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Let V be a vector space and let $\{v_1, \dots , v_n\}$ be a basis for $V$. Show that $\{v_1,v_1+v_2, \dots ,v_1+ \dots + v_n\}$ is also a basis for $V$.

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What have you tried and where did you get stuck? –  Vedran Šego Oct 15 '13 at 2:28
    
I know that for $\{v_1,v_1+v_2, \dots ,v_1+ \dots + v_n\}$ to be basis it need to be span $V$ and that spanning set needs to be linearly independent but I cant figure out how I can prove that –  shdsah Oct 15 '13 at 2:29
    
So, use linear independence of $\{v_1,\dots,v_n\}$ to prove linear independence of $\{v_1, v_1 + v_2, \dots, v_1 + \cdots + v_n \}$. Use the definition of the linear independence and induction by $n$. When you do that, you should have some idea on the span. Btw, -1 is not from me. –  Vedran Šego Oct 15 '13 at 2:33
    
how does the linear independence of $\{v_1, v_1 + v_2, \dots, v_1 + \cdots + v_n \}$ affect the linear independence of it spanning set? I have rough idea how I can prove that but how is that helpful? –  shdsah Oct 15 '13 at 2:35
    
I've put two hints as an answer. I hope they help you. –  Vedran Šego Oct 15 '13 at 2:47

2 Answers 2

Hint 1 (spanning):

Notice that $v_k = (v_1 + \cdots + v_k) - (v_1 + \cdots + v_{k-1})$.

Hint 2 (independence):

Notice that $v_1 + \cdots + v_k$ is linearly independeant of vectors $v_1, v_1 + v_2, \dots, v_1 + \cdots + v_{k-1}$. Start from $k = n$, and go down to $k = 1$.

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So basically since any vector in $\{v_1,v_1+v_2, \dots ,v_1+ \dots + v_n\}$ can expressed as any vector in $\{v_1, \dots , v_n\}$ and since we already know that $\{v_1, \dots , v_n\}$ spans V therefore $\{v_1,v_1+v_2, \dots ,v_1+ \dots + v_n\}$ should span V am I right? –  shdsah Oct 15 '13 at 3:10
    
No, it's the other way: every $v_i$ can be expressed as a linear combination of the vectors in $\{v_1, v_1+v_2, \dots, v_1+\cdots+v_n\}$, so $V = \operatorname{span}\{v_1,\dots,v_n\} \subseteq \operatorname{span}\{v_1, v_1+v_2, \dots, v_1+\cdots+v_n\}$. I've explained here how you get that. Of course, trivially, $\operatorname{span}\{v_1, v_1+v_2, \dots, v_1+\cdots+v_n\} \subseteq V$, so $\operatorname{span}\{v_1, v_1+v_2, \dots, v_1+\cdots+v_n\} = V$. –  Vedran Šego Oct 15 '13 at 11:03
    
Let me explain why your comment was wrong. Take, for example, $\{v_1, v_1+v_2\}$, but in a three-dimensional $V$ spanned by $\{v_1,v_2,v_3\}$. Every vector in $\{v_1, v_1+v_2\}$ can obviously be expressed as a linear combination of vectors in $\{v_1,v_2,v_3\}$, but $\operatorname{span}\{v_1, v_1+v_2\} \ne V$. –  Vedran Šego Oct 15 '13 at 11:07

Given that $\;\{v_1,...,v_n\}\;$ is a basis of $\;V\;$ , we deduce that a set with $\;n\;$ vectors is a basis here iff the set is linearly independent, and thus this is the only thing you have to prove.

Now, hint: for scalars $\;a_i\;$ in the definition field

$$0=a_1v_1+a_2(v_1+v_2)+\ldots+a_n(v_1+\ldots+v_n)=$$

$$=\left(a_1+\ldots+a_n\right)v_1+\left(a_2+\ldots+a_n\right)v_2+\ldots a_nv_n\implies$$

$$a_1+\ldots+a_n=a_2+\ldots+a_n=\ldots=a_n=0\implies \ldots\ldots$$

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