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Consider the following properties of the vector dot product:

Let $\vec{v},\vec{u}, \vec{w}$ $\in \mathbb{R}^n$ and let α $\in \mathbb{R}$. Then

  1. $\vec{u}$ · $\vec{v}$ =$\vec{v}$ · $\vec{u}$ (Commutativity)

  2. $\vec{u}$ · ($\vec{v}$ + $\vec{w}$) = $\vec{u}$ · $\vec{v}$ + $\vec{u}$ · $\vec{w}$ (Distributivity)

  3. α ∗ ($\vec{u}$ · $\vec{v}$)=(α ∗ $\vec{u}$) · $\vec{v}$ (Distributivity)

  4. $\vec{v}$ · $\vec{v}$ > 0 if $\vec{v}$ = 0 and $\vec{v}$ · $\vec{v}$ = 0 if $\vec{v}$ = 0

Using only the properties above (i.e. do not work with variable components of the vectors prove that $\vec{0}$ · $\vec{v}$ = 0 for any vector $\vec{v}$ ∈ $\mathbb{R}^n$.

What I tried on my own:

I considered a case where $\vec{v} = \vec{0}$ and used the second part of property four but got stuck when it came to a case where $\vec{v}$ is not $\vec{0}$.I tried to create another vector $\vec{0}$ and add it to the expression $\vec{0}$ · $\vec{v}$ and group them using the commutative property or add a random vector $\vec{u}$ and subtract it while grouping, but that didn't seem to work.

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Take $v \in \Bbb{R}^n$ and consider $v\cdot 0=v\cdot (0+0) = v \cdot 0 + v \cdot 0$

Which leads you to $v\cdot 0= v \cdot 0 + v \cdot 0$ and therefore you can conclude that $v \cdot 0 = 0$ by adding $-v\cdot 0$ on both sides. Since the $v\in \Bbb{R}^n$ you chose was arbitrary, it holds for any $v\in \Bbb{R}^n$

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  • $\begingroup$ Thank you; great reasoning. $\endgroup$ – Rloc Dec 5 '15 at 18:29

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