I am a high school student currently working through Axler's Linear Algebra Done Right by myself and I have stumbled upon this question in the second exercise.
Suppose that $(v_1,…,v_n)$ is linearly independent in $V$ and $w \in V$. Prove that if $(v_1 + w, … , v_n + w)$ is linearly dependent then $w \in \text{span}(v_1,…,v_n)$.
My proof is as follows:
Let $(v_1,…,v_n)$ be linearly independent in $V$ and $w\in V$. Suppose $(v_1 + w,…,v_n + w)$ is linearly dependent in $V$, then there exist $b_j \in \mathbb{R}$ such that
$$b_1(v_1+w)+…+b_n(v_n+w)=0$$
Where not all $b_j=0$
$$\implies b_1v_1+…+b_nv_n+(b_1+…+b_n)w=0$$
$$\implies w=-\frac{b_1v_1+…+b_nv_n}{(b_1+…+b_n)}$$
$\because$ $w$ can be written as a linear combination of vectors in $(v_1,…,v_n)$ it lies in the span of $(v_1,…,v_n)$ $\therefore w \in \text{span}(v_1,…,v_n)$ $$\tag*{$\blacksquare$}$$
However I feel my method was lackluster and my writing ambiguous. What changes can I make to this proof so that it is to an acceptable degree of rigor?
Best Answer
Suppose that $\{v_{1} + w, v_{2} + w, \ldots, v_{n} + w\}$ is LD.
Consequently, if one considers the linear combination \begin{align*} \alpha_{1}(v_{1} + w) + \alpha_{2}(v_{2} + w) + \ldots + \alpha_{n}(v_{n} + w) = 0 \end{align*}
there exists at least one $\alpha_{k}\neq 0$. WLOG we shall assume that $k = 1$. Therefore we may claim that \begin{align*} (\alpha_{1} + \alpha_{2} + \ldots + \alpha_{n})w = -\alpha_{1}v_{1} - \alpha_{2}v_{2} - \ldots - \alpha_{n}v_{n} \end{align*}
Notice that $\alpha_{1} + \alpha_{2} + \ldots + \alpha_{n} \neq 0$. If this were not the case, we could conclude that \begin{align*} \alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n} = 0 \Rightarrow \alpha_{1} = \alpha_{2} = \ldots = \alpha_{n} = 0 \end{align*}
which contradicts the fact that $\alpha_{1}\neq 0$.
Based on such considerations, we are able to infer that $w\in\operatorname{span}\{v_{1},v_{2},\ldots,v_{n}\}$, and we are done.
Hopefully this helps !