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- Thread starter judahs_lion
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CompuChip

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However, if you call any element of a vector space (that is, a set that satisfies certain axioms) a vector, then indeed they are.

Note by the way that there is always a bijection between n x m matrices and vectors in

- #3

judahs_lion

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I can tell the vector {4,4;4,4} would not be in the span because there is no multiple of d for({a, b;c,d} that would ever equal 4 because d is always 0.

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judahs_lion

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a,bsuch that [itex]a\,\mathbf v_1 + b\,\mathbf v_2 = \mathbf v_3[/itex] -- or show that no sucha,bexist.

Thanks so a = -7, b =1;

Span(v1,v2,v3) = {<a, -a; c , 0> : a, c belong to R} ?

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Fredrik

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Nope.Thanks so a = -7, b =1;

I suggest you use the standard definition of linear independence. [itex]\{v_1,\dots,v_k\}[/itex] is linearly independent if

[tex]\sum a_i v_i=0\Rightarrow a_i=0 \text{ for all }i[/tex]

(What DH said works too).

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Good. So what does that mean in terms of the problem? (You need to determine whether v1, v2, v3 are dependent or independent.)Thanks so a = -7, b =1;

That looks good.Span(v1,v2,v3) = {<a, -a; c , 0> : a, c belong to R} ?

- #8

judahs_lion

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Good. So what does that mean in terms of the problem? (You need to determine whether v1, v2, v3 are dependent or independent.)

That looks good.

That means they are dependent? v3 depends on v1 and v2

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Correct.

- #10

judahs_lion

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Correct.

Thank you

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