How to show that addition of a basis vector to all vectors in basis is still a basis?

Question

I am studying for an exam and I am struggling understanding how basis operates and in which conditions the basis holds.

Let's say I have basis vectors $B = \{ v_1, v_2, v_3, v_4 \}$ for some vector space $V$. Is $\{ v_1, v_2+v_1, v_3+v_1, v_4+v_1 \}$ also a basis for $V$? If so, how can I prove this?

Answer 1

Suppose that $\beta_{1}=\{v_{1},v_{2},v_{3},v_{4}\}$ is a basis for a vector space $V$. We will show that $\beta_{2}=\{v_{1},v_{2}+v_{1},v_{3}+v_{1},v_{4}+v_{1}\}$ is also a basis for $V$ showing by definition:

• ${\rm span}(\beta_{2})=V$.
• $\beta_{2}$ is linearly independent in $V$.

Since $\beta_{1}$ is a basis for $V$ so every vector in $V$ can written as $av_{1}+bv_{2}+cv_{3}+dv_{4}\in V$, then by definition of ${\rm span}$ we know $$av_{1}+bv_{2}+cv_{3}+dv_{4}=\alpha_{1}v_{1}+\alpha_{2}(v_{2}+v_{1})+\alpha_{3}(v_{3}+v_{1})+\alpha_{4}(v_{4}+v_{1})$$ Re-writing we get, $$av_{1}+bv_{2}+cv_{3}+dv_{4}=(\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4})v_{1}+(0\alpha_{1}+\alpha_{2}+0\alpha_{3}+0\alpha_{4})v_{2}+(0\alpha_{1}+0\alpha_{2}+\alpha_{3}+0\alpha_{4})v_{3}+(0\alpha_{1}+0\alpha_{2}+0\alpha_{3}+\alpha_{4})v_{4}$$ then the linear system is given by $$\begin{cases}\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=a,\\0\alpha_{1}+\alpha_{2}+0\alpha_{3}+0\alpha_{4}=b,\\ 0\alpha_{1}+0\alpha_{2}+\alpha_{3}+0\alpha_{4}=c,\\ 0\alpha_{1}+0\alpha_{2}+0\alpha_{3}+\alpha_{4}=d\end{cases} \iff \begin{bmatrix} \color{red}{1} & 1 & 1 & 1 & | & a,\\ 0 & \color{red}{1} & 0 & 0 & | & b,\\ 0 & 0 & \color{red}{1} & 0 & | & c\\ 0 & 0 & 0 & \color{red}{1} & | & d\end{bmatrix}$$ Hence consistent for $a,b,c,d\in \mathbb{R}$, then ${\rm span}(\beta_{2})=V$.

Now, $\beta_{2}$ is linearly independent because $$\alpha_{1}v_{1}+\alpha_{2}(v_{2}+v_{1})+\alpha_{3}(v_{3}+v_{1})+\alpha_{4}(v_{4}+v_{1})=0v_{1}+0v_{2}+0v_{3}+0v_{4}$$ the only solution for the linear system equation $$\begin{cases}\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=0,\\0\alpha_{1}+\alpha_{2}+0\alpha_{3}+0\alpha_{4}=0,\\ 0\alpha_{1}+0\alpha_{2}+\alpha_{3}+0\alpha_{4}=0,\\ 0\alpha_{1}+0\alpha_{2}+0\alpha_{3}+\alpha_{4}=0\end{cases} \iff \begin{bmatrix} \color{red}{1} & 1 & 1 & 1 & | & 0,\\ 0 & \color{red}{1} & 0 & 0 & | & 0,\\ 0 & 0 & \color{red}{1} & 0 & | & 0\\ 0 & 0 & 0 & \color{red}{1} & | & 0\end{bmatrix}$$ is given by $(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})=(0,0,0,0)$. Hence $\beta_{2}$ is linearly independent in $V$.

Therefore $\beta_{2}$ is a basis for $V$.