Let $1\leq k,n\in \mathbb{N}$ and $v_1, \ldots , v_k\in \mathbb{R}^n$.
Show that:
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For all $1\leq i\neq j\leq k$ and $\lambda\in \mathbb{R}$ it holds that $$\text{span}(v_1, \ldots , v_k)=\text{span}(v_1, \ldots , v_{j-1}, v_j+\lambda v_i, v_{j+1}, \ldots , v_k)$$
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Let $0_{\mathbb{R}}\neq \lambda\in \mathbb{R}$ and $1\leq i \leq k$ then it holds that $$\text{span}(v_1, \ldots , v_k)=\text{span}(v_1, \ldots , v_{i-1}, \lambda v_i, v_{i+1}, \ldots , v_k)$$
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Let's consider first the statement 1. To show this one do we have to show that $\text{span}(v_1, \ldots , v_k)\subseteq \text{span}(v_1, \ldots , v_{j-1}, v_j+\lambda v_i, v_{j+1}, \ldots , v_k)$ and that $\text{span}(v_1, \ldots , v_{j-1}, v_j+\lambda v_i, v_{j+1}, \ldots , v_k)\subseteq \text{span}(v_1, \ldots , v_k)$ so that the equality follows? Or should we show that in an other way?
If we do that in that way, I have done the following:
Let $i<j$.
Let $$y\in \text{span}(v_1, \ldots , v_{j-1}, v_j+\lambda v_i, v_{j+1}, \ldots , v_k)=\text{span}(v_1, \ldots , v_i, \ldots , v_{j-1}, v_j+\lambda v_i, v_{j+1}, \ldots , v_k)$$
Then we have that \begin{align*}y&=\alpha_1v_1+ \ldots \alpha_iv_i+\ldots +\alpha_{j-1} v_{j-1}+ \alpha_j(v_j+\lambda v_i)+ \alpha_{j+1}v_{j+1}+ \ldots +\alpha_k v_k \\ & = \alpha_1v_1+ \ldots \alpha_iv_i++ \ldots +\alpha_{j-1} v_{j-1}+ \alpha_jv_j+\alpha_j\lambda v_i+ \alpha_{j+1}v_{j+1}+ \ldots +\alpha_k v_k \\ & = \alpha_1v_1+ \ldots (\alpha_i+\alpha_j\lambda )v_i++ \ldots +\alpha_{j-1} v_{j-1}+ \alpha_jv_j+ \alpha_{j+1}v_{j+1}+ \ldots +\alpha_k v_k\end{align*} This is a linear combination of $v_1, \ldots , v_i, \ldots , v_{j-1}, v_j, v_{j+1}, \ldots , v_k$.
Therefore $y\in \text{span}(v_1, \ldots , v_i, \ldots , v_{j-1}, v_j, v_{j+1}, \ldots , v_k)=\text{span}(v_1, \ldots , v_k)$.
So it follows that $$\text{span}(v_1, \ldots , v_{j-1}, v_j+\lambda v_i, v_{j+1}, \ldots , v_k)\subseteq \text{span}(v_1, \ldots , v_k)$$ Is this direction correct and complete?
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Now we consider the other direction.
Let $y\in \text{span}(v_1, \ldots , v_k)$.
Could you give me a hint what we are suposed to do in this case?
Best Answer
Let $x\in \operatorname{span}\{ v_1,\ldots,v_k\}$ and $\lambda\neq 0$, then we have \begin{align*} x &=\sum_{\ell=1}^k a_\ell v_\ell \\ &=\sum_{\ell\in[[1,k]]\setminus\{ i,j\} }a_\ell v_\ell+a_iv_i+a_j v_j \\ &=\sum_{\ell\in[[1,k]]\setminus\{ i,j\} }a_\ell v_\ell+ a_i v_i+\color{blue}{a_j}(v_j +\color{blue}{\lambda v_i})\color{blue}{-\lambda a_j v_i} \\ &=\sum_{\ell\in[[1,k]]\setminus\{ i,j\} }a_\ell v_\ell +(a_i-\lambda a_j)v_i +a_j(v_j+\lambda v_i) \\ &\in \operatorname{span}\{v_1,\ldots,v_{j-1},v_j+\lambda v_i,v_{j+1},\ldots,v_k \}. \end{align*}
This looks really fancy, but there is nothing spectacular going on.