# If an inner product is non-zero then both arguments are non-zero

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## Contents

## Statement

Let [ilmath]((X,[/ilmath][ilmath]\mathbb{K} [/ilmath][ilmath]),\langle\cdot,\cdot\rangle)[/ilmath] be an inner product space, then:

- [ilmath]\forall x,y\in X[\langle x,y\rangle\neq 0\implies(x\neq 0\wedge y\neq 0)][/ilmath]

*Warning:*The converse *does not hold*: take [ilmath](1,0),\ (0,1)\in\mathbb{R}^2[/ilmath] and equip [ilmath]\mathbb{R}^2[/ilmath] with the dot-product (writing [ilmath]a\cdot b[/ilmath] as [ilmath]\langle a,b\rangle[/ilmath] though) we see:

- [ilmath]\langle(1,0),(0,1)\rangle\eq 1\cdot 0+0\cdot 1\eq 0+0\eq 0[/ilmath], yet [ilmath](1,0)\neq 0[/ilmath] and [ilmath](0,1)\neq 0[/ilmath]. This demonstrates the [ilmath]\impliedby[/ilmath] direction cannot be.
- See:
*orthogonal vectors*. In an inner product space two vectors are orthogonal if their inner-product is [ilmath]0[/ilmath].

- See:

## Proof

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Easy proof

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