Linear Algebra - Generators Proof

[mattmns]

Just curious if my proof is sufficient, again
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Let $A_{1},... ,A_{r}$ be generators of a subspace V of $R^n$. Let W be the set of all elements of $R^n$ which are perpendicular to $A_{1},... ,A_{r}$. Show that the vectors of W are perpendicular to every element of V.
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We must show that $v \cdot w = 0$, where $v \in V$, and $w \in W$
but since $A_{1},... ,A_{r}$ generate V, then v can be written as:
$$v = x_{1}A_{1} + ... + x_{r}A_{r}$$

so, $$v \cdot w = (x_{1}A_{1} + ... + x_{r}A_{r}) \cdot w$$
$$= (x_{1}A_{1}) \cdot w + ... + (x_{r}A_{r}) \cdot w$$
$$= x_{1}(A_{1} \cdot w) + ... + x_{r}(A_{r} \cdot w)$$
$$= x_{1}(0) + ... + x_{r}(0)$$
$$= 0$$

So, $$v \cdot w = 0$$, 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