# $\{ v_1,v_2,…,v_n\}$ is basis of $V$ if and only if $\{ v_1,v_1 + v_2,…,v_1 + v_2+…+v_n,\}$ is a basis of $V$

Let $V$ a vector space over a field $K$.

Is it true $\{ v_1,v_2,...,v_n\}$ is basis of $V$ if and only if $\{ v_1,v_1 + v_2,...,v_1 + v_2+...+v_n,\}$ is a basis of $V$ ?

I made some examples and got that , YES its true this claim but how can i prove it? and i tried to show that each of $v_1,\ldots,v_n$ can be obtained as linear combinations of $\{ v_1,v_1 + v_2,...,v_1 + v_2+...+v_n,\}$ but i stuck.

For $1 \leq k \leq n$, consider $$w_k := v_1 + \ldots +v_k = \sum_{i=1}^kv_i =0,$$ and let $\alpha_1, \ldots, \alpha_n \in K$ be such that $\sum_{k=1}^n\alpha_k w_k=0$, then $$0=\sum_{k=1}^n\alpha_k w_k = \sum_{k=1}^n\left(\alpha_k\sum_{i=1}^kv_i\right) = \sum_{k=1}^n v_k \left(\sum_{i = k}^n\alpha_i\right).$$ Since $v_1,\ldots,v_n$ are linearly independent this imply that $$\beta_k := \sum_{i = k}^n\alpha_i = 0, \quad \text{for every } 1 \leq k \leq n$$ In particular $$\alpha_n =\beta_n = 0 \Rightarrow \alpha_{n-1} = \beta_{n-1} -\alpha_n = 0 \Rightarrow \ldots \Rightarrow \alpha_1 = 0.$$ This shows that $w_1,\ldots,w_n$ are linearly independent. Now since $w_1,\ldots,w_n$ are $n$ linearly independent vectors in a space of dimension $n$ (because $v_1,\ldots,v_n$ is basis of it), it follows that it is a basis of $V$. Now if $w_1,\ldots,w_n$ is a basis of $V$, then let $\mu_1,\ldots,\mu_n$ be such that $\sum_{k=1}^n \mu_kv_k = 0$, in particular this imply (where $w_0 = 0$ and $\mu_0 = 0$ are introduced for compactness of the notations), $$0=\sum_{k=1}^n\mu_k v_k =\sum_{k=1}^n\mu_k (w_k-w_{k-1}) = \sum_{k=1}^n(\mu_k-\mu_{k-1})w_k.$$ Since $w_1,\ldots, w_n$ is a basis we get $\mu_k-\mu_{k-1} = 0$ for every $1 \leq k\leq n$, in particular this shows that $$0 = \mu_0=\mu_1 = \mu_2 = \ldots = \mu_n.$$ It follows that $v_1,\ldots,v_n$ are linearly independent and again using a dimension argument we may conclude that it is a basis of $V$.

Consider scalars $a_1,a_2,\ldots, a_n$ such the $a_1v_1+a_2(v_1+v_2)+\ldots+a_n(v_1+v_2+\ldots+v_n)=0$. Then you can show that the scalars have to be all zero, using the fact that $\lbrace v_1,v_2,\ldots, v_n\rbrace$ is a basis.

Yes the claim is true.

First we shall prove that $\{v_1,v_2,\dots,v_n\}$ is a basis $\implies \{v_1,v_1+v_2,\dots,v_1+v_2+\dots+v_n\}$ is a basis.

Let $a=c_1v_1+c_2v_2+\dots c_nv_n$. Then you have $a=m_1(v_1)+m_2(v_1+v_2)+\dots m_n(v_+v_2+\dots v_n)$ where $m_n=c_n, m_{n-1}=(c_{n-1}-m_n),\dots$.

To prove the converse, express $v_2$ in terms of the basis $\{v_1,v_1+v_2,\dots\,v_1+v_2+\dots +v_n\}$. Now replace $v_1+v_2$ in the basis by $v_2$. Now do this for every $v_i$.

We immediately have $$\mathrm{span}(v_1,v_2,\ldots,v_n) \supseteq \mathrm{span}(v_1,v_1+v_2,\ldots,v_1+v_2+\cdots+v_n)$$ and the other containment ($\subseteq$) follows since we have the linear combination $$v_i=(v_1+v_2+\cdots+v_i)-(v_1+v_2+\cdots+v_{i-1})$$ for all $i$.

So the two spans are equal, and the result follows from the Dimension Theorem.