Showing a set is the basis of a vector space

Question

The set $(v_1,v_2,...,v_n)$ is a basis for the $\mathbb R$-vector space $V$. Now $u_1=v_1$, $u_2=v_1+v_2$, $u_n=v_1+v_2+,...,+v_n$. Show that the set $U:=(u_1,u_2,..u_n)$ is also a basis of $V$.

Proof : $U$ is a basis of $V$, if it's both linearly independent and such that it spans $V$. To show linear independence, it should be sufficient to look at the matrix with columns $$(u_1,u_2,u_3,...,u_n)$$

Which is obviously in row echelon form and there are no free variables. As such the homogeneous equation $Ux=0$ has only the trivial solution and $U$ is linearly independent. Showing that $U$ spans $V$ is the problem. I believe that, since the dimension of $V$ is $n$ and is isomorphic to $\Bbb{R^n}$, every set that spans $V$ must have exactly $n$ vectors, which are linearly independent. This is the case with $U$ - what else must be shown in order to verify that $U$ indeed spans $V$?

Answer 1

It is enough to prove linear indepndence or spanning property. Not necessary to prove both.

A simple argument here is to note that any element of $V$ has the form $\sum a_kv_k=\sum a_n (u_k-u_{k-1})$ (interpreting $u_0$ as $0$) so $u_i$' span $V$ and there is no need to worry about independence.

Answer 2

Note that if $$A=\begin{bmatrix}v_1&v_2&\cdots& v_n\end{bmatrix}$$ and $$B=\begin{bmatrix}u_1&u_2&\cdots& u_n\end{bmatrix}$$then since $A$ spans $V$, for any $y\in V$, there exists a unique $x$ such that $$Ax=y$$since $$B=AM$$where $$M=\begin{bmatrix}1&1&\cdots& 1\\0&1&\cdots& 1\\\vdots\\0&0&\cdots&1\end{bmatrix}$$then also $B$ spans $V$ since $M$ is invertible.