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Suppose $v_1,...,v_m$ is linearly independent in $V$ and $w\in V$. Prove that $$ \dim (\operatorname{span}(v_1+w,...,v_m+w)) \ge m-1$$

It's an exercise in the book Linear Algebra Done Right.

I'm wondering if I can write $U_1 =\operatorname{span}(v_1,...,v_m)$ and $U_2=\operatorname{span}(w)$ then write $$ \dim(\operatorname{span}(v_1+w,...,v_m+w)) = \dim(U_1+U_2)$$ Would you please help me with this problem, I really want a rigorous proof, thanks.

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    $\begingroup$ $dim(U_1 + U_2) = dim(U_1) + dim(U_2) - dim(U_1 \cap U_2); dim(U_1 \cap U_2) = 0?$ $\endgroup$ – IAmNoOne May 6 '15 at 1:32
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    $\begingroup$ If the set of vectors is linearly independent, then adding another vector can at most cause but one of those vectors to vanish upon addition. (Vector spaces admit unique additive inverses). So, if the original set has $m$ vectors, then the span is of dimensionality $m$. Adding $w$ to each vector in the span will at most reduce the number of nontrivial vectors in the span by 1. Hence, the new span is of, at most, dimensionality $m-1$. I'm on my phone, else I'd make this sketch into an answer. Whoops. $\endgroup$ – kbh May 6 '15 at 1:33
  • $\begingroup$ NB, if by whatever time I get to my laptop there isn't an answer, I'll write it out rigorously. Cheers. $\endgroup$ – kbh May 6 '15 at 1:47
  • $\begingroup$ thanks! By the way, NB = nota bene? $\endgroup$ – When May 6 '15 at 1:50
  • $\begingroup$ @Mr.When NB = note bene indeed, but regardless it appears you have an answer (more elegant even, at that!). $\endgroup$ – kbh May 6 '15 at 2:50
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If we subtract the first vector from the rest, we get: $$\mathbf v_2-\mathbf v_1, \mathbf v_3-\mathbf v_1,\dots,\mathbf v_m - \mathbf v_1 \in \operatorname{span}(\mathbf v_1+\mathbf w,...,\mathbf v_m+\mathbf w)$$

So we have:

$$\dim\operatorname{span}(\mathbf v_1+\mathbf w,...,\mathbf v_m+\mathbf w)\geq\dim\operatorname{span}(\mathbf v_2-\mathbf v_1,\dots,\mathbf v_m - \mathbf v_1)$$

Since $$\dim \operatorname{span}(\mathbf v_1,\dots, \mathbf v_m)=\dim\operatorname{span}(\mathbf v_1,\mathbf v_2-\mathbf v_1,\dots,\mathbf v_m-\mathbf v_1)=m$$

We have $$\dim\operatorname{span}(\mathbf v_2-\mathbf v_1,\dots,\mathbf v_m-\mathbf v_1)\geq m-1$$

$$\dim\operatorname{span}(\mathbf v_1+\mathbf w,...,\mathbf v_m+\mathbf w)\geq m-1$$

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  • $\begingroup$ Cool proof! Appreciate it! $\endgroup$ – When May 6 '15 at 2:54

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