Linear Algebra - Generators Proof

[mattmns]

Just curious if my proof is sufficient, again
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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.

 

[Hurkyl]

Use [ itex ] instead of [ tex ] inside of paragraphs.

What sort of thing is

(x1A1,...,xrAr)

?

 

[mattmns]

Use [ itex ] instead of [ tex ] inside of paragraphs.

What sort of thing is

(x1A1,...,xrAr)

?

Thanks, I copied what I wrote originally, and forgot to change the commas to plusses

 

[mattmns]

20 minutes later, and I finally got the latex correct