# Prove. Let {v1,v2,v3} be a basis for a vector space V. Show that {u1,u2,u3} is also a basis, where u1=v1, u2=v1 +v2, and u3=v1+v2+v3

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?

• It is sufficient to show that the vectors $u1$, $u2$ and $u3$ are linearly independent. – russoo Mar 17 '15 at 15:16

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 $V.$
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 $\alpha_1=\alpha_2=\alpha_3=0$.