more advanced classes, we learn that there are other possible definitions of an inner product on a vector space. But if we want to call $$\left$$ an inner product, it has to obey certain conditions, one of which is that $$\left\ge 0$$, with $$\left=0$$ if and only if $$x=0$$.

By the way, the definition of positive definiteness that you give in your question is garbled. It should be something like:

Positive definiteness: $$\langle\mathbf{a}, \mathbf{a} \rangle \ge 0$$ for all $$\mathbf{a}$$, and the necessary and sufficient condition for $$\langle\mathbf{a}, \mathbf{a}\rangle = 0$$ is $$\mathbf{a} = \mathbf{0}$$.

• Thanks! You're right, the only definition of the inner product in my head was the relatively elementary one. The definition is a rough translation from Korean, thanks for cleaning it up it's much clearer. So if I'm correct, we may also define our own inner products by sticking to these rules? – Seankala Mar 19 '20 at 13:10
• Yes, that's right. For instance, if $V$ is the vector space of continuous real-valued functions on $[0,1]$, we could define the inner product $\langle f,g\rangle=\int_0^1 f(x)g(x)\;dx$. – TonyK Mar 19 '20 at 13:15

In addition to Jose's answer, if $$A$$ is a positive definite matrix, then $$⟨x,y⟩:=x^TAy$$ defines an inner product. The norm and the distance induced by an inner product give a metric $$d(x,y)$$, which satisfies $$d(x,y)=0$$ if and only if $$x=y$$ because the inner product is "positive definite". And this property is very natural for any metric.