# Showing a set is the basis of a vector space

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 $$V$$?

• Nothing. A linearly independent set of vectors with $\dim V$ elements is a basis. – Bernard Jan 27 '20 at 11:58
• Do you mean 'must have a subset of exactly $n$ vectors that are linearly independent'? – John Smith Kyon Jan 27 '20 at 11:59
• What is the implication if the wording is different? I'm still a bit confused about this subject. – variations Jan 27 '20 at 12:00
• @juhani Let $A$ be a subset of $V$ that spans $V$. It's kind of unclear if you mean that $A$ itself must be $A=\{v_1, ..., v_n\}$ or if you mean only that $A \supseteq \{v_1, ..., v_n\}$. For example $\mathbb R$ spans itself, but is not composed of only $n=1$ vector. – John Smith Kyon Jan 27 '20 at 12:14

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.

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.