If $B=\{u_{1},\dots,u_{n}\}$ is a basis of space $V$, then $\{u_{1},u_{1}+u_{2},\dots,u_{1}+\dots+u_{n}\}$ is also a basis

Question

Prove that if $B=\{u_{1},\dots,u_{n}\}$ is a basis of the space $V$, then $$\{u_{1},u_{1}+u_{2},\dots,u_{1}+\dots+u_{n}\}$$ is also a basis.

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Is it valid to verify linear independence for the set $\{ u_{1}, u_{1} + u_{2},\dots, u_{1} + \dots + u_{n} \}$ writing

$$ (\gamma_{1} + \dots + \gamma_{n}) u_{1} + \dots + \gamma_{n} u_{n} = 0 $$

and saying this system leads to a matrix of rank $n$ with $n$ variables, so there is only one solution, which must be $\gamma_{1}=\dots=\gamma_{n}=0$?

Answer 1

Your approach is good, but there's a simpler way, for this particular case. If you set $$ v_1=u_1,\quad v_2=u_1+u_2,\quad \dots,\quad v_n=u_1+\dots+u_n $$ then you have $$ u_1=v_1,\quad u_2=v_2-v_1,\quad \dots,\quad u_n=v_n-v_{n-1} $$ so $\{u_1,\dots,u_n\}$ and $\{v_1,\dots,v_n\}$ span the same subspace. As the former set is a basis by assumption, it follows that the latter is as well.

Answer 2

Your approach indeed works. You can let $A$ have $a_k = \sum_{i=1}^k u_i$ as its $k$-th column. Then $A=UT$ where $T=(t_{ij})$ with $$t_{ij}= \begin{cases}1 &\text{if}~j\geq i,\\ 0&\text{otherwise,} \end{cases}$$ and $U$ has $u_k$ as its $k$-th column. Since $\{u_k\}$ is a linearly independent set, it follows that $U$ is invertible, so you only need to prove that $T$ is also invertible, which should be easy.