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Can a linearly dependent set A = {(1,0,0), (0,1,0), (0,0,1), (1,2,3), (3,4,5)} span? Since columns 4 and 5 are linear combinations of 1,2 and 3, would spanA equals the span of columns 1,2, and 3?

Thanks! Sea

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    $\begingroup$ Can it span what? You don't just span, you span something. And the answer is yes, the span of the whole set is the same of the first three elements. $\endgroup$ – JustAskin Sep 29 '14 at 17:26
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Notice that the first three vectors of $A$ form a basis for $\Bbb R^3$ so

$$\operatorname{span}(A)=\Bbb R^3$$

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You can consider the vector subspace spanned by any set of vectors, linearly independent or not. The vector subspace spanned consists of all vectors obtained by linear combinations of vectors in the given set.

Of course, the vector subspace spanned by a set of vectors is the same as the spanned by any maximal subset of linearly independent vectors.

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Yes. Since $v_{4} = 1*v_{1} + 2*v_{2}+3*v_{3} $, we can conclude that $v_{4} \in\ Span\big\{v_{1},v_{2},v_{3} \big\}$ because it's a linear combination of the three vectors. Same goes for $v_{5}$.

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If $A = \{v_1,\ldots,v_n\}$ you can define $$Span(A) = \{v \in V \ | \ \exists \alpha_1, \ldots, \alpha_n \in K \land v = \sum_{i=0}^n \alpha_i v_i\}$$ In that sense $Span(A) = Span(\{(1,0,0), (0,1,0), (0,0,1)\}) = \mathbb{R}^3$

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