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. $$\langle(x_1,y_1),(x_2,y_2)\rangle = d(P,Q)$$

UPDATE: What i've already made

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

$ \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]

$ \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.

  • 1
    $\begingroup$ What are the properties of an inner product? How would you verify them in this case? $\endgroup$ Nov 29, 2018 at 20:43
  • $\begingroup$ What have you tried? Show us your work and maybe we can help you. $\endgroup$ Nov 29, 2018 at 20:54
  • $\begingroup$ Bilinearity,symmetry and positivity. I've tried and got the symmetry and positivity, but the linearity (even making each linearity separated) i do not. $\endgroup$ Nov 29, 2018 at 21:52
  • $\begingroup$ Have you considered that it might not be an inner product? Try a few examples and see if bilinearity holds for those examples. $\endgroup$
    – Kevin Long
    Nov 29, 2018 at 22:23
  • $\begingroup$ 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. $\endgroup$ Nov 29, 2018 at 22:27

1 Answer 1


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 $kd(P,\,Q)=0$.


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