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

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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].

Proof

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References

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Dzung Minh Ha's book has it, can't remember where... see chapter on inner products, near the start, right hand page near the bottom in grey box Alec (talk) 23:40, 7 April 2017 (UTC)