Linear Algebra problem on basis

Question

The question is "Let $x_k$ denote the vector in $\mathbb R^n$ whose first $k-1$ coordinates are $0$ and the last $n-k+1$ coordinates are $1$. Show that the set $\{x_1,x_2,\ldots,x_n\}$ is a basis in $\mathbb R^n$". I am a beginner and I am able to do this only for sets that are explicitly defined and not infinite sets like these. Any help on how to solve this is appreciated. Thank You.

Answer 1

The matrix $A=(x_1,\ldots ,x_n)$ is lower-triangular with only $1$'s on the diagonal. Hence $\det(A)=1$, and hence the column vectors $x_i$ form a basis.

Answer 2

Hint: $x_k-x_{k+1}=e_{k+1}$ where $\{e_1,..., e_n\}$ is the standard basis.

Answer 3

Since $\mathscr{B}:=\{x_1,\ldots,x_n\}$ is a set of $n$ vectors and $\mathbb{R}^n$ is an $n$-dimensional vector space, then we have that $\mathscr{B}$ is a basis iff $\mathscr{B}$ is linearly independent. Thus we only need to show that $\mathscr{B}$ is linearly independent. Let $c_1,\ldots,c_n$ be real numbers such that $$ c_1x_1+\cdots+c_nx_n=0. $$ Since $$0 = c_1x_1+\cdots+c_nx_n = \begin{pmatrix} c_1 \\ c_1+c_2 \\ \vdots \\ c_1+\cdots+c_n \end{pmatrix}, $$ we deduce that $c_1=c_2=\cdots=c_n=0$. Therefore $\mathscr{B}$ is linearly independent, and consequently is a basis for $\mathbb{R}$.