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Prove: Let $\{v_1,v_2,v_3\}$ be a basis for a vector space $V$. Show that $\{u_1,u_2,u_3\}$ is also a basis, where $u_1=v_1$, $u_2=v_1 +v_2$, and $u_3=v_1+v_2+v_3$.

I am not really sure where to start on this one. Would I want to start with a linear combination?

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    $\begingroup$ It is sufficient to show that the vectors $u1$, $u2$ and $u3$ are linearly independent. $\endgroup$ – russoo Mar 17 '15 at 15:16
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Hint: $\sum c_iu_i=0 \Rightarrow (c_1+c_2+c_3)v_1+(c_2+c_3)v_2+c_3v_3=0 \Rightarrow c_i=0, \forall i=1, 2, 3.$

The only thing that remains to show that the set $\{u_1,u_2,u_3\}$ generates the vector space $V.$

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Consider the linear map $f\colon V\to V$ such that $$ f(v_1)=u_1,\quad f(v_2)=u_2,\quad f(v_3)=u_3 $$ The matrix of $f$ with respect to the basis $\{v_1,v_2,v_3\}$ is $$ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} $$ which clearly has rank $3$. Thus $\{u_1,u_2,u_3\}$ spans $V$ and so it is a basis.

Alternative proof without linear maps. Suppose you have $$ \alpha_1u_1+\alpha_2u_2+\alpha_3u_3=0 $$ Then you get $$ (\alpha_1+\alpha_2+\alpha_3)v_1+(\alpha_2+\alpha_3)v_2+\alpha_3v_3=0 $$ which implies $$ \begin{cases} \alpha_1+\alpha_2+\alpha_3=0\\ \alpha_2+\alpha_3=0\\ \alpha_3=0 \end{cases} $$ Since the matrix of the system is the same as the one above and has rank $3$, the system has a unique solution, which is $\alpha_1=\alpha_2=\alpha_3=0$.

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