Given is a vector space $(\mathbb{R}^n,+, \cdot)$ over the field $(\mathbb{R},+, \cdot)$. For arbitrary elements $v,w \in \mathbb{R}^n$, we have the mapping $$\left \langle v,w \right \rangle= \text{max}(v_i \cdot w_i), \text{ where } 1\leq i\leq n$$

Does this mapping define a scalar product?

I think what I need to show that each of this condition is true

  1. $\left \langle x,y \right \rangle = \left \langle y,x \right \rangle \forall x,y \in \mathbb{R}^n$

  2. $\left \langle x_1+x_2,y \right \rangle=\left \langle x_1,y \right \rangle+\left \langle x_2,y \right \rangle \forall x_1,x_2,y \in \mathbb{R}^n$

  3. $\left \langle \lambda \cdot x,y \right \rangle= \lambda \cdot \left \langle x,y \right \rangle \forall x,y \in \mathbb{R}^n, \lambda \in \mathbb{R}$

  4. $\left \langle x,x \right \rangle \geq 0 \forall x \in \mathbb{R}^n$ and $\left \langle x,x \right \rangle=0$ if and only if $x=0$

But the problem is I don't even know what is meant by "max"? Maybe the limit is meant by that? If so, the limit could be either $1$ or $\infty$ because we have that $i \geq 1$? But no, $i$ just seems to be an index.. :s

  • 1
    $\begingroup$ Max$(v_i\cdot w_i)$ is just the largest number in the sequence $v_1\cdot w_1,v_2\cdot w_2,\ldots,v_n\cdot w_n$. $\endgroup$ – Lord Shark the Unknown Apr 21 '17 at 14:00

By max we mean the biggest number in the set $\{ x_1 \cdot y_1 , x_2 \cdot y_2 , \dots, x_n \cdot y_n \}$.

Here is an example. Take $n=3$, and $x = (2, -2, 1)^T , y=(-2,2,1)^T, z=(1,1,1)^T$.

Then you get $\langle x,z \rangle = \max\{2,-2,1\}=2, \langle y,z \rangle = \max\{-2,2,1\} = 2$, but

$4 = \langle x,z \rangle + \langle y,z \rangle \neq \langle x+y , z \rangle = \langle (0,0,2)^T , (1,1,1)^T\rangle = \max\{0,0,2\} = 2,$

and hence the second property is not satisfied.


$i$ is an index.

No, $$\langle v,w\rangle=\max(v_i\cdot w_i)$$ doesn't necessarily satisfy the second condition. E.g. $$\langle(1,0),(1,1)\rangle+\langle(0,1),(1,1)\rangle=1+1=2\neq 1=\langle(1,1),(1,1)\rangle.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.