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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.

When
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    $dim(U_1 + U_2) = dim(U_1) + dim(U_2) - dim(U_1 \cap U_2); dim(U_1 \cap U_2) = 0?$ – qqo May 06 '15 at 01:32
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    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. – kbh May 06 '15 at 01:33
  • NB, if by whatever time I get to my laptop there isn't an answer, I'll write it out rigorously. Cheers. – kbh May 06 '15 at 01:47
  • thanks! By the way, NB = nota bene? – When May 06 '15 at 01:50
  • @Mr.When NB = note bene indeed, but regardless it appears you have an answer (more elegant even, at that!). – kbh May 06 '15 at 02:50

2 Answers2

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

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

Community
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Kitegi
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  • Cool proof! Appreciate it! – When May 06 '15 at 02:54
  • why dim span($v_1$, $v2-v1$,..., $v_m-v1$) = m-1? – Hamilton Feb 06 '21 at 14:39
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    @YarenZhang that's not what i said. i said that $\dim \operatorname{span}(\mathbf v_1,\dots, \mathbf v_m)=\dim\operatorname{span}(\mathbf v_1,\mathbf v_2-\mathbf v_1,\dots,\mathbf v_m-\mathbf v_1)=m$

    you can show that by proving that each element that belongs to the first span also belongs to the second one, and vice versa

     – Kitegi Feb 06 '21 at 16:29
  • @Hamilton $v_1,v_2-v_1,\dots,v_m-v_1$ is linearly independent because "spanning list of the right length is a basis". So, $v_2-v_1,\dots,v_m-v_1$ is linearly independent too. – 佐武五郎 Oct 09 '24 at 06:08
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I am reading "Linear Algebra Done Right Fourth Edition" by Sheldon Axler.

This problem is Exercise 8 in Exercises 2C on p.49 in this book.

Exercise 8:
Suppose $v_1,\dots,v_m$ is linearly independent in $V$ and $w\in V$. Prove that $$\dim\operatorname{span}(v_1+w,\dots,v_m+w)\geq m-1.$$

I solved this exercise as follows.

$c_1v_1+\dots+c_mv_m=c_1(v_1+w)+\dots+c_m(v_m+w)-(c_1+\dots+c_m)w$.
So, $\operatorname{span}(v_1,\dots,v_m)\subset\operatorname{span}(v_1+w,\dots,v_m+w)+\operatorname{span}(w)$.
$\dim\operatorname{span}(v_1,\dots,v_m)\leq\dim\left(\operatorname{span}(v_1+w,\dots,v_m+w)+\operatorname{span}(w)\right)$.
$\dim\left(\operatorname{span}(v_1+w,\dots,v_m+w)+\operatorname{span}(w)\right)\leq\dim\operatorname{span}(v_1+w,\dots,v_m+w)+\dim\operatorname{span}(w)$.
So, $\dim\operatorname{span}(v_1,\dots,v_m)\leq\dim\operatorname{span}(v_1+w,\dots,v_m+w)+\dim\operatorname{span}(w)$.
And $\dim\operatorname{span}(w)\leq 1$.
So, $\dim\operatorname{span}(v_1,\dots,v_m)-1\leq\dim\operatorname{span}(v_1+w,\dots,v_m+w)$.
And $\dim\operatorname{span}(v_1,\dots,v_m)=m$.
So, $m-1\leq\dim\operatorname{span}(v_1+w,\dots,v_m+w)$.

佐武五郎
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