Showingthat $\mathbb{R}^2$ with operation $\langle (a,b), (c,d) \rangle = ac - bd$ is not an inner product space

Question

We define the "inner product" as $\langle (a,b),(c,d)\rangle = ac-bd$ on $R^2$

We want to verify if this is an inner product space. I'm saying it's not because it does not satisfy the positivity property of inner products, that is:

An inner product space $\langle x,x\rangle > 0$ if $x\not=0$.

I tried to prove this by letting $x=(a,b)=(c,d)=(5,5)$ so that we have the inner product $\langle x,x\rangle = \langle (5,5),(5,5)\rangle=5\times 5-5\times 5 = 0$ contradicting the positivity property of inner products.

Does this proof make sense?

Thank you!

Answer 1

Yes, your proof is correct.

Also, you might notice that $\langle (1,10), (1,10) \rangle < 0$, which also contradicts nonnegativity.

As an aside, if you rephrase your question as a question and write your proof as an answer, we could come along and upvote it to indicate that it is an answer. Writing yes, you're correct always feels like a somewhat unhelpful answer.

Answer 2

You are correct: this is not an inner product. Your argument is sufficient.

In fact, this is a particular instance of a metric which appears in General and Special Relativity. We say that $v$ is a spacelike vector if $\langle v,v \rangle > 0$ or $v=0$, $v$ is lightlike if $\langle v,v\rangle = 0$ but $v \neq 0$, and we say $v$ is timelike if $\langle v,v\rangle < 0$. People work in $4$ dimensions: $$\langle (x_1,y_1,z_1,t_1),(x_2,y_2,z_2,t_2)\rangle = x_1x_2+y_1y_2+z_1z_2-t_1t_2,$$ usually.

In this terminology, $(5,5)$ is a lightlike vector. There is a whole area worth studying about these bilinear forms.