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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

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    $\begingroup$ 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
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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.

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$i$ is an index.

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

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