Is there a simpler way to prove the following statement?
If $\{\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}\}$ is a spanning set for $\mathbb{R}^n$, then $\{\mathbf{u_1}, \mathbf{u_1}+\mathbf{u_2}, \mathbf{u_1}-\mathbf{u_3}\}$ also spans $\mathbb{R}^n$.
This is what I've come up with:
To save time I will only prove for $n=2$ because the proofs for $n=1$ and $n=3$ use similar logic
Case $n=2$
let $\mathbf{u_1}=\begin{bmatrix}a\\b\end{bmatrix}$, $\mathbf{u_2}=\begin{bmatrix}c\\d\end{bmatrix}$, $\mathbf{u_3}=\begin{bmatrix}e\\f\end{bmatrix}$ for arbitrary $a,b,c,d,e,f\in \mathbb{R}$
so the augmented matrix for $\{\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}\}$ is $\begin{bmatrix}a & c & e\\b & d & f\end{bmatrix}$ To put this in echelon form I will multiply row 1 by $-\frac{b}{a}$ then add it to row 2 to get $\begin{bmatrix}a & c & e\\0 & -\frac{bc}{a}+d & -\frac{be}{a}+f\end{bmatrix}$ And since $\{\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}\}$ spans $\mathbb{R}^2$, row 2 in the echelon matrix must have a pivot, meaning $-\frac{bc}{a}+d \not=0$ or $-\frac{be}{a}+f \not= 0$. Now if I create the augmented matrix for $\{\mathbf{u_1}, \mathbf{u_1}+\mathbf{u_2}, \mathbf{u_1}-\mathbf{u_3}\}$ I get $\begin{bmatrix}a & a+c & a-e\\b & b+d & b-f\end{bmatrix}$ To put this in echelon form I will once again multiply row 1 by $-\frac{b}{a}$ then add it to row 2 to get $\begin{bmatrix}a & a+c & a-e\\0 & -\frac{bc}{a}+d & -\left(-\frac{be}{a}+f\right)\end{bmatrix}$ and since I proved above that $-\frac{bc}{a}+d \not=0$ or $-\frac{be}{a}+f \not= 0$, then there must be a pivot in row 2, meaning that $\{\mathbf{u_1}, \mathbf{u_1}+\mathbf{u_2}, \mathbf{u_1}-\mathbf{u_3}\}$ also spans $\mathbb{R}^2$
Best Answer
If $(u_1, u_2, u_3)$ spans $\mathbb{R}^n$, then for any $x\in \mathbb{R}^n$, there exist $a_1, a_2, a_3\in \mathbb{R}$ such that $x=a_1 u_1+a_2 u_2+a_3 u_3$. With very little algebra, we obtain $x=(a_3-a_2+a_1)u_1+a_2 (u_1+u_2)-a_3(u_1-u_3)$, which means $x\in span(u_1, u_1+u_2, u_1-u_3)$. Hence, $(u_1, u_1+u_2, u_1-u_3)$ spans $\mathbb{R}^n$.