Question: Let $\{v_1,\cdots,v_n\}$ be a basis for vector space $V$ over $\mathbb{R}$. Suppose $w$ is any vector in $V$, then, for some choice of sign $\pm$, $\{v_1\pm w, v_2,\cdots,v_n\}$ is a basis for $V$.
Attempt: This question is asked here: Let $\{v_1,v_2,\cdots,v_n\}$ is a basis for vector space $V$. Let $w \in V$, prove that $W =\{v_1+w,v_2+w,\cdots,v_n+w\}$ is a basis for $V$, but I want to take the following approach:
We have that $\{v_1\pm w, \cdots, v_n\}$ is a spanning set of $V$ for some choice $\pm$. Suppose $v_1+w,v_2,\cdots,v_n$ is linearly independent. Then, $b_1(v_1+w)+\dots+b_nv_n=0$ and $a_1(v_1-w)+\dots+a_nv_n=0$ where $b_1,a_1\neq 0$. Then, we may write $a_1b_1(v_1+w)+\dots+a_1b_nv_n=0$ and $a_1b_1(v_1-w)+\dots+a_nb_1v_n=0$. Then, $2a_1b_1v_1+\dots+(a_1b_n+a_nb_1)v_n=0$. I want to get a contradiction, but I am a bit stuck. Any help is greatly appreciated! Thank you.
Suppose that $\{v_1-w,v_2,\ldots,v_n\}$ is linearly dependent. Then there are scalars $a_1,a_2,\ldots,a_n$, not all of which are $0$, such that$$a_1(v_1-w)+a_2v_2+\cdots+a_nv_n=0.\tag1$$Since $\{v_1,v_2,\ldots,v_n\}$ is linearly independent, we cannot have $a_1=0$. It follows then from $(1)$ that$$w=v_1+\frac{a_2}{a_1}v_2+\cdots+\frac{a_n}{a_1}v_n$$and that therefore$$v_1+w=2v_1+\frac{a_2}{a_1}v_2+\cdots+\frac{a_n}{a_1}v_n.\tag2$$So, if you define $f\colon V\longrightarrow V$ such that $f(v_1)$ equal to the RHS of $(2)$ and $f(v_k)=v_k$ when $k>1$, $\det f=2\ne0$. So, $\{f(v_1),f(v_2),\ldots,f(v_n)\}$ is also a basis of $V$. But$$\{f(v_1),f(v_2),\ldots,f(v_n)\}=\{v_1+w,v_2,\ldots,v_n\}.$$