Just curious if my proof is sufficient, again(adsbygoogle = window.adsbygoogle || []).push({});

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Let [itex]A_{1},... ,A_{r}[/itex] be generators of a subspace V of [itex]R^n[/itex]. Let W be the set of all elements of [itex]R^n[/itex] which are perpendicular to [itex]A_{1},... ,A_{r}[/itex]. Show that the vectors of W are perpendicular to every element of V.

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We must show that [itex]v \cdot w = 0[/itex], where [itex]v \in V[/itex], and [itex]w \in W[/itex]

but since [itex]A_{1},... ,A_{r}[/itex] generate V, then v can be written as:

[tex]v = x_{1}A_{1} + ... + x_{r}A_{r}[/tex]

so, [tex]v \cdot w = (x_{1}A_{1} + ... + x_{r}A_{r}) \cdot w[/tex]

[tex]= (x_{1}A_{1}) \cdot w + ... + (x_{r}A_{r}) \cdot w[/tex]

[tex]= x_{1}(A_{1} \cdot w) + ... + x_{r}(A_{r} \cdot w)[/tex]

[tex]= x_{1}(0) + ... + x_{r}(0)[/tex]

[tex]= 0[/tex]

So, [tex]v \cdot w = 0[/tex], and every vector of W is perpendicular to every element of V.

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# Linear Algebra - Generators Proof

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