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Prove that if $(v_1,\ldots,v_n)$ spans $V$, then so does the list $(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n).$

Proof: Suppose that $V = \text{span } (v_1,\ldots,v_n).$ Then for any $v \in V$, there exist $a_1,\ldots,a_n \in \mathbb{F}$ such that

\begin{align} v &= a_1v_1+\cdots +a_nv_n \\ &= a_1v_1-a_1v_2 + a_1v_2+a_2v_2+a_3v_3+\cdots+ a_nv_n \\ &= a_1v_1-a_1v_2+a_1v_2-a_1v_3+a_2v_2-a_2v_3+a_1v_3+a_2v_3+a_3v_3+a_4v_4+\cdots+a_nv_n \\ &= a_1(v_1-v_2)+(a_1+a_2)(v_2-v_3)+(a_1+a_2+a_3)v_3+a_4v_4+\cdots+a_nv_n \\ &=\sum_{i=1}^{n-1} \left[\left(\sum_{k=1}^{i}a_k\right)(v_i-v_{i+1})\right] + a_nv_n \in \text{span}(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n). \end{align}

Now let $u \in \text{span}(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n).$ Then there exist $b_1,\ldots,b_n \in \mathbb{F}$ such that

\begin{align} u &= b_1(v_1-v_2)+ b_2(v_2-v_3)+\cdots +b_{n-1}(v_{n-1}-v_n) + b_nv_n \\ &= b_1v_1 + (b_2-b_1)v_2+(b_3-b_2)v_3+\cdots+(b_n-b_{n-1})v_n \in \text{span}(v_1,\ldots,v_n) \end{align} completing the proof. $\hspace{125mm} \Box$

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    $\begingroup$ The problem can't be right the way you have written it, since one of your vectors is $v_2-v_2$ which is the zero vector. Maybe you mean $v_2-v_3$. $\endgroup$ – Gerry Myerson Oct 6 '13 at 2:18
  • $\begingroup$ Oops, my mistake. $\endgroup$ – St Vincent Oct 6 '13 at 2:21
  • $\begingroup$ At first you show that $\forall v \in V$ we have that $v \in \text{span} (v_1 - v_2, )\dots$ which leads to $V \subset \text{span} (v_1 - v_2, \dots )$. Why must you aswell show that $\forall u \in \text{span} (v_1 - v_2, \dots)$ we have that $u \in V$? What am I missing? $\endgroup$ – Olba12 Sep 15 '16 at 23:05
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Just for fun you can prove this more general result.

Let $V$ be an $n-$dimensional vector space, and let $\vec{v}\in V$. Assume that $\left( \vec{\beta}_1,\ldots ,\vec{\beta}_n\right)$ is a basis for $V$. If $\vec{v}=c_1\vec{\beta}_1 +\cdots+ c_n \vec{\beta}_n$, then for $c_j \ne 0$, we can replace one of the $\beta_j$'s with $v$ so that $\left( \vec{\beta}_1,\ldots , \vec{\beta}_{j-1},\vec{v},\vec{\beta}_{j+1},\ldots, \vec{\beta}_{n}\right)$ is a basis for $V$.

So you can use it $n$ times for each element in your list.

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  • $\begingroup$ Wouldn't some $c_i \neq 0$ imply linear dependence? I think I'm a bit confused on what you mean exactly. If it's linear dependent, it cannot be a basis of $V$. $\endgroup$ – St Vincent Oct 6 '13 at 3:34
  • $\begingroup$ Let $B = (\beta_1,\ldots,\beta_n)$ be a basis for $V$. Since $B$ is a linearly independent list that spans $V$, the list of all linear combinations of B, denoted $\text{span}(\beta_1,\ldots,\beta_ {n}) \in V$. We can rearrange $B$ for $c_j \neq 0$, such that $$\begin{align} \frac{v}{c_j}-\frac{c_1}{c_j}\beta_1-\cdots-\frac{c_{j-1}}{c_j}\beta_{j-1} = \beta_j. \end{align}$$ Wouldn't this mean $\beta_j \in \text{span}(\beta_1,\ldots,\beta_{j-1})$? $\endgroup$ – St Vincent Oct 6 '13 at 3:36
  • $\begingroup$ I had a some mistakes in the notation. I have already corrected it. $\endgroup$ – ILikeMath Oct 6 '13 at 3:48
  • $\begingroup$ is the correction clear? $\endgroup$ – ILikeMath Oct 6 '13 at 3:53
  • $\begingroup$ Well, I can see that $v \in \text{span}(\beta_1,\ldots,\beta_{i-1},\beta_i,\beta_{i+1},\ldots, \beta_n)$. I find the the fact that the notation mixes $i$'s and $n$'s a bit mystifying. $\endgroup$ – St Vincent Oct 6 '13 at 4:18
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Show that the second collection of vectors contains the first in its span. Should be a little quicker

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