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Show $S = \text{span} \space \{\overrightarrow{v_1} + ... + \overrightarrow{v_{i-1}} + \overrightarrow{v_{i+1}} + ... + \overrightarrow{v_k} \} \subseteq \text{span} \space \{\overrightarrow{v_1} + ... + \overrightarrow{v_k} \} = T$, where $1 \le i \le k$ and $\overrightarrow{v} \in \mathbb{R}^n$

Let $\overrightarrow{x} = c_1v_1 + ... + c_{i-1}v_{i-1} + c_{i+1}v_{i+1} + ... + c_{k}v_k \in S$

I need to show $\overrightarrow{x} \in T$ as well.

How do I show this formally?

What I am thinking: This may be weird because I was thinking of rewriting the span as a vector set in the form of $\overrightarrow{x}$ and letting coefficient of $v_{i} = 1$. But is that formal?

2 Answers 2

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What you have written is false, as DonAntonio points out in their answer. However, you might have meant to write $S=\operatorname{span}(v_1,\ldots,v_{i-1},v_{i+1},\ldots,v_k)$ and $T=\operatorname{span}(v_1,\ldots,v_k)$, in which case the claim $S\subset T$ is true. To establish it, use the hint below.

Hint: Suppose $x\in S$. Then, $$ x=c_{1}v_{1}+\cdots+c_{i-1}v_{i-1}+c_{i+1}v_{i+1}+\cdots+c_{k}v_{k} $$ for some scalars $c_{1},\ldots,c_{i-1},c_{i+1},\ldots,c_{k}$. Define a new scalar $c_{i}=0$. Then, $$ x=c_{1}v_{1}+\cdots+c_{k}v_{k}. $$ What does this tell you?

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    I think this is wrong: the question speaks of the span of **one single vector** in both sides, namely $\;v_1+v_2+...+v_k\;$ on the right side, and the same with one vector less on the left. You seem to have written about the span of $\;v_1\color{red},\,v_2\color{red},\,...\color{red},v_k\;$ , $\;k\;$ elements... – 2017-01-09
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    Ah. My brain saw what it wanted to see; editing. – 2017-01-09
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    @parsiad, your brain saw what I saw =) Sorry for confusion, thanks – 2017-01-10
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If I understood correctly, the claim is false: in the real plane $\;\Bbb R^2\;$

$$\text{Span}\,\left\{\;\binom10\;\right\}\rlap{\;\,/}\subset\text{Span}\,\left\{\;\binom10+\binom01=\binom11\;\right\}$$

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    MY FAULT! I meant it with the commas – 2017-01-10
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    @user405307 Fine, but then, for whoever may ever want to read this post, you should change that in your question. – 2017-01-10
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    @user405307: I agree with DonAntonio. Please edit your question. – 2017-01-10