# Inner product of distance between two vectors

We can define a 'distance' between two points $$P = (x_1,y_1)$$ and $$Q=(x_2,y_2)$$ of the plane by $$d(P,Q) = |x_2 - x_1| + |y_2 - y_1|$$. Verify if the sentence below is a inner product in the plane. $$\langle(x_1,y_1),(x_2,y_2)\rangle = d(P,Q)$$

Positivity
$$\langle P,P\rangle \geq 0$$
The distance from a point to a point is 0. [check]

Symmetry
$$\langle P,Q \rangle = \langle Q,P \rangle$$
The distance from a point P to a point Q is equal to distance from a point Q to the point Q.[check]

Bilinearity
$$\langle\lambda P,Q\rangle = \lambda\langle P,Q\rangle \\ \langle(\lambda x_1,\lambda y_1),(x_2,y_2)\rangle = |x_2 - \lambda x_1| + | y_2 - \lambda y_1 |\\$$
I'm stuck here, i don't know how get out with this.

• What are the properties of an inner product? How would you verify them in this case? Nov 29 '18 at 20:43
• What have you tried? Show us your work and maybe we can help you. Nov 29 '18 at 20:54
• Bilinearity,symmetry and positivity. I've tried and got the symmetry and positivity, but the linearity (even making each linearity separated) i do not. Nov 29 '18 at 21:52
• Have you considered that it might not be an inner product? Try a few examples and see if bilinearity holds for those examples. Nov 29 '18 at 22:23
• Yeah, the answer from J.G. helped me. I did not try before because the answer from my professor was that is a inner product, but i was confusing how do this. Nov 29 '18 at 22:27

It's not an inner product because it isn't linear. For example, the choice $$P=Q=(1,\,0)$$ implies $$d(P,\,kQ)=|k-1|$$, which for $$k\ne 1$$ differs from $$kd(P,\,Q)=0$$.